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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
 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.
 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.
 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.
 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.
 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.
 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
 If the rate of change of current of 2A / s generates an emf of 10mV in a solenoid, what would be the selfinductance of the solenoid?
Ans: Selfinductance is given by, L=dI/dt=10×1032= 5×103 H
Hence, selfinductance of the solenoid is 5 x 103 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.
 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.
 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.
 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
 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+XLXC2
But, provided that, XC=XL
That implies at minimum, Z= R
Hence, the bulb’s brightness will turn into maximum.
 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 5×104 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=5×104T
Angle of dip, =30°
The magnetic field of Earth’s vertical component,
BV=Bsin
BV=5×104sin30°
2.5×104T
So, voltage variation between the ends of the jet’s wing,
e=(Bv)x I x v
e=2.5×104x 25 x 500
3.125V.
Therefore, the voltage variation produced between the ends of the wings of the jet is 3.125V.
 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=I2I1=200=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.
 A horizontal straight cable 10m long expanding from east to west is falling with a speed of 5.0ms1, at particular angles to the horizontal component of the magnetic field of the Earth 0.30×104 Wbm2.
(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=0.3×1014Wbm2
Emf induced in the wire, e=Blv
e=0.3×104x5x10
e=1.5×103V
Therefore, immediate emf induced is 1.5×103V.
(b) What is the emf’s direction?
Ans: The direction of the emf induced is from West to East as of Fleming’s righthand rule.
(c) Which of the wire’s ends is at the higher electrical potential?
Ans: The wire’s eastern end is at higher potential.
 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, deadbeat 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 eastwest 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
16104m²
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=2.4104 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 2.4104 V that will last for 2s.
(b) Emf developed, e=Bbv
0.30.020.01=0.6104V
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 0.6104V that will last for 8s.
Q. 0A of current in a circuit drop to 0.0A in 0.1s. Estimate the circuit’s selfinductance 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=I1I2=5A
The time it took for the change, t=0.1s
Average emf, e=200V
For the relation for the average emf for the selfinductance (L) can be represented as
e=LdidtS
L=rdidt
∴ 20050.1=4H
Hence, 4H will be the coil’s selfinduction.
(a) Define selfinductance and mention its SI units.
(b) Create an expression for the selfinductance of a long solenoid with a crosssectional area of A and N turns. The length of the solenoid is l.
Ans. (a) Selfinductance of a coil
Since we know that flux =LI
emf induced = ddt= Ldldt
Here, L is the coefficient of the selfinductance or the selfinduction.
When a given amount of current flows through a coil, its selfinductance 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=crosssectional 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
∴ Selfinductance 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 selfinductance. Make an expression of a long solenoid selfinductance having a crosssectional area denoted with “A”, “n” denoting turns per unit and length denoted with “I”.
Ans. Coefficient of selfinduction. 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 selfinduction.
If I=1, then = L
Hence, when a unit current passes through a coil, its selfinductance is numerically equal to the magnetic flux associated with the coil.
Henry is the selfinductance SI unit (H). The term “selfinductance” is: Think about a long solenoid with a crosssectional 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 selfinductance 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 selfinductance. Identify the expression for the magnetic energy held in a selfinducting L inductor that allows a current I to flow through it.
Ans. Selfinductance: 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 selfinduction. 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 selfinductance is a constant of proportionality L. It is also known as the coil’s selfinduction 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 selfinduced emf. For circuits with basic geometries, the selfinductance can be calculated. Let’s determine the selfinductance of a long solenoid with n turns per unit length, crosssectional 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 selfinductance 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 selfinductance 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 selfinduced emf. The selfinductance 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
Q1For a car moving on a plane road, the induced emf in the axle connecting the two wheels is maximum when
opt
ait moves at poles
bit moves at the equator
cit remains stationary
dit moves in the tropical regions
ansit moves at poles
Q2 Find the self induced emf in a 0.4H coil when the current in it is changing at the rate of 800 A/S.
ansSelf induced emf is given by
$\mathrm{\mu}=\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}}=0.4\u2014800=320\mathrm{V}$Please register to view this section
FAQs (Frequently Asked Questions)
1. What is the SI unit of magnetic flux?
Weber is the SI unit of magnetic flux.
2. What is Faraday’s Law of EMI?
Faraday’s Law states that “When magnetic flux changes through a circuit, an emf is induced in it which lasts only as long as the change in the magnetic flux through the circuit continues”.
Average emf = NDf/Dt
Where N = number of turns in the coil.
3. What is a transformer?
A transformer effectively raises or lowers AC voltages by utilising Faraday’s Law and the ferromagnetic characteristics of an iron core. Naturally, it cannot increase power, therefore if the voltage is increased, the current will decrease accordingly, and vice versa.
4. What are eddy currents?
A circular current known as an eddy current flows in a solid conductor, such as metal sheets or rods. Electromagnetic induction happens when such a conductor is exposed to a shifting magnetic field. Eddy currents are the result of the currents flowing in circles because the charges are not constrained by a narrow conductor.
5. Why is EMI regulated?
EMI is regulated to prevent interference from other electronic devices and the equipment itself from impairing the operation of today’s sensitive equipment, which is necessary for it to work correctly. For reliable radio frequency communications, the EMI spectrum must be maintained because it is a finite natural resource. Future electronic devices will be able to function as intended, in the intended environment, without incurring any performance deterioration owing to interference and without interfering with the operation of other equipment, thanks to the effective regulation of EMI interference.