case study questions electromagnetic induction

Class 12 Physics Case Study Questions Chapter 6 Electromagnetic Induction

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In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Downloads of CBSE Class 12 Physics Chapter 6 Electromagnetic Induction Case Study and Passage-Based Questions with Answers were Prepared Based on the Latest Exam Pattern. Students can solve NCERT Class 12 Physics Case Study Questions Electromagnetic Induction  to know their preparation level.

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In CBSE Class 12 Physics Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Electromagnetic Induction Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 12 Physics  Chapter 6 Electromagnetic Induction

Case Study/Passage-Based Questions

Case Study 1: Currents can be induced not only in conducting coils but also in conducting sheets or blocks. Current is induced in solid metallic masses when the magnetic flux threading through them changes. Such currents flow in the form of irregularly shaped loops throughout the body of the metal. These currents look like eddies or whirlpools in water so they are known as eddy currents. Eddy currents have both undesirable effects and practically useful applications. For example, it causes unnecessary heating and wastage of power in electric motors, dynamos, and in the cores of transformers. (I) The working of speedometers of trains is based on

Answer: (b) eddy currents

(ii) Identify the wrong statement

Answer: (a) Eddy currents are produced in a steady magnetic field

(iii) Which of the following is the best method to reduce eddy currents?

Answer: (a) Laminating core

(iv) The direction of eddy currents is given by

Answer: (c) Lenz’s law

(v) Eddy currents can be used to heat localized tissues of the human body. This branch of medical therapy is called 

Answer: (c) Inductothermy

Case Study 2: Self Induction. When a current I flows through a coil, the flux linked with it is φ = LI, where L is a constant known as the self-inductance of the coil.

Any change in current sets up an induced emf in the coil. Thus, the self-inductance of a coil is the induced emf set up in it when the current passing through it changes at the unit rate. It is a measure of the opposition to the growth of the decay of current flowing through the coil. Also, the value of self-inductance depends on the number of turns in the solenoid, its area of cross-section, and the permeability of its core material.

(i) The inductance in a coil plays the same role as (a) inertia in mechanics (b) energy in mechanics (c) momentum in mechanics (d) force in mechanics

Answer: (a) inertia in mechanics

(ii) A current of 2.5 A flows through a coil of inductance 5 H. The magnetic flux linked with the coil is (a) 0.5 Wb (b) 12.5 Wb (c) zero (d) 2 Wb

Answer: (b) 12.5 Wb

The inductance L of a solenoid depends upon its radius R as (a) L ∝ R (b) L ∝ 1/R (c) L ∝ R 2 (d) L ∝ R 3

Answer: (c) L ∝ R2

(iv) The unit of self-inductance is

(a) Weber ampere (b) Weber -1  ampere (c) Ohm second (d) Farad

Answer: (c) Ohm second

(v) The induced emf in a coil of 10 henry inductance in which current varies from 9 A to 4 A in 0.2 second is (a) 200 V (b) 250 V (c) 300 V (d) 350 V

Answer: (b) 250 V

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Case Study on Electromagnetic Induction Class 12 Physics PDF

Better preparation of Case Study on Electromagnetic Induction Class 12 Physics can help students score good marks in the CBSE Class 12 Board examination. Additionally, it helps build confidence and enables students to deepen their knowledge of Electromagnetic Induction. Because case-based questions are equally important for learning and board exam preparation, our team has prepared Case Study on Electromagnetic Induction Class 12 Physics in a PDF file for free distribution among students.

Links to download the PDF file of the Electromagnetic Induction Case Study for Class 12 Physics free of cost are mentioned on this page.

Electromagnetic Induction Case Study for Class 12 Physics with Solutions in PDF

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4 Tips to Answer Class 12 Physics Electromagnetic Induction Case Study Questions?

There are 4 solid tips to answer Class 12 Physics Electromagnetic Induction Case Study questions that we are discussing in this section.

• Read the Case Carefully: To start gathering insights from the given case-based questions, it is vital to read the Electromagnetic Induction case carefully and identify the key facts, figures, and units of measurement. Pay attention to any diagrams or graphs related to Electromagnetic Induction provided, as they may contain important information.
• Identify the Problem: While reading the Electromagnetic Induction case it is essential to consider the possible causes and effects. This will help you determine the appropriate approach to solving the problem in Class 12 Physics Electromagnetic Induction.
• Use Elimination Methods Too: Since case study questions of Class 12 Physics Electromagnetic Induction, are often framed in Multiple choice questions, students should have the knowledge of elimination methods in MCQs to better answer the questions.
• Before All, Master the Concept of Electromagnetic Induction: If the above two methods are not working for you to answer Case Study on Electromagnetic Induction Class 12 Physics then you need to revisit the lesson and master the concepts explained in the Electromagnetic Induction Class 12 Physics.

What is the Benefit of Practising Class 12 Physics Electromagnetic Induction Case Study Questions?

When a student decides to practise Class 12 Physics Electromagnetic Induction case study questions before the board examination they generally get these 3 benefits:

• Quick Conceptual Revision: Nothing is better for revision than solving relevant questions and therefore, those who involve in solving the Electromagnetic Induction Case study questions before the Class 12 Physics exam are able to better revise their conceptual learnings from the lesson.
• Better Board Exam Preparation: No doubt, the more you practise Electromagnetic Induction case study questions the better your exam preparation will be so, solving Case-based questions prior to the board examination helps a lot in the preparation. It also enables the students to know what are the Electromagnetic Induction questions which have the possibility to be asked in the board examination.
• Confident in Using Analytical or Critical Thinking Skills: The Case-based questions on Electromagnetic Induction are all about using analytical or critical thinking skills where students are required to solve problems based on the situations or data given. Thus, practising Case Study on Electromagnetic Induction Class 12 Physics benefits students to feel confident and comfortable in using analytical skills.

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20.3 Electromagnetic Induction

Section learning objectives.

By the end of this section, you will be able to do the following:

• Explain how a changing magnetic field produces a current in a wire
• Calculate induced electromotive force and current

Teacher Support

The learning objectives in this section will help your students master the following standards:

• (G) investigate and describe the relationship between electric and magnetic fields in applications such as generators, motors, and transformers.

In addition, the OSX High School Physics Laboratory Manual addresses content in this section in the lab titled: Magnetism, as well as the following standards:

Section Key Terms

Changing magnetic fields.

In the preceding section, we learned that a current creates a magnetic field. If nature is symmetrical, then perhaps a magnetic field can create a current. In 1831, some 12 years after the discovery that an electric current generates a magnetic field, English scientist Michael Faraday (1791–1862) and American scientist Joseph Henry (1797–1878) independently demonstrated that magnetic fields can produce currents. The basic process of generating currents with magnetic fields is called induction ; this process is also called magnetic induction to distinguish it from charging by induction, which uses the electrostatic Coulomb force.

When Faraday discovered what is now called Faraday’s law of induction, Queen Victoria asked him what possible use was electricity. “Madam,” he replied, “What good is a baby?” Today, currents induced by magnetic fields are essential to our technological society. The electric generator—found in everything from automobiles to bicycles to nuclear power plants—uses magnetism to generate electric current. Other devices that use magnetism to induce currents include pickup coils in electric guitars, transformers of every size, certain microphones, airport security gates, and damping mechanisms on sensitive chemical balances.

One experiment Faraday did to demonstrate magnetic induction was to move a bar magnet through a wire coil and measure the resulting electric current through the wire. A schematic of this experiment is shown in Figure 20.33 . He found that current is induced only when the magnet moves with respect to the coil. When the magnet is motionless with respect to the coil, no current is induced in the coil, as in Figure 20.33 . In addition, moving the magnet in the opposite direction (compare Figure 20.33 with Figure 20.33 ) or reversing the poles of the magnet (compare Figure 20.33 with Figure 20.33 ) results in a current in the opposite direction.

Virtual Physics

Faraday’s law.

Try this simulation to see how moving a magnet creates a current in a circuit. A light bulb lights up to show when current is flowing, and a voltmeter shows the voltage drop across the light bulb. Try moving the magnet through a four-turn coil and through a two-turn coil. For the same magnet speed, which coil produces a higher voltage?

• The sign of voltage will change because the direction of current flow will change by moving south pole of the magnet to the left.
• The sign of voltage will remain same because the direction of current flow will not change by moving south pole of the magnet to the left.
• The sign of voltage will change because the magnitude of current flow will change by moving south pole of the magnet to the left.
• The sign of voltage will remain same because the magnitude of current flow will not change by moving south pole of the magnet to the left.

Induced Electromotive Force

If a current is induced in the coil, Faraday reasoned that there must be what he called an electromotive force pushing the charges through the coil. This interpretation turned out to be incorrect; instead, the external source doing the work of moving the magnet adds energy to the charges in the coil. The energy added per unit charge has units of volts, so the electromotive force is actually a potential. Unfortunately, the name electromotive force stuck and with it the potential for confusing it with a real force. For this reason, we avoid the term electromotive force and just use the abbreviation emf , which has the mathematical symbol ε . ε . The emf may be defined as the rate at which energy is drawn from a source per unit current flowing through a circuit. Thus, emf is the energy per unit charge added by a source, which contrasts with voltage, which is the energy per unit charge released as the charges flow through a circuit.

To understand why an emf is generated in a coil due to a moving magnet, consider Figure 20.34 , which shows a bar magnet moving downward with respect to a wire loop. Initially, seven magnetic field lines are going through the loop (see left-hand image). Because the magnet is moving away from the coil, only five magnetic field lines are going through the loop after a short time Δ t Δ t (see right-hand image). Thus, when a change occurs in the number of magnetic field lines going through the area defined by the wire loop, an emf is induced in the wire loop. Experiments such as this show that the induced emf is proportional to the rate of change of the magnetic field. Mathematically, we express this as

where Δ B Δ B is the change in the magnitude in the magnetic field during time Δ t Δ t and A is the area of the loop.

Note that magnetic field lines that lie in the plane of the wire loop do not actually pass through the loop, as shown by the left-most loop in Figure 20.35 . In this figure, the arrow coming out of the loop is a vector whose magnitude is the area of the loop and whose direction is perpendicular to the plane of the loop. In Figure 20.35 , as the loop is rotated from θ = 90° θ = 90° to θ = 0° , θ = 0° , the contribution of the magnetic field lines to the emf increases. Thus, what is important in generating an emf in the wire loop is the component of the magnetic field that is perpendicular to the plane of the loop, which is B cos θ . B cos θ .

This is analogous to a sail in the wind. Think of the conducting loop as the sail and the magnetic field as the wind. To maximize the force of the wind on the sail, the sail is oriented so that its surface vector points in the same direction as the winds, as in the right-most loop in Figure 20.35 . When the sail is aligned so that its surface vector is perpendicular to the wind, as in the left-most loop in Figure 20.35 , then the wind exerts no force on the sail.

Thus, taking into account the angle of the magnetic field with respect to the area, the proportionality E ∝ Δ B / Δ t E ∝ Δ B / Δ t becomes

Another way to reduce the number of magnetic field lines that go through the conducting loop in Figure 20.35 is not to move the magnet but to make the loop smaller. Experiments show that changing the area of a conducting loop in a stable magnetic field induces an emf in the loop. Thus, the emf produced in a conducting loop is proportional to the rate of change of the product of the perpendicular magnetic field and the loop area

where B cos θ B cos θ is the perpendicular magnetic field and A is the area of the loop. The product B A cos θ B A cos θ is very important. It is proportional to the number of magnetic field lines that pass perpendicularly through a surface of area A . Going back to our sail analogy, it would be proportional to the force of the wind on the sail. It is called the magnetic flux and is represented by Φ Φ .

The unit of magnetic flux is the weber (Wb), which is magnetic field per unit area, or T/m 2 . The weber is also a volt second (Vs).

The induced emf is in fact proportional to the rate of change of the magnetic flux through a conducting loop.

Finally, for a coil made from N loops, the emf is N times stronger than for a single loop. Thus, the emf induced by a changing magnetic field in a coil of N loops is

The last question to answer before we can change the proportionality into an equation is “In what direction does the current flow?” The Russian scientist Heinrich Lenz (1804–1865) explained that the current flows in the direction that creates a magnetic field that tries to keep the flux constant in the loop. For example, consider again Figure 20.34 . The motion of the bar magnet causes the number of upward-pointing magnetic field lines that go through the loop to decrease. Therefore, an emf is generated in the loop that drives a current in the direction that creates more upward-pointing magnetic field lines. By using the right-hand rule, we see that this current must flow in the direction shown in the figure. To express the fact that the induced emf acts to counter the change in the magnetic flux through a wire loop, a minus sign is introduced into the proportionality ε ∝ Δ Φ / Δ t . ε ∝ Δ Φ / Δ t . , which gives Faraday’s law of induction.

Lenz’s law is very important. To better understand it, consider Figure 20.36 , which shows a magnet moving with respect to a wire coil and the direction of the resulting current in the coil. In the top row, the north pole of the magnet approaches the coil, so the magnetic field lines from the magnet point toward the coil. Thus, the magnetic field B → mag = B mag ( x ^ ) B → mag = B mag ( x ^ ) pointing to the right increases in the coil. According to Lenz’s law, the emf produced in the coil will drive a current in the direction that creates a magnetic field B → coil = B coil ( − x ^ ) B → coil = B coil ( − x ^ ) inside the coil pointing to the left. This will counter the increase in magnetic flux pointing to the right. To see which way the current must flow, point your right thumb in the desired direction of the magnetic field B → coil, B → coil, and the current will flow in the direction indicated by curling your right fingers. This is shown by the image of the right hand in the top row of Figure 20.36 . Thus, the current must flow in the direction shown in Figure 4(a) .

In Figure 4(b) , the direction in which the magnet moves is reversed. In the coil, the right-pointing magnetic field B → mag B → mag due to the moving magnet decreases. Lenz’s law says that, to counter this decrease, the emf will drive a current that creates an additional right-pointing magnetic field B → coil B → coil in the coil. Again, point your right thumb in the desired direction of the magnetic field, and the current will flow in the direction indicate by curling your right fingers ( Figure 4(b) ).

Finally, in Figure 4(c) , the magnet is reversed so that the south pole is nearest the coil. Now the magnetic field B → mag B → mag points toward the magnet instead of toward the coil. As the magnet approaches the coil, it causes the left-pointing magnetic field in the coil to increase. Lenz’s law tells us that the emf induced in the coil will drive a current in the direction that creates a magnetic field pointing to the right. This will counter the increasing magnetic flux pointing to the left due to the magnet. Using the right-hand rule again, as indicated in the figure, shows that the current must flow in the direction shown in Figure 4(c) .

Faraday’s Electromagnetic Lab

This simulation proposes several activities. For now, click on the tab Pickup Coil, which presents a bar magnet that you can move through a coil. As you do so, you can see the electrons move in the coil and a light bulb will light up or a voltmeter will indicate the voltage across a resistor. Note that the voltmeter allows you to see the sign of the voltage as you move the magnet about. You can also leave the bar magnet at rest and move the coil, although it is more difficult to observe the results.

• Yes, the current in the simulation flows as shown because the direction of current is opposite to the direction of flow of electrons.
• No, current in the simulation flows in the opposite direction because the direction of current is same to the direction of flow of electrons.

Watch Physics

Induced current in a wire.

This video explains how a current can be induced in a straight wire by moving it through a magnetic field. The lecturer uses the cross product , which a type of vector multiplication. Don’t worry if you are not familiar with this, it basically combines the right-hand rule for determining the force on the charges in the wire with the equation F = q v B sin θ . F = q v B sin θ .

Grasp Check

What emf is produced across a straight wire 0.50 m long moving at a velocity of (1.5 m/s) x ^ x ^ through a uniform magnetic field (0.30 T) ẑ ? The wire lies in the ŷ -direction. Also, which end of the wire is at the higher potential—let the lower end of the wire be at y = 0 and the upper end at y = 0.5 m)?

• 0.15 V and the lower end of the wire will be at higher potential
• 0.15 V and the upper end of the wire will be at higher potential
• 0.075 V and the lower end of the wire will be at higher potential
• 0.075 V and the upper end of the wire will be at higher potential

Worked Example

Emf induced in conducing coil by moving magnet.

Imagine a magnetic field goes through a coil in the direction indicated in Figure 20.37 . The coil diameter is 2.0 cm. If the magnetic field goes from 0.020 to 0.010 T in 34 s, what is the direction and magnitude of the induced current? Assume the coil has a resistance of 0.1 Ω. Ω.

Use the equation ε = − N Δ Φ / Δ t ε = − N Δ Φ / Δ t to find the induced emf in the coil, where Δ t = 34 s Δ t = 34 s . Counting the number of loops in the solenoid, we find it has 16 loops, so N = 16 . N = 16 . Use the equation Φ = B A cos θ Φ = B A cos θ to calculate the magnetic flux

where d is the diameter of the solenoid and we have used cos 0° = 1 . cos 0° = 1 . Because the area of the solenoid does not vary, the change in the magnetic of the flux through the solenoid is

Once we find the emf, we can use Ohm’s law, ε = I R , ε = I R , to find the current.

Finally, Lenz’s law tells us that the current should produce a magnetic field that acts to oppose the decrease in the applied magnetic field. Thus, the current should produce a magnetic field to the right.

Combining equations ε = − N Δ Φ / Δ t ε = − N Δ Φ / Δ t and Φ = B A cos θ Φ = B A cos θ gives

Solving Ohm’s law for the current and using this result gives

Lenz’s law tells us that the current must produce a magnetic field to the right. Thus, we point our right thumb to the right and curl our right fingers around the solenoid. The current must flow in the direction in which our fingers are pointing, so it enters at the left end of the solenoid and exits at the right end.

Let’s see if the minus sign makes sense in Faraday’s law of induction. Define the direction of the magnetic field to be the positive direction. This means the change in the magnetic field is negative, as we found above. The minus sign in Faraday’s law of induction negates the negative change in the magnetic field, leaving us with a positive current. Therefore, the current must flow in the direction of the magnetic field, which is what we found.

Now try defining the positive direction to be the direction opposite that of the magnetic field, that is positive is to the left in Figure 20.37 . In this case, you will find a negative current. But since the positive direction is to the left, a negative current must flow to the right, which again agrees with what we found by using Lenz’s law.

Magnetic Induction due to Changing Circuit Size

The circuit shown in Figure 20.38 consists of a U-shaped wire with a resistor and with the ends connected by a sliding conducting rod. The magnetic field filling the area enclosed by the circuit is constant at 0.01 T. If the rod is pulled to the right at speed v = 0.50 m/s, v = 0.50 m/s, what current is induced in the circuit and in what direction does the current flow?

We again use Faraday’s law of induction, E = − N Δ Φ Δ t , E = − N Δ Φ Δ t , although this time the magnetic field is constant and the area enclosed by the circuit changes. The circuit contains a single loop, so N = 1 . N = 1 . The rate of change of the area is Δ A Δ t = v ℓ . Δ A Δ t = v ℓ . Thus the rate of change of the magnetic flux is

where we have used the fact that the angle θ θ between the area vector and the magnetic field is 0°. Once we know the emf, we can find the current by using Ohm’s law. To find the direction of the current, we apply Lenz’s law.

Faraday’s law of induction gives

Solving Ohm’s law for the current and using the previous result for emf gives

As the rod slides to the right, the magnetic flux passing through the circuit increases. Lenz’s law tells us that the current induced will create a magnetic field that will counter this increase. Thus, the magnetic field created by the induced current must be into the page. Curling your right-hand fingers around the loop in the clockwise direction makes your right thumb point into the page, which is the desired direction of the magnetic field. Thus, the current must flow in the clockwise direction around the circuit.

Is energy conserved in this circuit? An external agent must pull on the rod with sufficient force to just balance the force on a current-carrying wire in a magnetic field—recall that F = I ℓ B sin θ . F = I ℓ B sin θ . The rate at which this force does work on the rod should be balanced by the rate at which the circuit dissipates power. Using F = I ℓ B sin θ , F = I ℓ B sin θ , the force required to pull the wire at a constant speed v is

where we used the fact that the angle θ θ between the current and the magnetic field is 90° . 90° . Inserting our expression above for the current into this equation gives

The power contributed by the agent pulling the rod is F pull v , or F pull v , or

The power dissipated by the circuit is

We thus see that P pull + P dissipated = 0 , P pull + P dissipated = 0 , which means that power is conserved in the system consisting of the circuit and the agent that pulls the rod. Thus, energy is conserved in this system.

Practice Problems

The magnetic flux through a single wire loop changes from 3.5 Wb to 1.5 Wb in 2.0 s. What emf is induced in the loop?

What is the emf for a 10-turn coil through which the flux changes at 10 Wb/s?

Check Your Understanding

• An electric current is induced if a bar magnet is placed near the wire loop.
• An electric current is induced if a wire loop is wound around the bar magnet.
• An electric current is induced if a bar magnet is moved through the wire loop.
• An electric current is induced if a bar magnet is placed in contact with the wire loop.
• Induced current can be created by changing the size of the wire loop only.
• Induced current can be created by changing the orientation of the wire loop only.
• Induced current can be created by changing the strength of the magnetic field only.
• Induced current can be created by changing the strength of the magnetic field, changing the size of the wire loop, or changing the orientation of the wire loop.

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Class 12 Physics Chapter 6 Case Study Question Electromagnetic Induction PDF Download

In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Download of CBSE Class 12 Physics Chapter 6 Electromagnetic Induction Case Study and Passage Based Questions with Answers were Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Physics Electromagnetic Induction  to know their preparation level.

In CBSE Class 12 Physics Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Electromagnetic Induction Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 12 Physics  Chapter 6 Electromagnetic Induction

Case Study/Passage Based Questions

Question: 1

Currents can be induced not only in conducting coils but also in conducting sheets or blocks. Current is induced in solid metallic masses when the magnetic flux threading through them changes. Such currents flow in the form of irregularly shaped loops throughout the body of the metal. These currents look like eddies or whirlpools in water so they are known as eddy currents. Eddy currents have both undesirable effects and practically useful applications. For example, it causes unnecessary heating and wastage of power in electric motors, dynamos, and in the cores of transformers. (I) The working of speedometers of trains is based on

Answer: (b) eddy currents

(ii) Identify the wrong statement

Answer: (a) Eddy currents are produced in a steady magnetic field

(iii) Which of the following is the best method to reduce eddy currents?

Answer: (a) Laminating core

(iv) The direction of eddy currents is given by

Answer: (c) Lenz’s law

(v) Eddy currents can be used to heat localized tissues of the human body. This branch of medical therapy is called 

Answer: (c) Inductothermy

Question: 2

Self Induction. When a current I flows through a coil, flux linked with it is φ = LI, where L is a constant known as self-inductance of the coil.

Any change in current sets up an induced emf in the coil. Thus, the self-inductance of a coil is the induced emf set up in it when the current passing through it changes at the unit rate. It is a measure of the opposition to the growth or the decay of current flowing through the coil. Also, the value of self-inductance depends on the number of turns in the solenoid, its area of cross-section, and the permeability of its core material.

(i) The inductance in a coil plays the same role as (a) inertia in mechanics (b) energy in mechanics (c) momentum in mechanics (d) force in mechanics

Answer: (a) inertia in mechanics

(ii) A current of 2.5 A flows through a coil of inductance 5 H. The magnetic flux linked with the coil is (a) 0.5 Wb (b) 12.5 Wb (c) zero (d) 2 Wb

Answer: (b) 12.5 Wb

The inductance L of a solenoid depends upon its radius R as (a) L ∝ R (b) L ∝ 1/R (c) L ∝ R 2 (d) L ∝ R 3

Answer: (c) L ∝ R2

(iv) The unit of self-inductance is

(a) Weber ampere (b) Weber -1  ampere (c) Ohm second (d) Farad

Answer: (c) Ohm second

(v) The induced emf in a coil of 10 henry inductance in which current varies from 9 A to 4 A in 0.2 second is (a) 200 V (b) 250 V (c) 300 V (d) 350 V

Answer: (b) 250 V

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Case Study Chapter 6 Electromagnetic Induction Class 12 Physics

Please refer to below Case Study Chapter 6 Electromagnetic Induction Class 12 Physics. These Case Study Questions Class 12 Physics will be coming in your examinations. Students should go through the Chapter 6 Electromagnetic Induction Case Study based questions in their Class 12 Physics CBSE, NCERT, KVS book as this will help them to secure more marks in upcoming exams.

Case Study Based Questions Physics Class 12 – Chapter 6 Electromagnetic Induction

Read the para given below and answer the questions that follow:

Motional emf from Lorentz Force. The emf induced across the ends of a conductor due to its motion in a magnetic field is called motional emf. It is produced due to the magnetic Lorentz force acting on the free electrons of the conductor. For a circuit shown in figure, if a conductor of length l moves with velocity V in a magnetic field B perpendicular to both its length and the direction of the magnetic filed, then all the induced parameters are possible in the circuit.

Question. Direction of current induced in a wire moving in a magnetic field is found using (a) Fleming’s left hand rule (b) Fleming’s right hand rule (c) Ampere’s rule (d) Right hand clasp rule

Question. A conducting road of length l is moving in a transverse magnetic field of strength B with velocity V. The resistance of the rod is R. The current in the rod is

Question. A 0.1 m long conductor carrying a current of 50A is held perpendicular to a magnetic field of 1.25 mT. The mechanical power required to move the conductor with a speed of 1 m s -1 is (a) 62.5 mW (b) 625 mW (c) 6.25 mW (d) 12.5 mW

Question. A bicycle generator creates 1.5 V at 15 km/hr. The emf generated at 10 km/hr is (a) 1.5 volts (b) 2 volts (c) 0.5 volts (d) 1 volt

Question. The dimensional formula for emf ε in MKS system will be (a) ML 2 T –3 A -1 (b) [ML 2 T -1 A] (c) ML 2 A (d) [MLT -2 A -2 ]

Lenz’s law states that direction of induced current in a circuit is such that it opposes the change which produces it. Thus, if the magnetic flux linked with a closed circuit increases, the induced current flows in such a direction that a magnetic flux is created in the opposite direction of the original magnetic flux. If the magnetic flux linked with the closed circuit decreases, the induced current flows in such a direction so as to create a magnetic flux in the direction of the original flux.

Question. Which of the following statement is correct? (a) The induced emf is not in the direction opposing the change in magnetic flux so as to oppose the cause which produces it. (b) The relative motion between the coil and magnet produces change in magnetic flux. (c) Emf is induced only if the magnet is moved towards coil. (d) Emf is induced only if the coil is moved towards magnet.

Question. The polarity of induced emf is given by (a) Ampere’s circuital law (b) Biot-Savart law (c) Lenz’s law (d) Fleming’s right hand rule

Question. Lenz’s law is a consequence of the law of conservation of (a) charge (b) mass (c) momentum (d) energy

Question. Near a circular loop of conducting wire as shown in the figure, an electron moves along a straight line. The direction of the induced current if any in the loop is (a) variable (b) clockwise (c) anticlockwise (d) zero

Question. Two identical coils A and B are kept in a horizontal tube side without touching each other. If the current in the coil A increases with time, in response, the coil B (a) is attracted by A (b) remains stationary (c) is repelled (d) rotates

Read the para given below and answer the  questions that follow:

Mutual inductance. Mutual inductance is the phenomenon of inducing emf in a coil, due to a change of current in the neighbouring coil. The amount of mutual inductance that links one coil to another depends very much on the relative positioning of two coils, their geometry and relative separation between them. Mutual inductance between the two coils increase μ r times if the coils are wound over an iron core of relative permeability μ r .

Question. A short solenoid of radius a, number of turns per unit length n 1 , and length L is kept coaxially inside a very long solenoid of radius b, number of turns per unit length n 2 . What is the mutual inductance of the system? (a) μ 0 πb 2 n 1 n 2 L (b) μ 0 πa 2 n 1 n 2 L 2 (c) μ 0 πa 2 n 1 n 2 L (d) μ 0 πb 2 n 1 n 2 L 2

Question. If a change in current of 0.01 A in one coil produces a change in magnetic flux of 2 × 10 -2 weber in another coil, then the mutual inductance between coils is (a) 0 (b) 0.5 H (c) 2 H (d) 3 H

Question. Mutual inductance of two coils can be increased by (a) decreasing the number of turns in the coils (b) increasing the number of turns in the coils (c) winding the coils on wooden cores (d) none of these.

Question. When a sheet of iron is placed in between the two co-axial, then the mutual inductance between the coils will (a) increase (b) decrease (c) remains same (d) cannot be predicted

Question. The SI unit of mutual inductance is (a) ohm (b) mho (c) henry (d) none of these

Eddy Current and their effects. Currents can be induced not only in conducting coils, but also in conducting sheets or blocks. Current is induced in solid metallic masses when the magnetic flux threading through them changes. Such current flow in the form of irregularly shaped loops throughout the body of the metal. These current look like eddies or whirlpools in water so they are known as eddy current. Eddy current have both undesirable effects and practically useful application. For example it causes unnecessary heating and wastage of power in electric motors, dynamos and in the cores of transformers.

Question. The working of speedometers of trains is based on (a) wattless currents (b) eddy currents (c) alternating currents (d) pulsating currents

Question. Identify the wrong statement. (a) Eddy current are produced in a steady magnetic field. (b) Induction furnace uses eddy currents to produce heat. (c) Eddy currents can be used to produce braking force in moving trains. (d) Power meters work on the principle of eddy currents.

Question. Which of the following is the best method to reduce eddy currents? (a) Laminating core (b) Using thick wires (c) By reducing hysteresis loss (d) None of these

Question. The direction of eddy currents is given by (a) Fleming’s left hand rule (b) Biot-Savart law (c) Lenz’s law (d) Ampere-circuital law

Question. Eddy current can be used to heat localised tissues of the human body. This branch of medical therapy is called (a) Hyperthermia (b) Diathermy (c) Inductothermy (d) None of these

Self Induction. When a current I flows through a coil, flux linked with it is φ = LI, where L is a constant known as self inductance of the coil.

Any charge in current sets up an induced emf in the coil. Thus, self inductance of a coil is the induced emf set up in it when the current passing through it changes at the unit rate. It is a measure of the opposition to the growth or the decay of current flowing through the coil. Also, value of self inductance depends on the number of turns in the solenoid, its area of cross-section and the permeability of its core material

Question. The inductance in a coil plays the same role as (a) inertia in mechanics (b) energy in mechanics (c) momentum in mechanics (d) force in mechanics

Question. A current of 2.5 A flows through a coil of inductance 5 H. The magnetic flux linked with the coil is (a) 0.5 Wb (b) 12.5 Wb (c) zero (d) 2 Wb

Question. The inductance L of a solenoid depends upon its radius R as (a) L ∝ R (b) L ∝ 1/R (c) L ∝ R 2 (d) L ∝ R 3

Question. The unit of self-inductance is (a) Weber ampere (b) Weber -1 ampere (c) Ohm second (d) Farad

Question. The induced emf in a coil of 10 henry inductance in which current varies from 9 A to 4 A in 0.2 second is (a) 200 V (b) 250 V (c) 300 V (d) 350 V

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12th Class Physics Electro Magnetic Induction Question Bank

Done case based (mcqs) - electromagnetic induction total questions - 30.

A) The induced e.m.f. is not in the direction opposing the change in magnetic flux so as to oppose the cause which prodocues it. done clear

B) The relative motion between the coil and magnet produces change in magnetic flux. done clear

C) Emf is induced only if the magnet is moved towards coil done clear

D) Emf is induced only if the coil is moved towards magnet. done clear

question_answer 2) The polarity of induced emf is given by

A) Ampere's circuital law done clear

B) Biot-Savart law done clear

C) Lenz's law done clear

D) Fleming's right and hand rule. done clear

question_answer 3) Lenz's law is a consequence of the law of conservation of

A) charge done clear

B) mass done clear

C) momentum done clear

D) energy done clear

A) variable done clear

B) clockwise done clear

C) anticlockwise done clear

D) zero done clear

question_answer 5) Two identical circular coils A and B are kept in a horizontal tube side by side without touching each other. If the current in the coil A increases with time, in response, the coil B

A) is attracted by A done clear

B) remains stationary done clear

C) is repelled done clear

D) rotates done clear

A) \[{{\mu }_{0}}\pi {{b}^{2}}{{n}_{1}}{{n}_{2}}L\] done clear

B) \[{{\mu }_{0}}\pi {{a}^{2}}{{n}_{1}}{{n}_{2}}{{L}^{2}}\] done clear

C) \[{{\mu }_{0}}\pi {{a}^{2}}{{n}_{1}}{{n}_{2}}L\] done clear

D) \[{{\mu }_{0}}\pi {{b}^{2}}{{n}_{1}}{{n}_{2}}{{L}^{2}}\] done clear

question_answer 7) If a change in current of \[0\centerdot 01A\] in one coil produces a change in magnetic flux of \[2\times {{10}^{-2}}\] weber in another coil, then the mutual inductance between coils is

A) 0 done clear

B) \[0\centerdot 5\,H\] done clear

C) 2H done clear

D) 3H done clear

question_answer 8) Mutual inductance of two coils can be increased by

A) decreasing the number of turns in the coils done clear

B) increasing the number of turns in the coils done clear

C) winding the coils on wooden cores done clear

D) none of these done clear

question_answer 9) When a sheet of iron is placed in between the two co-axial coils, then the mutual inductance between the coils will

A) increase done clear

B) decrease done clear

C) remains same done clear

D) cannot be predicted done clear

question_answer 10) The SI units of mutual inductance is

A) ohm done clear

B) mho done clear

C) henry done clear

D) none of these. done clear

A) wattless currents done clear

B) eddy currents done clear

C) alternating currents done clear

D) pulsating currents. done clear

question_answer 12) Identify the wrong statement

A) Eddy currents are produced in a steady magnetic field done clear

B) Induction furnace used eddy currents to produce heat done clear

C) Eddy currents can be used to produce braking force in moving trains done clear

D) Power meters work on the principle of Eddy currents. done clear

question_answer 13) Which of the following is the best method to reduce eddy currents?

A) Laminating core done clear

B) Using thick wires done clear

C) By reducing hysteresis loss done clear

D) None of these. done clear

question_answer 14) The direction of eddy currents is given by

A) Fleming's left hand rule done clear

D) Ampere-circuital law. done clear

question_answer 15) Eddy current can be used to heat localised tissues of the human body. This branch of medical therapy is called

A) Hyperthemia done clear

B) Diathermy done clear

C) Inductothermy done clear

A) inertia in mechanics done clear

B) energy in mechanics done clear

C) momentum in mechanics done clear

D) force in mechanics. done clear

question_answer 17) A current of \[2\centerdot 5\,A\] flows through a coil of inductance 5 H. The magnetic flux linked with the coil is

A) \[0\centerdot 5\,Wb\] done clear

B) \[12\centerdot 5\,Wb\] done clear

C) zero done clear

D) 2 Wb done clear

question_answer 18) The-inductance L of a solenoid depends upon its radius R as

A) \[L\propto R\] done clear

B) \[L\propto 1/R\] done clear

C) \[L\propto {{R}^{2}}\] done clear

D) \[L\propto {{R}^{3}}\] done clear

question_answer 19) The unit of self-inductance is

A) weber ampere done clear

B) weber\[^{-1}\] ampere done clear

C) ohm second done clear

D) farad done clear

question_answer 20) The induced e.m.f. in a coil of 10 henry inductance in which current varies from 9 A to 4 A in \[0\centerdot 2\] secondary is

A) 200V done clear

B) 250V done clear

C) 300V done clear

D) 350V done clear

A) moving towards the solenoid done clear

B) moving into the solenoid done clear

C) at rest inside the solenoid done clear

D) moving out of the solenoid done clear

question_answer 22) Two similar circular loops carry equal curents in the same direction. On moving the coils further apart, the electric current will

A) remain unaltered done clear

B) increases in one and decreases in the second done clear

C) increase in both done clear

D) decrease in both. done clear

question_answer 23) A closed iron ring is held horizontally and a bar magnetic is dropped through the ring with its length along the axis of the ring. The acceleration of the falling magnet is

A) equal to g done clear

B) less than g done clear

C) more than g done clear

D) depends on the diameter of the ring and length of magnet. done clear

question_answer 24) Whenever there is a relative motion between a coil and a magnet, the magnitude of induced e.m.f. set up in the coil does not depend upon the

A) relative speed between the coil and magnet done clear

B) magnetic moment of the coil done clear

C) resistance of the coil done clear

D) number of turn in the coil. done clear

question_answer 25) A coil of metal wire is kept stationary in a non-uniform magnetic field

A) an e.m.f. and current both are induced in the coil done clear

B) a current but no e.m.f. is induced in the coil done clear

C) an e.m.f. but no current is induced in the coil done clear

D) neither e.m.f. nor current is induced in the coil. done clear

B) Fleming's right hand rule done clear

C) Ampere's rule done clear

D) Right hand clasp rule. done clear

question_answer 27) A conducting rod of length l is moving in a transeverse magnetic field of strength B with velocity v. The resistance of the rod is R. The current in the rod is

A) \[\frac{Blv}{R}\] done clear

B) \[B/v\] done clear

D) \[\frac{{{B}^{2}}{{v}^{2}}{{l}^{2}}}{R}\] done clear

question_answer 28) \[A\,0\centerdot 1m\] long conductor carrying a current of 50 A is held perpendicular to a magnetic field of \[1\centerdot 25\,mT\]. The mechanical power required to move the conductor with a speed of \[1\,m\,{{s}^{-1}}\]is

A) \[62\centerdot 5\,mW\] done clear

B) \[625\,mW\] done clear

C) \[6\centerdot 25\,mW\] done clear

D) \[12\centerdot 5\,mW\] done clear

question_answer 29) A bicycyle generator creates \[1\centerdot 5\,V\] at \[15\,km/hr\]. The EMF gererated at \[10\,km/hr\] is 

A) \[1\centerdot 5\,Volts\] done clear

B) 2 volts done clear

C) \[0\centerdot 5\,Volts\] done clear

D) 1 volt done clear

question_answer 30) The dimensional formula for e.m.f. e is MKS system will be 

A) \[\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-1}} \right]\] done clear

B) \[\left[ M{{L}^{2}}{{T}^{-1}}A \right]\] done clear

C) \[\left[ M{{L}^{2}}A \right]\] done clear

D) \[\left[ ML{{T}^{-2}}{{A}^{-2}} \right]\] done clear

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Case Based (MCQs) - Electromagnetic Induction

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• Physics Concept Questions And Answers

Electromagnetic Induction Questions

The process of electromagnetic induction generates a voltage or Electromotive Force (EMF) across the electrical conductor due to a changing magnetic field. Electromagnetic induction is generally referred to as induction.

In 1831 electromagnetic induction was discovered by Michael Faraday, and later James Clerk Maxwell mathematically described it as Faraday’s law of induction. The formula gives electromagnetic induction:

e = N × dΦ / dt

e – induced voltage (in volts)

N – number of turns in the coil

Φ – magnetic flux (in Webbers)

t -time (in seconds)

The induction of an electromotive force by the motion of a conductor across a magnetic field or by a change in magnetic flux in a magnetic field is known as Electromagnetic Induction. This phenomenon takes place when a conductor is set in a moving magnetic field or when a conductor is always moving in a stationary magnetic field.

The concept of electromagnetic induction is used in generators, inductors, motors, and transformers. The magnetic flow meter is based on this concept too.

The first and second laws of electromagnetic induction are:

First law: Whenever a conductor is placed in a varying magnetic field, EMF induces, and this EMF is called an induced EMF. If the conductor is a closed circuit then the induced current flows through it.

Second law: The magnitude of the induced EMF is equal to the rate of change of flux linkages.

The amount of voltage induced in a coil is proportional to the number of turns of the coil and the rate of changing the magnetic field.

Lenz’s law of electromagnetic induction states that, when an emf induces according to Faraday’s law, the polarity (direction) of that induced emf is such that it opposes the cause of its production.

According to Lenz’s law

E = -N (dΦ/ dt) (volts)

A step-down transformer is the one in which the voltage is higher in the primary than the secondary voltage.

Important Electromagnetic Induction Questions with Answer

1. The direction of the induced field can be predicted by____

• Kepler’s law
• Newton’s law
• Lenz’s law
• Faraday’s law

Answer: c) Lenz’s law

Explanation: Lenz’s law states that the direction of the electric current induced in a conductor by a changing magnetic field is such that the magnetic field created by the induced current opposes changes in the initial magnetic field.

2. Fill in the blanks: Electric generator converts electrical energy into____

• Chemical energy
• Thermal energy
• Electrical energy
• Solar energy

Answer: c) Electrical energy.

3. The phenomenon called electromagnetic induction was first investigated by

Answer: c) Faraday

Explanation: Electromagnetic Induction was discovered by Michael Faraday in 1831

4. Magnetic flux is represented as

• \(\begin{array}{l}\Theta\end{array} \)
• \(\begin{array}{l}\epsilon\end{array} \)
• \(\begin{array}{l}\alpha\end{array} \)

Answer: d) Φ

5. Factors that affect the voltage generation in Faraday’s experiment is

• Number of Coils
• Changing Magnetic Field
• Changing Environment
• Option a) and b)

Answer: Option a) and b)

Explanation: Changing magnetic field and the number of coils affects the voltage production in Faraday’s experiment.

6. Define magnetic flux.

The number of magnetic field lines passing through a given closed surface gives magnetic flux.

7. Lenz’s law is derived from ____.

• Coulomb’s Law

Answer: d) Faraday’s law

8. State true or false: Eddy currents flow in closed loops in planes perpendicular to the magnetic field.

Answer: a) TRUE

9. The SI unit of magnetic flux is

Answer: b) Weber

Explanation: The SI unit of magnetic flux is Weber.

10. The induced voltage is found according to Faraday’s law using the formula____

• e = N × dΦdt
• e = N × dΦ/dt
• e = N + dΦdt
• e = N – dΦdt

Answer: a) e = N × dΦdt

Practice Questions

• Explain Faraday’s experiment.
• State Faraday’s law of electromagnetism.
• What is an electrical generator?
• What are the applications of electromagnetic induction?
• Define electromagnetic induction.

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NCERT Solutions For Class 12 Physics Chapter 6 Electromagnetic Induction

NCERT Solutions For Class 12 Physics Chapter 6: Score good marks in class 12th Board physics subjects with errorless NCERT Solutions For Class 12 Physics Chapter 6 Electromagnetic Induction.

February 9, 2024

Table of Contents

NCERT Solutions for class-12 Physics Chapter 6 Electromagnetic Induction is prepared by our senior and renown teachers of Physics Wallah primary focus while solving these questions of class-12 in NCERT textbook, also do read theory of this Chapter 6 Electromagnetic Induction while going before solving the NCERT questions. You can download and share NCERT Solutions  of Class 12 Physics from Physics Wallah.

NCERT Solutions For Class 12 Physics Chapter 6 Overview

NCERT Solutions for Class 12 Physics Chapter 6 contains all the important topics for the exams. Our experts created these questions for the students to ace the examination. This article contains all the important questions and their easy to understand answers for the better understanding.

Students are advised to go through these questions to clarify their concepts better. These questions are created to help students in the better understanding of the NCERT Solutions for Class 12 Physics Chapter 6 (Electromagnetic Induction).

NCERT Solutions For Class 12 Physics Chapter 6 PDF

To help students understand and practise chapter concepts, our team of experts at Physics Wallah has developed thorough solutions for NCERT Class 12 Physics, Chapter 6. The purpose of these questions is to make explanations easier to understand for students.

You can use the following link to obtain the NCERT Solutions for Class 12 Physics Chapter 6 Pdf.

NCERT Solutions for Class 12 Physics Chapter 6 PDF Download Link

NCERT Solutions For Class 12 Physics Chapter 6

Answer The Following Question Answer of NCERT Solutions For Class 12 Physics Chapter 6 Electromagnetic Induction:

Question 1. Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f ).

Solution : Lenz’s law shows the direction of induced current in a closed loop. In the given two figures they shows the direction of induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.

We can predict the direction induced current in different situation by using the Lenz’s rule.

(i) The direction of the induced current is along qrpq.

(ii) The direction of the induced current is along prqp.

(iii) The direction of the induced current is along yzxy.

(iv) The direction of the induced current is along zyxz.

(v) The direction of the induced current is along xryx.

(vi) No current is induced since the field lines are lying in the plane of the closed loop.

NCERT Solution For Class 12 Physics Chapter 4

Question 2. Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19:

(a) A wire of irregular shape turning into a circular shape;

(b) A circular loop being deformed into a narrow straight wire.

Solution : According to Lenz’s law, the direction of the induced emf is such that it tends to produce a current that opposes the change in the magnetic flux that produced it.

(a) When the shape of the wire changes, the flux piercing through the unit surface area increases. As a result, the induced current produces an opposing flux. Hence, the induced current flows along adcb.

(b) When the shape of a circular loop is deformed into a narrow straight wire, the flux piercing the surface decreases. Hence, the induced current flows along adcba

NCERT Solution For Class 12 Physics Chapter 5

Question 3. A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

Solution : Number of turns on the solenoid = 15 turns/cm = 1500 turns/m

Number of turns per unit length, n = 1500 turns

The solenoid has a small loop of area, A = 2.0 cm2 = 2 × 10−4 m2

Current carried by the solenoid changes from 2 A to 4 A.

Change in current in the solenoid, di = 4 − 2 = 2 A

Change in time, dt = 0.1 s

Induced emf in the solenoid is given by Faraday’s law as:

Ø= Induced flux through the small loop

= BA … (ii)

B = Magnetic field

μ0 = Permeability of free space

= 4π×10−7 H/m

Hence, equation (i) reduces to:

Hence, the induced voltage in the loop is 7.5 x 10 -6  v

Question 4. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s−1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

Solution : Length of the rectangular wire, l = 8 cm = 0.08 m

Width of the rectangular wire, b = 2 cm = 0.02 m

Hence, area of the rectangular loop,

= 0.08 × 0.02

= 16 × 10−4 m2

Magnetic field strength, B = 0.3 T

Velocity of the loop, v = 1 cm/s = 0.01 m/s

(a) Emf developed in the loop is given as:

= 0.3 × 0.08 × 0.01 = 2.4 × 10−4 V

Hence, the induced voltage is 2.4 × 10−4 V which lasts for 2 s.

(b) Emf developed, e = Bbv

= 0.3 × 0.02 × 0.01 = 0.6 × 10−4 V

Hence, the induced voltage is 0.6 × 10−4 V which lasts for 8 s.

Question 5. A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s−1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

Solution : Length of the rod, l = 1 m

Angular frequency,ω = 400 rad/s

Magnetic field strength, B = 0.5 T

One end of the rod has zero linear velocity, while the other end has a linear velocity of lω.

Emf developed between the centre and the ring,

Hence, the emf developed between the centre and the ring is 100 V.

Question 6. A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s−1 in a uniform horizontal magnetic field of magnitude 3.0×10−2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

Solution : Max induced emf = 0.603 V

Average induced emf = 0 V

Max current in the coil = 0.0603 A

Average power loss = 0.018 W

(Power comes from the external rotor)

Radius of the circular coil, r = 8 cm = 0.08 m

Area of the coil, A = πr2 = π × (0.08)2 m2

Number of turns on the coil, N = 20

Angular speed, ω = 50 rad/s

Magnetic field strength, B = 3 × 10−2 T

Resistance of the loop, R = 10 Ω

Maximum induced emf is given as:

= 20 × 50 × π × (0.08)2 × 3 × 10−2

The maximum emf induced in the coil is 0.603 V.

Over a full cycle, the average emf induced in the coil is zero.

Maximum current is given as:

= 0.603/10 = 0.0603

Average power loss due to joule heating:

= 0.603 x 0.0603/2 = 0.018W

The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent. It must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.

Question 7. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s−1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10−4 Wb m−2. 

(a) What is the instantaneous value of the emf induced in the wire?

(b) What is the direction of the emf?

(c) Which end of the wire is at the higher electrical potential?

Solution : Wire’s Length, l = 10 m

Speed of the wire with which it is falling, v = 5.0 m/s

Strength of magnetic field, B = 0.3×10−4Wbm −2

(a) EMF induced in the wire, e = Blv

=0.3×10 −4 ×5×10=1.5×10 −3 V

(b) We can determine the direction of the induced current by using the Fleming’s right hand thumb rule, here the current is flowing in the direction from West to East.

(c) In this case the eastern end of the wire will be having higher potential

Question 8. Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.

Solution : Initial current, I1 = 5.0 A

Final current, I2 = 0.0 A

Change in current,dl = I 1  – I 2  = 5A

Time taken for the change, t = 0.1 s

Average emf, e = 200 V

For self-inductance (L) of the coil, we have the relation for average emf as:

e = L di/dt

L = e/(di/dt)

200/(5/0.1) = 4H

Hence, the self induction of the coil is 4 H.

Question 9. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?

Given, a pair of adjacent coils.

Mutual inductance, M = 1.5 H

Current in the coil, I = 20 A

Time ,t = 0.5 s

Using formula, Φ = MI we get,

Φ = 1.5 x 20 = 30H

Hence, the change in the flux linkage is 30 Wb.

Question 10. Calculate the difference in voltage developed between two ends of a jet plane, when the jet plane is travelling towards west with a speed of 1800 km/h and the wing of the plane is having a span of 25 m. We are given that the earth’s magnetic field are the location has a magnitude of 5×10 −4 T and the dip angle is 30 0 .

Solution : Speed of the plane with which it is moving,  v = 1800 km/h = 500 m/s

Wing span of the jet, l = 25 m

Magnetic field strength by earth, B = 5 ×10 −4 T

Dip angle, δ=30 ∘

Vertical component of Earth’s magnetic field,

Bv=Bsin δ=5 ×10 −4  sin30 ∘=2.5 ×10 − 4 T

Difference in voltage between both the ends can be calculated as:

e=(Bv)× l × v = 2.5 × × 10 − 4  ×25 ×500 =3.123V

Hence, the voltage difference developed between the ends of the wings is 3.125 V.

Question 11. Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s−1. If the cut is joined and the loop has a resistance of 1.6 Ω how much power is dissipated by the loop as heat? What is the source of this power?

Solution : Sides of the rectangular loop are 8 cm and 2 cm.

Hence, area of the rectangular wire loop,

A = length × width

= 8 × 2 = 16 cm2

Initial value of the magnetic field, B’ = 0.3 T

Rate of decrease of the magnetic field, dB/dt = 0.02T/s

Emf developed in the loop is given as:

dΦ = Change in flux through the loop area = AB

∴ e = d(AB)/dt = AdB/dt

Resistance of the loop, R = 1.6 Ω

The current induced in the loop is given as:

= 0.32 x 10 -4 /1.6

Power dissipated in the loop in the form of heat is given as:

= (2 × 10 -5 ) x 1.6

= 6.4 x 10 -10  W

The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.

Question 12. A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s−1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10−3 T cm−1 along the negative x-direction (that is it increases by 10− 3 T cm−1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10−3 T s−1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.

Solution : Side of the square loop, s = 12 cm = 0.12 m

Area of the square loop, A = 0.12 × 0.12 = 0.0144 m2

Velocity of the loop, v = 8 cm/s = 0.08 m/s

Gradient of the magnetic field along negative x-direction,

And, rate of decrease of the magnetic field,

Resistance of the loop, R = 4.5Ω  = 4.5 x 10 – ³ Ω

Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:

Rate of change of the flux due to explicit time variation in field B is given as:

Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:

∴Induced current, i = e/R

Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.

Question 13. It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of magnet.

Solution : Area of the small flat search coil, A = 2 cm2 = 2 × 10−4 m2

Number of turns on the coil, N = 25

Total charge flowing in the coil, Q = 7.5 mC = 7.5 × 10−3 C

Total resistance of the coil and galvanometer, R = 0.50

Now, Initial flux/turnwhen coil is normal to the field ,Φ i  = BA

Hence, the field strength of the magnet is 0.75 T.

Question 14. Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.

(a) Suppose K is open and the rod is moved with a speed of 12 cm s−1 in the direction shown. Give the polarity and magnitude of the induced emf.

(b) Is there an excess charge built up at the ends of the rods when

K is open? What if K is closed? 

(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.

(d) What is the retarding force on the rod when K is closed?

(e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12 cm s−1) when K is closed? How much power is required when K is open?

(f) How much power is dissipated as heat in the closed circuit?

What is the source of this power?

(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

Solution : Length of the rod, l = 15 cm = 0.15 m

Magnetic field strength, B = 0.50 T

Resistance of the closed loop, R = 9 mΩ = 9 × 10−3 Ω

(a) Induced emf = 9 mV; polarity of the induced emf is such that end P shows positive while end Q shows negative ends.

Speed of the rod, v = 12 cm/s = 0.12 m/s

Induced emf is given as:

= 0.5 × 0.12 × 0.15

= 9 × 10−3 v

The polarity of the induced emf is such that end P shows positive while end Q shows negative ends.

(b) Yes; when key K is closed, excess charge is maintained by the continuous flow of current.

When key K is open, there is excess charge built up at both ends of the rods.

When key K is closed, excess charge is maintained by the continuous flow of current.

(c) Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod.

There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.

(d) Retarding force exerted on the rod, F = IBl

I = Current flowing through the rod

No power will be expended when the key K will be opened.

Speed of the rod, v = 12 cm/s = 0.12 m /s

Power, P = Fv

=75× 10  − 3 ×0.12 = 9× 10 − 3  W=9mW

When the key K is opened no power is expended.

(f) Power is provided by an external agent.

Power loss in the form of heat = I2R 12×9×10 −3 

There would be no emf induced in the coil. As the emf induces if the motion of the rod cuts the field lines. But in this case motion of the rod does not cut across the field lines.

Question 15. An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10−3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

Solution : Length of the solenoid, l = 30 cm = 0.3 m

Area of cross-section, A = 25 cm2 = 25 × 10−4 m2

Number of turns on the solenoid, N = 500

Current in the solenoid, I = 2.5 A

Current flows for time, t = 10−3 s

Average back emf, e = dΦ/dt

dΦ = Change in flux

= NAB … (2)

B = Magnetic field strength

μ0 = Permeability of free space = 4π × 10−7 T m A−1

Using equations (2) and (3) in equation (1), we get

Hence, the average back emf induced in the solenoid is 6.5 V.

Question 16. (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21.

(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s.

Calculate the induced emf in the loop at the instant when x = 0.2 m.

Take a = 0.1 m and assume that the loop has a large resistance.

Solution : (a) Take a small element dy in the loop at a distance y from the long straight wire (as shown in the given figure).

Magnetic flux associated with element

dØ = BdA = Bady

dA = Area of element dy = a dy

B = Magnetic field at distance y

B = μ0 I / 2πy

I = Current in the wire

y tends from x to a + x

for mutual inductunce M, the flux is given as :

Question 17. A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,

B = − B0 k (r ≤ a; a < R)

= 0 (otherwise)

What is the angular velocity of the wheel after the field is suddenly switched off?

r = Distance of the point within the wheel

Mass of the wheel = M

Radius of the wheel = R

At distance r,the magnetic force is balanced by the centripetal force i.e.,

It is the angular velocity of the wheel when the field is suddenly shut off.

NCERT Solutions For Class 12 Physics Chapter 6 FAQs

NCERT Solutions Class 12 Physics Chapter 6 Electromagnetic Induction.

Electromagnetic Induction or Induction is a process in which a conductor is put in a particular position and magnetic field keeps varying or magnetic field is stationary and a conductor is moving. This produces a Voltage or EMF (Electromotive Force) across the electrical conductor.

Determine the object's original velocity by dividing the time it took for the object to travel a given distance by the total distance. In the equation V = d/t, V is the velocity, d is the distance, and t is the time.

Initial velocity is the speed (along with direction) of the object with which it starts moving. On the other side, the final velocity is the speed (along with direction) of the same moving object once it has reached its final position.

Whenever a conductor is placed in a varying magnetic field, EMF induces and this emf is called an induced emf and if the conductor is a closed circuit than the induced current flows through it.

NCERT Solutions For Class 12 Physics Chapter 12 Atoms

NCERT Solutions For Class 12 Physics Chapter 9 Ray Optics and Opticals Instrument

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Electromagnetic Induction Class 12 Physics Important Questions

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Electromagnetic Induction Class 12 Physics Important Questions. We know Physics is tough subject within the consortium of science subjects physics is an important subject. But if you want to make career in these fields like IT Consultant, Lab Technician, Laser Engineer, Optical Engineer etc. You need to have strong fundamentals in physics to crack the exam.   myCBSEguide has just released Chapter Wise Question Answers for class 12 Physics. There chapter wise Extra Questions with complete solutions are available for download in  myCBSEguide   website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Physics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in  CBSE Class 12 Physics syllabus  and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.

Class 12 Physics Extra Questions

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CBSE Class 12 Physics Chapter 6 Extra Questions

Class – 12 Physics (Electromagnetic Induction)

Two coils {tex}{{\rm{C}}_{\rm{1}}}{/tex} , {tex}{{\rm{C}}_{\rm{2}}}{/tex} have {tex}{{\rm{N}}_{\rm{1}}}{/tex} and {tex}{{\rm{N}}_{\rm{2}}}{/tex} turns respectively. Current {tex}{{\rm{i}}_{\rm{1}}}{/tex} in coil {tex}{{\rm{C}}_{\rm{1}}}{/tex} is changing with time. The emf in {tex}{{\rm{C}}_{\rm{2}}}{/tex} is given by

• {tex}-\frac{di_1}{dt}{/tex}
• {tex}-M\frac{di_1}{dt}{/tex}
• {tex}-M\frac{di_2}{dt}{/tex}
• {tex}-M\frac{di_1^2}{dt}{/tex}

A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, change of flux linkage with the other coil is

According to Lenz’s law.

• The polarity of induced emf is such that it tends to produce a current which aids the change in magnetic flux that produced it.
• The induced emf is proportional rate of change in magnetic flux that produced it.
• The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
• The induced emf is proportional change in magnetic flux that produced it.

The magnetic field between the Horizontal poles of an electromagnet is uniform at any time, but its magnitude is increasing at the rate of 0.020 T/s. The area of a horizontal conducting loop in the magnetic field is 120cm 2 , and the total circuit resistance, including the meter, is {tex}{\rm{5 }}\Omega {/tex} . Induced emf and the induced current in the circuit are

• 0.18 mV, 0.048 mA
• 0.20 mV, 0.048 mA
• 0.24 mV, 0.048 mA
• 0.22 mV, 0.048 mA

When current changes from + 2 A to – 2 A in 0.05 sec, an emf of 8 V is induced in a coil. The coefficient of self inductance of the coil is:

A train is moving with uniform velocity from north to south. Will any induced emf appear across the ends of the axle?

A metallic piece gets hot when surrounded by a coil carrying high-frequency alternating current. Why?

Show that the energy stored in an inductor i.e. the energy required to build current in the circuit from zero to I is {tex}\frac{1}{2}L{I^2}{/tex} , where L is the self inductance of the circuit.

A long solenoid of 20 turns per cm has a small loop of area 4 cm 2 placed inside the solenoid normal to its axis. If the current by the solenoid changes steadily from 4 A to 6A in 0.25, what is the (average) induced emf in the loop while the current is changing?

• A toroidal solenoid with an air core has an average radius of 0.15 m, area of cross section {tex}12 \times {10^{ – 4}}{m^2}{/tex} and 1200 turns. Obtain the self inductance of the toroid. Ignore field variation across the cross section of the toroid.
• A second coil of 300 turns is wound closely on the toroid above. If the current in the primary coil is increased from zero to 2.0 A in 0.05 s, obtain the induced emf in the secondary coil.

A 1.0 m long conducting rod rotates with an angular frequency of 400rad s -1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

A rectangular loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 tesla directed normal to the loop. What is the voltage developed across the cut if velocity of loop is {tex}1 \; cms^{-1}{/tex} in a direction normal to the (i) longer side (ii) shorter side of the loop? For how long does the induced voltage last in each case?

A current of 10 A is flowing in a long straight wire situated near a rectangular coil. The two sides, of the coil, of length 0.2 m are parallel to the wire. One of them is at a distance of 0.05 m and the other is at a distance of 0.10 m from the wire. The wire is in the plane of the coil. Calculate the magnetic flux through the rectangular coil. If the current uniformly to zero in 0.02 s, find the emf induced in the coil and indicate the direction in which the induced current flows.

Class – 12 Physics (Electromagnetic Induction) Answers

• ​​​​ {tex}-M\frac{di_1}{dt}{/tex} Explanation: {tex}{N_2}{\phi _2} \propto {i_1}{/tex} {tex}{N_2}{\phi _2} = M{i_1}{/tex} {tex}{e_2} = – {N_2}\frac{{d{\phi _2}}}{{dt}} = – \frac{{d({N_2}{\phi _2})}}{{dt}}{/tex} {tex}{e_2} = – M\frac{{d{i_1}}}{{dt}}{/tex}
• 30 Wb Explanation: {tex}\Delta \phi = M\Delta i = 1.5 \times 20 = 30{\rm{Wb}}{/tex}
• The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. Explanation: The direction of an induced emf, or the current, in any circuit is such as to oppose the cause that produces it. If the direction of the induced current were such as not to oppose then we would be obtaining electrical energy continuously without doing work, which is impossible.
• 0.24 mV, 0.048 mA Explanation: e = Rate of change of magnetic field {tex} \times {/tex} area {tex} = 0.02 \times 120 \times {10^{ – 4}} = 0.24{\rm{mV}}{/tex} {tex}i = {e \over R} = {{0.24} \over 5} = 0.48{\rm{mA}}{/tex}
• 0.1 H Explanation: {tex}L = – \frac{e}{{\frac{{\Delta i}}{{\Delta t}}}}{/tex} {tex}{{\Delta i} \over {\Delta t}} = – {4 \over {0.05}} = – 80{/tex} e = 8 volt {tex}{\rm{L}} = – {8 \over { – 80}} = 0.1{\rm{H}}{/tex}
• Yes. The vertical component of earth’s magnetic field shall induce emf.
• It happens due to production of eddy currents.
• ring 1 is clockwise.
• ring 2 is anti-clockwise.
• Energy spent by the source to increase current from i to i + di in time dt in an inductor. {tex} = L\frac{{di}}{{dt}} \times i \times dt{/tex} = Li di Energy required to increase current from 0 to I {tex}E = \int\limits_0^I {Li} \,di = L\left[ {\frac{{{i^2}}}{2}} \right]_0^I{/tex} {tex}E = L\left[ {\frac{{{I^2}}}{2} – 0} \right] = \frac{1}{2}L{I^2}{/tex}
• The induced emf, {tex}\varepsilon = – \frac{{d\phi }}{{dt}}{/tex} {tex}\varepsilon = \frac{d}{{dt}}(BA\cos \phi )(\because \cos \phi = 1){/tex} {tex} = \frac{{20 \times 100 \times 4 \times {{10}^{ – 4}}}}{{0.2}}{/tex} = 4 volt.
• {tex}B = {\mu _0}{n_1}I = \frac{{{\mu _0}{N_1}I}}{1} = \frac{{{\mu _0}{N_1}I}}{{2\pi r}}{/tex} Total magnetic flux, {tex}{\phi _B} = {N_1}BA = \frac{{{\mu _0}N_1^2IA}}{{2\pi r}}{/tex} But {tex}{\phi _B} = LI{/tex} {tex}\therefore L = \frac{{{\mu _0}N_1^2A}}{{2\pi r}}{/tex} Or {tex}L = \frac{{4\pi \times {{10}^{ – 7}} \times 1200 \times 1200 \times 12 \times {{10}^{ – 4}}}}{{2\pi \times 0.15}}H{/tex} {tex} = 2.3 \times {10^{ – 3}}H{/tex} = 2.3 mH
• {tex}\left| E \right| = \frac{d}{{dt}}({\phi _2}){/tex} where {tex}{\phi _2}{/tex} is the total magnetic flux linked with the second coil. {tex}\left| E \right| = \frac{d}{{dt}}({N_2}BA) = \frac{{d}}{{dt}}\left[ {{N_2}\frac{{{\mu _0}{N_1}I}}{{2\pi r}}A} \right]{/tex} or {tex}\left| E \right| = \frac{{{\mu _0}{N_1}{N_2}A}}{{2\pi r}}\frac{{dI}}{{dt}}{/tex} {tex}\left| E \right| = \frac{{4\pi \times {{10}^{ – 7}} \times 1200 \times 300 \times 12 \times {{10}^{ – 4}} \times 2}}{{2\pi \times 0.15 \times 0.05}}V{/tex} = 0.023 V
• Here, l = 1 m, {tex}\omega = 400{s^{ – 1}}{/tex} B = 0.5 T, e = ? Note that linear velocity of one end of rod is zero and linear velocity of other end is {tex}\left( {1\omega } \right){/tex} . Average linear velocity {tex}v = \frac{{0 + 1\omega }}{2} = \frac{{1\omega }}{2}\left( {\because v = r\omega } \right){/tex} {tex}\therefore e = Blv = Bl\frac{{\left( {1\omega } \right)}}{2} = \frac{{B{l^2}\omega }}{2} = \frac{{0.5 \times {1^2} \times 400}}{2}=100v{/tex}
• The magnitude of induced emf, {tex}\varepsilon= B.l.v{/tex} {tex}=0.3\times 8 \times 10^{-2}\times 10 ^{-2}{/tex} {tex}=2.4\times 10^{-4} \;V{/tex} Time for which induced emf will last is equal to the time taken by the coil to move outside the field is {tex}I = \frac{{dis\tan ce\;travelled}}{{velocity}} = \frac{{2 \times {{10}^{ – 2}}}}{{{{10}^{ – 2}}}}\;m{/tex} = 2 sec.
• The conductor is moving outside the field normal to the shorter side. {tex}b=2\times 10 ^{-2} \; m{/tex} The magnitude of induced emf is {tex}\varepsilon= B.b.v{/tex} {tex}=0.3\times 2 \times 10^{-2}\times 10 ^{-2}{/tex} {tex}=0.6\times 10^{-4} \; V{/tex} Time, {tex}t = \frac{{dis\tan ce\;travelled}}{{velocity}} = \frac{{8 \times {{10}^{ – 2}}}}{{{{10}^{ – 2}}}}\;sec=8 \; sec{/tex}

Chapter Wise Extra Questions of Class 12 Physics Part I & Part II

• Electric Charges and Fields
• Electrostatic Potential and Capacitance
• Current Electricity
• Moving Charges and Magnetism
• Magnetism and Matter
• Electromagnetic Induction
• Alternating Current
• Electromagnetic Waves
• Ray Optics and Optical
• Wave Optics
• Dual Nature of Radiation and Matter
• Electronic Devices

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Class 12th Physics - Electromagnetic Induction Case Study Questions and Answers 2022 - 2023

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 12th Physics Subject - Electromagnetic Induction, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Electromagnetic induction case study questions with answer key.

12th Standard CBSE

Final Semester - June 2015

(ii) The polarity of induced emf is given by

(iii) Lenz's law is a consequence of the law of conservation of

(v) Two identical circular coils A and B are kept in a horizontal tube side by side without touching each other. If the current in coil A increases with time, in response, the coil B.

(ii) If a change in current of 0.01 A in one coil produces a change in magnetic flux of 2 x l0 -2 weber in another coil, then the mutual inductance between coils is

(iii) Mutual inductance of two coils can be increased by

(iv) When a sheet of iron is placed in between the two co-axial coils, then the mutual inductance between the coils will

(v) The SI unit of mutual inductance is

Currents can be induced not only in conducting coils, but also in conducting sheets or blocks. Current is induced in solid metallic masses when the magnetic flux threading through them changes. Such currents flow in the form of irregularly shaped loops throughout the body of the metal. These currents look like eddies or whirlpools in water so they are known as eddy currents. Eddy currents have both undesirable effects and practically useful applications. For example it causes unnecessary heating and wastage of power in electric motors, dynamos and in the cores of transformers. (I) The working of speedometers of trains is based on

(ii) Identify the wrong statement

(iii) Which of the following is the best method to reduce eddy currents?

(iv) The direction of eddy currents is given by

(v) Eddy currents can be used to heat localised tissues of the human body. This branch of medical therapy is called 

(ii) A current of 2.5 A flows through a coil of inductance 5 H. The magnetic flux linked with the coil is

(iii) The inductance L of a solenoid depends upon its radius R as

(iv) The unit of self-inductance is

(v) The induced e.m.f in a coil of 10 henry inductance in which current varies from 9 A to 4 A in 0.2 second is

(ii) Two similar circular loops carry equal currents in the same direction. On moving the coils further apart, the electric current will

(iii) A closed iron ring is held horizontally and a bar magnet is dropped through the ring with its length along the axis of the ring. The acceleration of the falling magnet is

(iv) Whenever there is a relative motion between a coil and a magnet, the magnitude of induced emf set up in the coil does not depend upon the

(v) A coil of metal wire is kept stationary in a non-uniform magnetic field

(ii) A conducting rod of length I is moving in a transverse magnetic field of strength B with velocity v. The resistance of the rod is R. The current in the rod is

(iii) A 0.1 m long conductor carrying a current of 50 A is held perpendicular to a magnetic field of 1.25 mT. The mechanical power required to move the conductor with a speed of 1 m s -1 is

(iv) A bicycle generator creates 1.5 V at 15 km/hr. The EMF generated at 10 km/hr is

(v) The dimensional formula for emf E in MKS system will be

A very small circular loop of area 5 x 10 -4 cm 2 with resistance 4 \(\Omega\)  is placed such that it is concentric and in the same plane with another loop of radius 10 cm. A constant current of 2A is passed in the bigger loop, which is rotated with angular speed of  \(\omega\)  rad S -1 , about its diameter. The magnetic field produced due to the bigger loop provides magnetic flux, which is linked with smaller loop. (i) Determine magnetic field at the centre. (ii) Determine magnetic flux linked with smaller loop. (iii) What is the value of induced emf and current in the smaller loop as a function of time.

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Physics MCQs for Class 12 with Answers Chapter 6 Electromagnetic Induction

• Last modified on: 3 years ago
• Reading Time: 28 Minutes

MCQs based on Electromagnetic Induction:

Q.1. Whenever the magnetic flux linked with an electric circuit changes, an emf is induced in the circuit. This is called (a) electromagnetic induction (b) lenz’s law (c) hysteresis loss (d) kirchhoff’s laws

Q.2. In electromagnetic induction, the induced charge is independent of (a) change of flux (b) time. (c) resistance of the coil (d) None of these

Q.3. An induced e.m.f. is produced when a magnet is plunged into a coil. The strength of the induced e.m.f. is independent of (a) the strength of the magnet (b) number of turns of coil (c) the resistivity of the wire of the coil (d) speed with which the magnet is moved

Q.4. According to Faraday’s law of electromagnetic induction (a) electric field is produced by time varying magnetic flux. (b) magnetic field is produced by time varying electric flux. (c) magnetic field is associated with a moving charge. (d) None of these

Q.5. A moving conductor coil produces an induced e.m.f. This is in accordance with (a) Lenz’s law (b) Faraday’s law (c) Coulomb’s law (d) Ampere’s law

Q.6. A coil of insulated wire is connected to a battery. If it is taken to galvanometer, its pointer is deflected, because (a) the induced current is produced (b) the coil acts like a magnet (c) the number of turns in the coil of the galvanometer are changed (d) None of these

Q.7. The polarity of induced emf is given by (a) Ampere’s circuital law (b) Biot-Savart law (c) Lenz’s law (d) Fleming’s right hand rule

Q.8. The self inductance of a coil is a measure of (a) electrical inertia (b) electrical friction (c) induced e.m.f. (d) induced current

Q.9. The coils in resistance boxes are made from doubled insulated wire to nullify the effect of (a) heating (b) magnetism (c) pressure (d) self induced e.m.f.

Q.10. Two pure inductors each of self inductance L are connected in series, the net inductance is (a) L (b) 2 L (c) L/2 (d) L/4

Q.11. Lenz’s law is a consequence of the law of conservation of (a) charge (b) mass (c) energy (d) momentum

Q.12. A magnet is moved towards a coil (i) quickly (ii) slowly, then the induced e.m.f. is (a) larger in case (i) (b) smaller in case (i) (c) equal to both the cases (d) larger or smaller depending upon the radius of the coil

Q.13. The laws of electromagnetic induction have been used in the construction of a (a) galvanometer (b) voltmeter (c) electric motor (d) generator

Q.14. Two coils are placed closed to each other. The mutual inductance of the pair of coils depends upon (a) the rate at which currents are changing in the two coils. (b) relative position and orientation of two coils. (c) the material of the wires of the coils. (d) the currents in the two coils.

Q.15. Two identical coaxial circular loops carry a current i each circulating in the same direction. If the loops approach each other, you will observe that the current in (a) each increases (b) each decreases (c) each remains the same (d) one increases whereas that in the other decreases

Q.16. When current in a coil changes from 5 A to 2 A in 0.1 s, average voltage of 50 V is produced. The self-inductance of the coil is (a) 1.67 H (b) 6 H (c) 3 H (d) 0.67 H

Q.17. The self inductance associated with a coil is independent of (a) current (b) induced voltage (c) time (d) resistance of a coil

Q.18. A coil having 500 sq. loops of side 10 cm is placed normal to magnetic flux which increases at a rate of 1 T/s. The induced emf is (a) 0.1 V (b) 0.5 V (c) 1 V (d) 5 V

Q.19. A coil of 100 turns carries a current of 5 mA and creates a magnetic flux of 10 -5  weber. The inductance is (a) 0.2 mH (b) 2.0 mH (c) 0.02 mH (d) 0.002 H

Q.20. The north pole of a long bar magnet was pushed slowly into a short solenoid connected to a short galvanometer. The magnet was held stationary for a few seconds with the north pole in the middle of the solenoid and then withdrawn rapidly. The maximum deflection of the galvanometer was observed when the magnet was (a) moving towards the solenoid (b) moving into the solenoid (c) at rest inside the solenoid (d) moving out of the solenoid

Q.22. In a coil of self-induction 5 H, the rate of change of current is 2 As-1. Then emf induced in the coil is (a) 10 V (b) -10 V (c) 5 V (d) -5 V

Q.24. Two identical coaxial coils P and Q carrying equal amount of current in the same direction are brought nearer. The current in (a) P increases while in Q decreases (b) Q increases while in P decreases (c) both P and Q increases (d) both P and Q decreases

Q.25. Faraday’s laws are consequence of the conservation of (a) charge (b) energy (c) magnetic field (d) both (b) and (c)

Q.26. Direction of current induced in a wire moving in a magnetic field is found using (a) Fleming’s left hand rule (b) Fleming’s right hand rule (c) Ampere’s rule (d) Right hand clasp rule

Q.27. Which of the following statements is not correct? (a) Whenever the amount of magnetic flux linked with a circuit changes, an emf is induced in circuit. (b) The induced emf lasts so long as the change in magnetic flux continues. (c) The direction of induced emf is given by Lenz’s law. (d) Lenz’s law is a consequence of the law of conservation of momentum.

Q.28. Lenz’s law is a consequence of the law of conservation of (a) charge (b) energy (c) induced emf (d) induced current

Q.29. A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, the current will (a) increase (b) decrease (c) remain same (d) first increase then decrease

Q.30. There is a uniform magnetic field directed perpendicular and into the plane of the paper. An irregular shaped conducting loop is slowly changing into a circular loop in the plane of the paper. Then (a) current is induced in the loop in the anti-clockwise direction. (b) current is induced in the loop in the clockwise direction. (c) ac is induced in the loop. (d) no current is induced in the loop.

Q.32. Which of the following does not use the application of eddy current? (a) Electric power meters (b) Induction furnace (c) LED lights (d) Magnetic brakes in trains

Q.33. The north pole of a bar magnet is rapidly introduced into a solenoid at one end (say A). Which of the following statements correctly depicts the phenomenon taking place? (a) No induced emf is developed. (b) The end A of the solenoid behaves like a south pole. (c) The end A of the solenoid behaves like north pole. (d) The end A of the solenoid acquires positive potential.

Q.34. A metal plate can be heated by (a) passing either a direct or alternating current through the plate. (b) placing in a time varying magnetic field. (c) placing in a space varying magnetic field, but does not vary with time. (d) both (a) and (b) are correct.

Q.35. Identify the wrong statement. (a) Eddy currents are produced in a steady magnetic field. (b) Eddy currents can be minimized by using laminated core. (c) Induction furnace uses eddy current to produce heat. (d) Eddy current can be used to produce braking force in moving trains.

Q.36. If number of turns in primary and secondary coils is increased to two times each, the mutual inductance (a) becomes 4 times (b) becomes 2 times (c) becomes A times (d) remains unchanged 4

Q.37. When the rate of change of current is unity, the induced emf is equal to (a) thickness of coil (b) number of turns in coil (c) coefficient of self inductance (d) total flux linked with coil

Q.38. Two inductors of inductance .L each are connected in series with opposite? magnetic fluxes. The resultant inductance is (Ignore mutual inductance) (a) zero (b) L (c) 2L (d) 3L

Q.39. A square of side L metres lies in the x-y plane in a region, where the magnetic field is given by B = B 0 {li + 3j + 4k) T, where Bo is constant. The magnitude of flux passing through the square is [NCERT Exemplar] (a) 2B o L² Wb. (b) 3B o L² Wb. (c) 4B o L² Wb. (d) √29 B o L² Wb.

Q.40. A loop, made of straight edges has six comers at A(0, 0, 0), B(L, 0, 0) C(L, L, 0), D(0, L, 0), E(0, L, L) and F(0,0, L). A magnetic field B = B o  ( i+k )T is present in the region. The flux passing through the loop ABCDEFA (in that order) is [NCERT Exemplar] (a) B o L² Wb. (b) 2B o L² Wb. (c) √2B o L² Wb. (d) 4B o L² Wb.

Q.41. An e.m.f is produced in a coil, which is not connected to an external voltage source. This is not due to (a) the coil being in a time varying magnetic field. (b) the coil moving in a time varying magnetic field. (c) the coil moving in a constant magnetic field. (d) the coil is stationary in external spatially varying magnetic field, which does not change with time.

Q.45. Eddy currents do not cause (a) damping (b) heating (c) sparking (d) loss of energy

You may also check the MCQs on other chapters from the links provided below:

• MCQs on CBSE Class 12 Physics Chapter 1 Electric Charges and Fields
• MCQs on CBSE Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance
• MCQs on CBSE Class 12 Physics Chapter 3 Current Electricity
• MCQs on CBSE Class 12 Physics Chapter 4 Moving Charges and   Magnetism.
• MCQs on CBSE Class 12 Physics Chapter 5 Magnetism and Matter
• MCQs on CBSE Class 12 Physics Chapter 6 Electromagnetic Induction
• MCQs on CBSE Class 12 Physics Chapter 7 Alternating Current
• MCQs on CBSE Class 12 Physics Chapter 8 Electromagnetic Waves
• MCQs on CBSE Class 12 Physics Chapter 9 Ray Optics and Optical Instruments
• MCQs on CBSE Class 12 Physics Chapter 10 Wave Optics
• MCQs on CBSE Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter
• MCQs on CBSE Class 12 Physics Chapter 12 Atoms
• MCQs on CBSE Class 12 Physics Chapter 13 Nuclei
• MCQs on CBSE Class 12 Physics Chapter 14 Semiconductor Electronics

Also Check:

• NCERT Class 12 Physics Solutions
• Revision Notes for Class 12 Physics Chapter 6 Electromagnetic Induction
• Multiple Choice Questions on Electromagnetic Induction
• Important Very Short Answer Type Questions on Electromagnetic Induction
• Download Concept Map
• Electromagnetic Induction Study Notes for JEE
• Electromagnetic Induction Study Notes for NEET
• H C Verma Concept of Physics Solution for Chapter 38 Electromagnetic Induction

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MCQ Questions for Class 12 Physics Chapter 6 Electromagnetic Induction with Answers

We have compiled the NCERT MCQ Questions for Class 12 Physics Chapter 6 Electromagnetic Induction with Answers Pdf free download covering the entire syllabus. Practice MCQ Questions for Class 12 Physics with Answers on a daily basis and score well in exams. Refer to the Electromagnetic Induction Class 12 MCQs Questions with Answers here along with a detailed explanation.

Electromagnetic Induction Class 12 MCQs Questions with Answers

Answer: (a) clockwise

Question 2. Which of the following gives the direction of the induced e.m.f.? (a) Faraday’s law (b) Lenz’s law (c) Ampere (d) Biot-Savart’s law.

Answer: (b) Lenz’s law

Question 3. Which of the following is based on the law of conservation of energy? (a) Faraday’s law (b) Lenz’s law (c) Ampere (d) Biot-Savart’s law.

Question 4. The electric current in a circuit varies from + 2A to – 2A in a time 10 -2 s. Another coil of resistance 20 Ω and inductance 2H is placed hear it. What will be the induced current in the second coil? (a) 4A (b) 8A (c) 20A (d) 40A

Answer: (d) 40A

Question 5. Which of the following instrument do not make use of eddy currents? (a) Electrical brakes (b) Dead beat galvanometer (c) inductor motor (d) Transformer.

Answer: (d) Transformer.

Question 6. The motional e.m.f. is the induced e.m.f. (a) in a circuit due to variation in its own current. (b) in a circuit due to variation of current in the neighbouring circuit. (c) in a coil due to the motion of the magnet near it. (d) across the ends of a wire moving in a magnetic field.

Answer: (d) across the ends of a wire moving in a magnetic field.

Question 7. Whaf voltage is developed across the axle of the wheels of a train, when it moves with a speed of 72 kmh -1 ? The horizontal component of earth’s magnetic field is 0.40 × 10 -4 T and the angle of dip is 30°. The length of the axle is 1.5 m. (a) 0.8 mv (b) 0.4 mv (c) 0.2 mv (d) 0.7 mv

Answer: (d) 0.7 mv

Question 8. Eddy currents produced in a conductor are responsible for: (a) damping (b) loss of energy (c) heating (d) All of the above

Answer: (d) All of the above

Question 9. A car moves up on a plane road. The induced e.m.f. in the axle connecting the two wheels is maximum, when it moves: (a) At the poles (b) At equator (c) remains stationary (d) No e.m.f. is induced at all.

Answer: (a) At the poles

Question 10. When the current through a solenoid increases at a constant rate, the induced current: (a) is a constant and is in the direction of inducing current. (b) is a constant and is opposite to the direction of the inducing current. (c) increases with time and is in the direction of the inducing current. (d) increases with time and is opposite to the direction of the inducing current.

Answer: (b) is a constant and is opposite to the direction of the inducing current.

Question 11. Which of the following is/are equal to Henry? (a) Volt second/ampere. (b Volt(second)²/coulomb. (c) Joule (second)²/(coulomb)² (d) All of these

Answer: (d) All of these

Question 12. A coil of metal wire is kept stationary in a non-uniform magnetic field: (a) An e.m.f. and current both are induced in the coil. (b) An e.m.f. but not the current is induced iii the coil. (c) Current but not the e.m.f is induced in the coil. (d) Neither e.m.f. nor current is induced in the coil.

Answer: (d) Neither e.m.f. nor current is induced in the coil.

Question 13. Two different coils have self-inductances L 1 = 8 mH, L 2 = 2 mH. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coils is the same. At that time, the current, the induced voltage and energy stored in the two coils are I 1 , V 1 , U 1 , and I 2 , V 2 , U 2 respeactively. (a) \(\frac {V_2}{V_1}\) = \(\frac {1}{4}\) (b) \(\frac {I_1}{I_2}\) = \(\frac {1}{4}\) (c) \(\frac {U_2}{U_1}\) = 4 (d) All of the Above

Answer: (d) All of the Above

Question 14. Induced e.m.f. produced in a coil rotating in a magnetic field will be maximum when the angle between the axis of coil and direction of magnetic field is: (a) 90° (b) 45° (c) 0° (d) 180°

Answer: (a) 90°

Question 15. A copper rod moves parallel to the horizontal direction. The induced e.m.f. developed across its ends due to earth’s magnetic field will be maximum at the: (a) poles (b) equator (c) latitude 30° (d) latitude 60°.

Answer: (a) poles

Question 16. A coil having 500 square loops each of side 10 cm is placed normal to the magnetic field which increases at a rate of 1.0 Wb s -1 . The induced e.m.f. is; (a) 0.1V (b) 5 V (c) 0.5 V (d) 1.0 V

Answer: (b) 5 V

Question 17. When current I is passed through an inductor of coefficient of self-inductance L, energy stored in it is \(\frac {1}{2}\) Lt². This energy stored is in the form of: (a) Voltage (b) Current (c) Magnetic field (d) Electric field.

Answer: (c) Magnetic field

Fill in the Blanks

Question 1. If a core of soft iron is introduced into a coil its coefficient of self induction gets ……………….

Answer: increased.

Answer: Zero.

Question 3. The coils in the resistance boxes are made from ……………….

Answer: double wire.

Question 4. When the rate of change of current through a closed circuit is unity, then the induced e.m.f. produced in it is equal to ……………….

Answer: Coefficient of self induction or self inductance L.

Question 5. A copper rod of length l is rotated about one end perpendicular to the uniforfh field B with constant angular velocity to. The induced e.m.f. between its two end is ……………….

Answer: \(\frac {1}{2}\) B ωl²

Question 6. If a coil is removed from a magnetic field: (a) slowly, (b) rapidly, then work done will be more when the coil is removed from the magnetic field ……………….

Answer: rapidly.

Question 7. A coil having area A, total number of turns N is rotated in a uniform magnetic field \(\vec{B}\) with an angular velocity ω. The maximum e.m.f. induced in the coil is equal to ……………….

Answer: NBAω.

Question 8. A coil of copper wire is being pulled with a constant velocity \(\vec{v}\) in a magnetic field \(\vec{B}\). If its ohmic resistance is increased, it will be ………………. to pull it.

Answer: easier.

Question 9. If in Q. 8 above, the magnetic field be doubled, then the force required to pull the coil will become ………………. the initial value.

Answer: four times

Question 10. A coil of resistance R and inductance L is connected to a battery of e.m.f. E volts. The final current in the coil is ……………….

Answer: \(\frac {E}{R}\)

Question 11. A 50 mH coil carries a current of 2A. The energy stored in it is ………………. J.

Answer: 0.1

Question 12. A device that rely upon the relationship between the electric current and the magnetic field is ………………. and ……………….

Answer: Galvanometer, Generator, electric motor.

Question 13. When the number of turns in a coil is doubled without any change in the length of the coil, its self inductance becomes ……………….

Question 14. The induced e.m.f. produced in a wire of length l moving in a magnetic field \(\vec{B}\) with a constant velocity v is given by the induced current in a loop of this wire having resistance R is ……………….

Answer: Bvl = \(\frac {Bvl}{R}\)

Question 15. The normal drawn to the surface of a conductor makes an angle θ with the direction of magnetic field \(\vec{B}\). The magnetic flux ø passing through the area A is ……………….

Answer: o = \(\vec{B}\). \(\vec{A}\)

Question 16. The magnetic energy stored in a solenoid in terms of B, A and length of solenoid is ……………….

Answer: \(\frac {1}{2µ_0}\) B²Al

Question 17. The magnetic energy per unit volume is ……………….

Answer: \(\frac {B^2}{2µ_0}\)

Question 18. A circular coil of N turns and radius r is placed in a uniform magnetic field \(\vec{B}\). Initially the plane of coil is perpendicular to the magnetic field. The coil is then rotated by 90°. If the resistance of the oil is R, then the charge passing through the coil is given by ……………….

Answer: \(\frac {nBπr^2}{R}\)

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