Important Questions Class 12 Physics Chapter 6

Important Questions for CBSE Class 12 Physics Chapter 6 – Electromagnetic Induction

In these Class 12 Physics Chapter 6 Important Questions, you will learn the answers to the important questions on the topic of electromagnetic induction. These Class 12 Physics Chapter 6 Important Questions will help you to understand the pattern of questions in the CBSE examination. Moreover, these Class 12 Physics Chapter 6 Important Questions are made in accordance with the CBSE syllabus and CBSE sample papers.

CBSE Class 12 Physics Chapter 6 Important Questions

Study Important Questions for Class 12 Physics Chapter 6 – Electromagnetic Induction

CBSE Class 12 Physics Chapter 6 Important Questions contain formulas and CBSE extra questions. Here are some important questions from class 12 physics chapter 6, that students can study to better prepare them for their examinations. Moreover, when you study these Class 12 Physics chapter 6 important questions you will be able to solve CBSE past years’ question papers.

Very Short Answer Questions

1. A metallic cable coil is stationary in a non–uniform magnetic field. What is the emf generated in the coil?

Ans: No emf is generated in the coil as there is no shift in the magnetic flux connected with the secondary coil.

2. Why does a metallic piece turn very hot when it is encircled by a coil having a high frequency (H.F) alternating current?

Ans: When a metallic piece is covered by a coil giving high frequency (H.F) alternating current, it turns hot because eddy currents are generated, which in turn creates the joule’s heating effect.

3. An electrical component X, when linked to an alternating source of voltage, has current running through it, leading the voltage by 2 radians. Identify component X and write the expression for its reactance.

Ans: The X is an immaculately capacitive circuit. The expression for capacitive reactance is

XC=1C=12c.

4. A transformer levels up 220V to 2200V. What is the ratio of this transformation?

Ans: The transformation ratio, K=NsNP= EsEP=2200220=10

Hence, the ratio of transformation is 10.

5. The induced emf is also known as back emf. Why?

Ans: It is called back emf because induced emf generated in a circuit always resists the cause which creates it.

6. Why are a copper disc’s oscillations in a magnetic field lightly damped?

Ans: Copper discs oscillate due to the generation of eddy currents which resist its oscillating motion to result in the damped motion.

Short Answer Questions 

1. If the rate of change of current of 2A / s generates an emf of 10mV in a solenoid, what would be the self-inductance of the solenoid?

Ans: Self-inductance is given by, L=dI/dt=10×10-32= 5×10-3 H

Hence, self-inductance of the solenoid is 5 x 10-3 H.

2. A circular copper disc of 10cm radius rotates at a speed of 2 rad/sec around an axis through its centre and perpendicular to the disc. A uniform magnetic field of 0.2T functions perpendicular to the disc.

a) Calculate the possible difference created between the axis of the disc and the rim.

Ans: Provided, radius 10cm, B= 0.2T,  =2rad/s

Bwr2

=12x 0.2x2x(0.1)2

= 0.00628volts

Hence, the possible difference developed is 0.00628volts.

b) What is the generated current if the disc’s resistance is 2?

Ans: I=R0.0628

I= 0.0314A

If the resistance is 2, the produced current is found to be 0.0314A.

3. An exemplary inductor uses no electric power in a.c. circuit. Explain.

Ans: Power consumed by the inductor in a.c. circuit, P= ErmsIrmscos

But for an exemplary inductor, =2

P =0

Hence, the exemplary inductor in a.c. circuit doesn’t consume power.

4. Why does DC gets blocked by the capacitor?

Ans: The capacitive reactance, XC=1C=12c

For d.c., =0

XC=

Since the capacitor provides infinite resistance to d.c. flow, d.c. can’t pass through it.

5. Why is the emf zero, when most magnetic lines of force pass through the coil?

Ans: The upright position of the coil provides maximum magnetic flux.

But as the coil rotates, ddt=0

Therefore, generated emf, =ddt=0

6. A L inductor of XL reactance is linked in series with a bulb B to an a.c. source as illustrated in the figure.

Briefly explain the process of the brightness of the bulb changing when

(a) the number of turns of the inductor is reduced.

Ans: We know, Z=R2+XL2

When the number of turns of the inductor gets lessened XL and Z reduces and gradually increases.

Because of this, the bulb would glow more brightly.

(b) A capacitor of reactance XC=XL is inclusive within a series in the same circuit.

Ans: When the capacitor is inclusive within the circuit, Z=R2+XL-XC2

But, provided that, XC=XL

That implies at minimum, Z= R

Hence, the bulb’s brightness will turn into maximum.

7. A jet plane is flying towards the west at a speed of 1800 km/hr. What is the voltage variation generated between the ends of the wing having a span of 25m, if the Earth’s magnetic field at the location has a magnitude of 5×10-4 T and the dip angle is 30°?

Ans: Given that the jet plane speed, v 1800km / h 500m / s

Jet plane’s wingspan, l= 25m

Magnetic field strength of Earth, B=5×10-4T

Angle of dip, =30°

The magnetic field of Earth’s vertical component,

BV=Bsin

BV=5×10-4sin30°

2.5×10-4T

So, voltage variation between the ends of the jet’s wing,

e=(Bv)x I x v

e=2.5×10-4x 25 x 500

3.125V.

Therefore, the voltage variation produced between the ends of the wings of the jet is 3.125V.

8. A pair of adjoining coils has a mutual inductance of 1.5H. If the current in a coil varies from 0 to 20A in 0.5s, what is the flux change linkage with the second coil?

Ans: Provided that the mutual inductance of these coils, =1.5H

Initial current, I1=0A

Final current, I2=20A

Variation in current, dI=I2-I1=20-0=20A

Time taken for the variation, dt =0.5s

Induced emf, e=ddt… (1)

Where d is the variation in the flux linkages with the coil.

Emf is related with mutual inductance as, e=dldt…(2)

Equating (1) and (2),

ddt=dldt

d=1.5x(20)

d= 3Wb

Therefore, the variation in the flux linkage is 30Wb.

9. A horizontal straight cable 10m long expanding from east to west is falling with a speed of 5.0ms-1, at particular angles to the horizontal component of the magnetic field of the Earth 0.30×10-4 Wbm-2.

(a) What is the immediate value of the emf induced in the cable?

Ans: Given that, the wire’s length,  l= 10m

The wire’s falling speed, v 5.0m / s

Magnetic field strength, B=0.3×10-14Wbm-2

Emf induced in the wire, e=Blv

e=0.3×10-4x5x10

e=1.5×10-3V

Therefore, immediate emf induced is 1.5×10-3V.

(b) What is the emf’s direction?

Ans: The direction of the emf induced is from West to East as of Fleming’s right-hand rule.

(c) Which of the wire’s ends is at the higher electrical potential?

Ans: The wire’s eastern end is at higher potential.

10. A 1.0ms long metallic pole is turned with an angular frequency of 400rads about an axis normal to the rod passing through its one edge. The rod’s other end is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5T parallel to the axis exists everywhere. Compute the emf produced between the centre and the ring.

Ans: From the given information, the rod’s length, l=1m

Angular frequency, = 400rad / s

Strength of magnetic field, B 0.5T

An end of the rod has 0 linear velocity, while the other end has a linear velocity of I(t)

The rod’s average linear velocity, v=l(t)+02=l(t)2

Emf generated between the ring and the centre,

e=Blv=Bll(t)+02=l(t)2

e=0.5x(l)2×4002=100V

Therefore, the emf generated between the ring and the centre is 100V.

Q. How does a pair of coils’ mutual inductance change when

• the distance between the coils has increased?

• there is an increase in each coil’s turn count.

• other elements remain the same, except a thin iron sheet is sandwiched between two coils. In each instance, 

Ans. (a) The flux associated with the secondary coils diminishes as the existing separation between the coils increases. The mutual induction thus declines.

(b) as mutual inductance, M=0N1N2A1

Hence, the mutual induction will increase when N1 and N2 will increase.

(c) As mutual inductance, M ∝ r (relative permeability of material), hence, the mutual induction will increase.

Q. Answer the following questions.

(i) The galvanometer exhibits brief deflection when primary coil P is shifted in the direction of secondary coil S (as shown in the figure). What may be done to increase the galvanometer’s deflection while using the same battery?

(ii) Also mention the related law.

(i) Coil P must be brought closer to coil S more quickly so that the rate of change of magnetic flux is greater in order to have a larger deflection in the galvanometer with the same battery.

(ii) Faraday’s second law of electromagnetic induction, which states that induced emf is created in a circuit when magnetic flux associated with it changes, is the related law guiding this occurrence. The rate of change of magnetic flux has a direct relationship with the strength of the induced emf.

Q. The same magnetic field rotates two similar loops made of copper and aluminum at the same angular speed. Compare the following.

(i) The induced emf

(ii) the current that the two coils produce 

Ans. (i) The induced emf in a coil is = NBA sin t. Due to the identical angular speed in both loops, the induced emf will likewise be the same.

(ii)  The current that will be induced in a loop is I=R=Apl. More current is induced in copper because its resistance is lower.

Q.  What do you mean by eddy currents? Mention any two use cases of eddy current.

Ans. Eddy currents are created when there is an induced current in the volume (or bulk) of a material as a result of a change in magnetic flux. It can be utilised in devices like electric brakes, induction furnaces, dead-beat galvanometers, and more, and it results in energy loss while producing heat, making it a necessary evil in an arrangement.

(b) (i) A change in area occurs as the arm RS of length l is moved at a constant speed. It is provided by dA = Idx = Ivdt

The emf induced, e=- ddt

Here, refers to the magnetic flux

e = BdAdt= -Blv

(ii) I= eR= BlvR

Here, r refers to the ner resistance is the network with arm RS.

Hence, the force experienced is

F=IB= B2l2vR

(iii) power required for the

Movement, P=F.v=Fv=B2l2vR2

Q.  Define Lenz’s law. Allowing a metallic rod to descend due to gravity while being held horizontally in an east-west direction At its ends, will an emf be generated? Explain your response.

Ans. According to Lenz’s law, “induced emf has such a polarity that it tends to produce a current that opposes the change in magnetic flux that induced it.”

Yes, since the horizontal component of the earth’s field, the rod’s speed, and its length are all perpendicular to one another, an induced electromagnetic field will exist. As the metallic rod descends, the magnetic flux caused by the Earth’s vertical magnetic field changes.

Q. Determine the primary current drawn by a transformer’s steps where 200V is reduced to 20V in order to power a 20 gadget. Consider the transformer efficiency to be 80%.

Ans. The information is given that, =80%

EP=200V

ES=20V

Z=20

IS=ESZ=2020=1A

Now, we have,

=ESISEPIP

If we substitute the values that are given,

=80200= 201200IP

IP=200080200

∴ IP=0.125A

Hence, the 0.125A will be the current drawn by the primary of the transformer.

Q. A 0.3T uniform magnetic field directed normally to a rectangular wire loop with sides of 8 cm and 2 cm is travelling away from the loop. If the loop moves at 1 cm/s in a direction normal to the cut, what is the emf that will be developed across the cut? Find the following.

(a) its longer side

(b) the shorter side. 

Also how long will the induced voltage last in each case? 

Ans. (a) The length of the rectangular wire, I= 8cm= 0.08m

The rectangular wire’s width is. b= 2cm = 0.02m

So, area of the rectangular loop,

A=lb= 0.090.02

1610-4m²

Magnetic field strength, B=0.3T

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

Emf created in the loop can be represented by

e= Blv

0.30.080.01=2.410-4 V

The time that will be taken across the width will be

∴  t=Distamce travelledVelocity=bv=0.020.01=2s

Hence, the induced voltage will be 2.410-4 V that will last for 2s.

(b) Emf developed, e=Bbv

0.30.020.01=0.610-4V

The time that will be taken to travel along the length will be

∴  t=Distamce travelledVelocity=1v=0.080.01=8s

Hence, the induced voltage will be 0.610-4V that will last for 8s.

Q. 0A of current in a circuit drop to 0.0A in 0.1s. Estimate the circuit’s self-inductance if an emf 200V is induced.

Ans. It is given that the initial current or I1 is 5.0A

Final current, IS= 0.0A

The change in the current, dl=I1-I2=5A

The time it took for the change, t=0.1s

Average emf, e=200V

For the relation for the average emf for the self-inductance (L) can be represented as

e=LdidtS

L=rdidt

∴ 20050.1=4H

Hence, 4H will be the coil’s self-induction.

(a) Define self-inductance and mention its SI units.

(b) Create an expression for the self-inductance of a long solenoid with a cross-sectional area of A and N turns. The length of the solenoid is l.

Ans. (a) Self-inductance of a coil

Since we know that flux =LI

emf induced = -ddt= -Ldldt

Here, L is the coefficient of the self-inductance or the self-induction.

When a given amount of current flows through a coil, its self-inductance is mathematically equal to the magnetic flux associated with the coil. Henry is its S.I. unit.

(b) Consider a long solenoid with n turns per unit length and dimensions of length l, radius r, and r<<1. The magnetic field inside the coil is nearly constant and is given by if a current I passes through it.

B=0nl

Magnetic flux that is linked with each turn=BA= 0nlA

Here, A=r²=cross-sectional area of the solenoid

∴ Magnetic flux that is linked with entire solenoid is

= Flux that is linked with each turn Total number of turns

= 0nIA nl= 0n2IAl

But = LI

∴ Self-inductance of the long solenoid is

L= 0n2IA

If N represents to the total number of turns in the solenoid

Then n=Nl

∴  L=0N2Al

Q. Write the S.I unit of a coil’s self-inductance. Make an expression of a long solenoid self-inductance having a cross-sectional area denoted with “A”, “n” denoting turns per unit and length denoted with “I”. 

Ans. Coefficient of self-induction. Imagine a coil L which has a current I that flows through the coil at any given instance. The magnetic flux that is licked with the coil and the current passing through it at a given instance will be directly proportional to each other.

∴ ∝ I

= LI

Here, L refers to the coefficient of self-induction.

If I=1, then = L

Hence, when a unit current passes through a coil, its self-inductance is numerically equal to the magnetic flux associated with the coil.

Henry is the self-inductance SI unit (H). The term “self-inductance” is: Think about a long solenoid with a cross-sectional area of A and n turns per unit length. It has a length Z.

B is the magnetic field produced by the solenoid’s current flow. It can be represented by the following equation

B=0nl

The total flux that is linked with the solenoid is

=(nl) (0nl) A= 0n2 Al

Here, nl is the total number of turns

Thus, the self-inductance is, I

= 0n2AlI1= 0n2 Al

So if you have the solenoid’s inside is filled relative permeability r then

L= r0n²Al

Q. Give an explanation of the word solenoid self-inductance. Identify the expression for the magnetic energy held in a self-inducting L inductor that allows a current I to flow through it.

Ans. Self-inductance: When the current through a single isolated coil is changed, the flux through the coil changes, causing an emf to be generated. This occurrence is known as self-induction. In this instance, flux linkage via a coil with N turns is proportional to the coil’s current and is represented by the following equation:

NB ∝ I

NB=LI

The coil’s self-inductance is a constant of proportionality L. It is also known as the coil’s self-induction coefficient. An emf is produced in the coil when the current is changed, which also causes the flux connected to the coil to change. The induced emf is calculated using the equation above by

=d(NB)dt or =dldt

As a result, every alteration (increase or reduction) of current in the coil is always opposed by the self-induced emf. For circuits with basic geometries, the self-inductance can be calculated. Let’s determine the self-inductance of a long solenoid with n turns per unit length, cross-sectional area A, and length Z. A current I passing through the solenoid produces the magnetic field B is 0nl (edge effects are neglected, as before). The total flux that will be linked to the solenoid is

NB=(nl) (0nl) (A)= 0n2 Al

Hence, the self-inductance is, L=NBl= 0n2 Al

Soft iron, which has a high relative permeability value, is an example of a material with relative permeability that can be used to fill the interior of a solenoid.

L=r0n² Al

The coil’s self-inductance is influenced by both its geometry and the medium’s permeability. The back emf, which resists any change in the current in a circuit, is another name for the self-induced emf. The self-inductance simulates inertia physically. It is the equivalent of mass in electromagnetics. Therefore, in order to ascertain the current, one must operate against the back emf (). Magnetic potential energy is used to store this work. The rate of work done in a circuit for current I at a given moment is

= -L dldt dWdt=LI dldt

So, the total work done to establish the current I will be

W=∫dW= 01 LI dl= 12 LI²

Hence, the magnetic energy that is needed to build up the current I is.

The current I is, W=12 LI²