NCERT Solutions Class 12 Physics Chapter 6 - Electromagnetic Induction

Electromagnetic induction (EMI) is the sixth chapter of Physics in 12th-grade and is one of the most important advances in the history of sciences. The chapter discusses both aspects of current producing magnetic fields as well as generating current from a changing magnetic field. It is the first time that both concepts have been presented in the same chapter and will give the student an appreciation of both fields of study that are now directly related to each other through laws of mathematics. The contributions made by Michael Faraday and Joseph Henry to the understanding of how to calculate the induced emf due to a change in a magnetic field will be included.

Students who understand the concept of EMI will have a solid understanding of how electricity is generated in the modern age, including the creation of electricity from large hydroelectric dams all the way to charging our smartphones with no physical connection. EMI will continue to be a major focus of study in CBSE Board Exams, JEE, and NEET.

The NCERT Solutions for EMI (Chapter 6 in Class 12 Physics) were written to provide students with an opportunity to master the concepts related to magnetic flux and the direction of induced current. They also bridge the gap between mechanical energy and electrical energy. In addition, the solutions make the interaction between geometry and magnetism easy to understand.

1.0Class 12 Physics Chapter 6 Electromagnetic Induction: Key Concepts

Class 12 Physics Chapter 6, Electromagnetic Induction, explains how a changing magnetic field can produce an induced emf and current in a conductor. The main concepts covered in this chapter include:

Magnetic Flux: Understand magnetic flux and how it changes with the strength and direction of a magnetic field and the area through which the field passes.
Faraday’s Law of Electromagnetic Induction: Learn how the induced emf is related to the rate of change of magnetic flux through a circuit.
Lenz’s Law: Understand how to determine the direction of induced current. The induced current always opposes the change in magnetic flux that produces it.
Motional EMF: Study the emf produced when a conductor moves through a magnetic field.
Eddy Currents: Learn about the circulating currents produced inside bulk conductors when the magnetic flux through them changes, along with their applications in magnetic braking and induction heating.
Self-Inductance and Mutual Inductance: Understand how a changing current in a coil produces an induced emf in the same coil or in a nearby coil.
AC Generator: Learn the basic working of an AC generator and how electromagnetic induction is used to convert mechanical energy into electrical energy.

2.0NCERT Solutions Class 12 Physics Chapter 6 Electromagnetic Induction : Detailed Solutions

SOLVED EXAMPLES

Consider Faraday's Experiment

Blue diagram of Faraday’s experiment showing a coil connected to a galvanometer to detect induced current.

(a) What would you do to obtain a large deflection of the galvanometer? (b) How would you demonstrate the presence of an induced current in the absence of a galvanometer?

Sol.

(a) To obtain a large deflection, one or more of the following steps can be taken: (i) Use a rod made of soft iron inside the coil $C_{2}$ , (ii) Connect the coil to a powerful battery, and (iii) Move the arrangement rapidly towards the test coil $C_{1}$ . (b) Replace the galvanometer by a small bulb, the kind one finds in a small torch light. The relative motion between the two coils will cause the bulb to glow and thus demonstrate the presence of an induced current.

In experimental physics one must learn to innovate. Michael Faraday who is ranked as one of the best experimentalists ever, was legendary for his innovative skills.

A square loop of side 10 cm and resistance $0.5Ω$ is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time-interval. Sol. The angle $θ$ made by the area vector of the coil with the magnetic field is 45°. The initial magnetic flux is

$Φ = BA cos θ = \frac{0.1 \times 1 0 ^{- 2}}{2} Wb$

Final flux, $Φ_{min} = 0$ The change in flux is brought about in 0.70 s. The magnitude of the induced emf is given by

$ε = \frac{∣ Δ Φ _{B} ∣}{Δ t} = \frac{∣ ( Φ - 0 ) ∣}{Δ t} = \frac{1 0 ^{- 3}}{2 \times 0.7} = 1.0 mV$

And the magnitude of the current is

$I = \frac{ε}{R} = \frac{1 0 ^{- 3} V}{0.5Ω} = 2 mA$

Note that the earth's magnetic field also produces a flux through the loop. But it is a steady field (which does not change within the time span of the experiment) and hence does not induce any emf.

A circular coil of radius 10 cm, 500 turns and resistance $2Ω$ is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through $18 0^{\circ}$ in 0.25 s. Estimate the magnitudes of the emf and current induced in the coil. Horizontal component of the earth's magnetic field at the place is $3.0 \times 1 0^{- 5} T$ . Sol. Initial flux through the coil, $Φ_{B (initial)} = BA cos θ$

$= 3.0 \times 1 0^{- 5} \times (π \times 1 0^{- 2}) \times cos 0^{\circ} = 3 π \times 1 0^{- 7} Wb$

Final flux after the rotation,

$Φ_{B (final)} = 3.0 \times 1 0^{- 5} \times (π \times 1 0^{- 2}) \times cos 18 0^{\circ} = - 3 π \times 1 0^{- 7} Wb$

Therefore, estimated value of the induced emf is,

$ε = N \frac{ΔΦ}{Δ t} = \frac{500 \times ( 6 π \times 1 0 ^{- 7} )}{0.25} = 3.8 \times 1 0^{- 3} V I = \frac{ε}{R} = 1.9 \times 1 0^{- 3} A$

Note that the magnitudes of $ε$ and I are the estimated values. Their instantaneous values are different and depend upon the speed of rotation at the particular instant.

Figure shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader. Determine the direction of induced current in each loop using Lenz's law.

Planar loops of different shapes moving into or out of a magnetic field, illustrating induced current directions using Lenz’s law.

Sol.

(i) The magnetic flux through the rectangular loop abcd increases, due to the motion of the loop into the region of magnetic field. The induced current must flow along the path bcdab so that it opposes the increasing flux. (ii) Due to the outward motion, magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacb, so as to oppose the change in flux. (iii) As the magnetic flux decreases due to motion of the irregular shaped loop abcd out of the region of magnetic field, the induced current flows along cdabc, so as to oppose change in flux.

Note that there are no induced current as long as the loops are completely inside or outside the region of the magnetic field.

(a) A closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets? (b) A closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop (i) When it is wholly inside the region between the capacitor plates. (ii) When it is partially outside the plates of the capacitor? The electric field is normal to the plane of the loop. (c) A rectangular loop and a circular loop are moving out of a uniform magnetic field region (figure) to a field-free region with a constant velocity $v$ . In which loop do you expect the induced emf to be constant during the passage out of the field region? The field is normal to the loops.

Rectangular and circular loops moving out of a uniform magnetic field into a field-free region.

(d) Predict the polarity of the capacitor in the situation described by figure.

Diagram showing two bar magnets with opposite poles facing a conducting loop connected to a capacitor.

Sol.

(a) No, However strong the magnet may be, current can be induced only by changing the magnetic flux through the loop. (b) No current is induced in either case. Current cannot be induced by changing the electric flux.

(c) The induced emf is expected to be constant only in the case of the rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of the field region is not constant, hence induced emf will vary accordingly. (d) The polarity of plate 'A' will be positive with respect to plate 'B' in the capacitor.

A metallic rod of 1 m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular to the plane of the ring (figure). A constant and uniform magnetic field of 1 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?

Rotating metallic rod extending from the centre to the rim of a circular ring in a uniform magnetic field.

Sol. As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached. Using Eq., the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by $d ε =$ Bvdr. Hence,

$ε = \int d ε = \int_{0}^{R} Bvdr = \int_{0}^{R} B ω rdr = \frac{B ω R ^{2}}{2}$

Note that we have used $v = ω r$ . This gives

$ε = \frac{1}{2} \times 1.0 \times 2 π \times 50 \times (1^{2}) = 157 V$

A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth's magnetic field $H_{E}$ at a place. If $H_{E} = 0.4 G$ at the place, what is the induced emf between the axle and the rim of the wheel? Note that $1 G = 1 0^{- 4} T$ . Sol. Induce $emf = (1/2) ω BR^{2}$

$= (1/2) \times 4 π \times 0.4 \times 1 0^{- 4} \times (0.5)^{2} = 6.28 \times 1 0^{- 5} V$

The number of spokes is immaterial because the spokes are in parallel.

Two concentric circular coils, one of small radius $r_{1}$ and the other of large radius $r_{2}$ , such that $r_{1} ≪ r_{2}$ , are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. Sol. Let a current $I_{2}$ flow through the outer circular coil. The field at the centre of the coil is $B_{2} = \frac{μ _{0} I _{2}}{2 r _{2}}$ . Since the other co-axially placed coil has a very small radius. $B_{2}$ may be considered constant over its cross-sectional area. Hence.

$Φ_{1} = π r_{1}^{2} B_{2} = \frac{μ _{0} π r _{1}^{2}}{2 r _{2}} I_{2} = M_{12} I_{2}$

Thus, $M_{12} = \frac{μ _{0} π r _{1}^{2}}{2 r _{2}}$ From Eq.

$M_{12} = M_{21} = \frac{μ _{0} π r _{1}^{2}}{2 r _{2}}$

Note that we calculated $M_{12}$ from an approximate value of $Φ_{1}$ , assuming the magnetic field $B_{2}$ to be uniform over the area $π r_{1}^{2}$ . However, we can accept this value because $r_{1} ≪ r_{2}$ .

(a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B , area A and length $l$ of the solenoid. (b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor?

Sol.

(a) The magnetic energy is

$U_{B} = \frac{1}{2} L I^{2} = \frac{1}{2} L (\frac{B}{μ _{0} n})^{2}$

(since $B = μ_{0} nI$ , for a solenoid)

$= \frac{1}{2} (μ_{0} n^{2} A l) (\frac{B}{μ _{0} n})^{2} = \frac{1}{2 μ _{0}} B^{2} A l$

(b) The magnetic energy per unit volume is,

$u_{B} = \frac{U _{B}}{V}$

(where V is volume that contains flux)

$u_{B} = \frac{U _{B}}{A l} = \frac{B ^{2}}{2 μ _{0}}$

We have already obtained the relation for the electrostatic energy stored per unit volume in a parallel plate capacitor

$u_{E} = \frac{1}{2} ε_{0} E^{2}$

In both the cases energy is proportional to the square of the field strength.

Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a 100 turn coil of area $0.10 m^{2}$ . The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil? Sol. Here, $v = 0.5 Hz; N = 100$ ,

$A = 0.1 m^{2} and B = 0.01 T .$

Maximum voltage

$ε_{0} = NBA (2 π v) = 100 \times 0.01 \times 0.1 \times 2 \times 3.14 \times 0.5 = 0.314 V$

The maximum voltage is 0.314 V We urge you to explore such alternative possibilities for power generation.

EXERCISE QUESTIONS WITH SOLUTIONS

Predict the direction of induced current in the situations described by the following figures (a) to (f).

Six diagrams showing different situations for determining the direction of induced current using Lenz’s law.

Sol. (a) When the magnet approaches the coil, an induced current is set up in the coil. According to Lenz's law, a south pole is developed at the end q so as to oppose the motion of the magnet towards the coil. The direction of the induced current (as seen from magnet side of the coil) is clockwise, i.e., along qrpq. (b) As discussed in (a), according to Lenz's law, the coil pq develops a south pole at the end q (as south pole of the magnet moves towards it) and the coil xy also develops a south pole at the end x (as the north pole moves away from it). Hence, induced current flows along prqp in coil pq and along yzxy in the coil xy. (c) When the tapping key is just closed in the left loop, the current in it grow in the direction shown. Due to mutual induction, the induced current in right loop flows in the opposite direction, i.e., yzxy to oppose this growth. (d) When the rheostat setting is changed as shown, the current in the right loop increases (due to decreasing resistance of the rheostat). Due to mutual induction, an induced current in the left loop flows in the opposite direction zyxz to oppose the increase in current in the right loop. (e) As the tapping key is just released, the current in the left solenoid decreases from maximum to zero (i.e., it decays). Due to mutual induction, an induced current is set up in the right solenoid in a direction which tries to prevent the decay. This is possible only if the induced current in the right solenoid flows in the same direction as in the left solenoid, i.e., in the direction xryx. (f) The direction $B$ due to wire is in the plane of the circular coil. As such $ϕ_{B} = 0$ , hence no induced current in the coil.

Use Lenz's law to determine the direction of induced current in the situations described by figure (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire.

Diagrams showing a wire changing shape and a circular loop deforming to illustrate induced current direction using Lenz’s law.

Sol. (a) When the wire of irregular shape changes into a circular shape, its area increases and consequently magnetic flux linked with it also increases $(ϕ_{B} = BA)$ . To produce opposing flux (Lenz's law), an induced current is set up in the circular wire in the anticlockwise direction (i.e., adcba) (so that magnetic field due to it is directed upwards as the original magnetic field is directed downwards.) (b) When the circular loop (abcda) is deformed into a narrow straight wire loop (a'b'c'd'a'), its area decreases and consequently magnetic flux linked with it also decreases ( $ϕ_{B} = BA$ ). To produce supporting flux (Lenz's law), an induced current is set up in the straight wire loop $a^{'} b^{'} c^{'} d^{'} a^{'}$ in the anticlockwise direction, i.e., a'd'c'b'a' so that its magnetic field is directed upwards because the original magnetic field is also directed upwards.

A long solenoid with 15 turns per cm has a small loop of area $2.0 cm^{2}$ 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 ? Sol. Here; $n = 15 turns / cm = 1500 turns / m$ ,

$A = 2.0 cm^{2} = 2.0 \times 1 0^{- 4} m^{2} \frac{dI}{dt} = \frac{4.0 - 2.0}{0.1} = 20 A / s$

Induced emf, $ε = \frac{d ϕ _{B}}{dt} = \frac{d}{dt} (ϕ_{B}) = \frac{d}{dt} (BA)$

$= \frac{d}{dt} (μ_{0} nIA) = μ_{0} nA \frac{dI}{dt} = 4 π \times 1 0^{- 7} \times 1500 \times 2.0 \times 1 0^{- 4} \times 20 = 7.5 \times 1 0^{- 6} V$

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? Sol. Here; length of the loop, $ℓ = 8 cm = 0.08 m$ , breadth of the loop, $b = 2 cm = 0.02 m$ , Magnetic field, $B = 0.3 T$ , velocity of the loop, $v = 1 cm / s = 1 0^{- 2} m / s$ (a) If the loop moves perpendicular to its longer side then induced emf.

$ε = v B ℓ = \frac{1}{100} \times 0.3 \times \frac{8}{100} = 2.4 \times 1 0^{- 4} V$

Rectangular loop with a small cut moving out of a magnetic field, illustrating motional emf for motion perpendicular to its longer side.

It will last till the edge PS comes out of field. Time after which emf disappear = time taken by loop to pass the field

$t_{1} = \frac{distance}{velocity} = \frac{2}{1} = 2 s$

(b) If the loop moves perpendicular to its shorter side then emf is induced only in side RS which also appears across the cut. Induced em,

$ε = vB ℓ = \frac{1}{100} \times 0.3 \times \frac{2}{100} = 0.6 \times 1 0^{- 4} V$

Rectangular wire loop moving out of a uniform magnetic field, showing the loop dimensions, velocity direction, and magnetic field.

It will last till the edge PQ comes out of field

$t_{2} = \frac{distance}{velocity} = \frac{8}{1} = 8 s$

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.

Sol. Length of the rod, $ℓ = 1.0 m$ Angular frequency,

$ω = 400 rad / s$

Magnetic field strength,

$B ε = 0.5 T = \frac{1}{2} B ω ℓ^{2} = \frac{1}{2} \times 0.5 \times 400 \times (1)^{2} = 100 V$

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

A horizontal straight wire 10 m long extending from east to west is falling with a speed of $5.0 ms^{- 1}$ , at right angles to the horizontal component of the earth's magnetic field, $0.30 \times 1 0^{- 4} Wb m^{- 2}$ . (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of emf ? (c) Which end of the wire is at the higher electrical potential? Sol. (a) Induced emf,

$ε = B_{H} v ℓ (B ⊥ v ⊥ ℓ) = 0.3 \times 1 0^{- 4} \times 5 \times 10 = 1.5 mV$

(b) Direction of induced emf ⇒ west to east. (c) Eastern end is at higher electric potential.

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. Sol. Initial current,

$I_{1} = 5.0 A; Final current, I_{2} = 0.0 A$

Change in current, $dI = I_{1} - I_{2} = 5 A$ Time taken for the change, $dt = 0.1 s$ Average emf, $ε = 200 V$ Self-inductance (L) of the coil,

$L = \frac{ε}{( \frac{dI}{dt} )} = \frac{200}{( \frac{5}{0.1} )} = 4 H$

Hence, the self-inductance of the coil is 4 H .

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? Sol. Here;

$M = 1.5 H, dI = 20 - 0 = 20 A ε = - \frac{d ϕ _{B}}{dt} = - M \frac{dI}{dt} \frac{d ϕ _{B}}{dt} = M (\frac{dI}{dt}) \Rightarrow d ϕ_{B} = MdI = 1.5 \times 20 = 30 Wb$

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Get NCERT Solutions for Class 12 Physics covering all chapters, with exercise answers, useful formulas, and clear explanations to help students learn concepts and solve questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 6

Clear Use of Lenz’s Law: The solutions explain how to find the direction of induced current using Lenz’s law and cover common types of questions based on it.
Step-by-Step Derivations: Important derivations, such as motional emf in a moving or rotating conductor and mutual inductance of two long coaxial solenoids, are explained step by step.
Work and Energy: The solutions explain the relationship between mechanical work and electrical energy during electromagnetic induction, such as when a magnet is moved near a coil.
Self-Inductance Calculations: The solutions explain the factors that affect the self-inductance of a solenoid, including the number of turns, length, cross-sectional area, and nature of the core.
Eddy Currents and Their Uses: The solutions explain how eddy currents produce heat and cover their practical applications, such as magnetic braking and induction heating.

Frequently Asked Questions

NCERT Solutions Class 12 Physics Chapter 6 - Electromagnetic Induction

1.0Class 12 Physics Chapter 6 Electromagnetic Induction: Key Concepts

Class 12 Physics Chapter 6, Electromagnetic Induction, explains how a changing magnetic field can produce an induced emf and current in a conductor. The main concepts covered in this chapter include:

Magnetic Flux: Understand magnetic flux and how it changes with the strength and direction of a magnetic field and the area through which the field passes.
Faraday’s Law of Electromagnetic Induction: Learn how the induced emf is related to the rate of change of magnetic flux through a circuit.
Lenz’s Law: Understand how to determine the direction of induced current. The induced current always opposes the change in magnetic flux that produces it.
Motional EMF: Study the emf produced when a conductor moves through a magnetic field.
Eddy Currents: Learn about the circulating currents produced inside bulk conductors when the magnetic flux through them changes, along with their applications in magnetic braking and induction heating.
Self-Inductance and Mutual Inductance: Understand how a changing current in a coil produces an induced emf in the same coil or in a nearby coil.
AC Generator: Learn the basic working of an AC generator and how electromagnetic induction is used to convert mechanical energy into electrical energy.

2.0NCERT Solutions Class 12 Physics Chapter 6 Electromagnetic Induction : Detailed Solutions

SOLVED EXAMPLES

Consider Faraday's Experiment

(a) What would you do to obtain a large deflection of the galvanometer? (b) How would you demonstrate the presence of an induced current in the absence of a galvanometer?

Sol.

In experimental physics one must learn to innovate. Michael Faraday who is ranked as one of the best experimentalists ever, was legendary for his innovative skills.

A square loop of side 10 cm and resistance $0.5Ω$ is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time-interval. Sol. The angle $θ$ made by the area vector of the coil with the magnetic field is 45°. The initial magnetic flux is

$Φ = BA cos θ = \frac{0.1 \times 1 0 ^{- 2}}{2} Wb$

Final flux, $Φ_{min} = 0$ The change in flux is brought about in 0.70 s. The magnitude of the induced emf is given by

$ε = \frac{∣ Δ Φ _{B} ∣}{Δ t} = \frac{∣ ( Φ - 0 ) ∣}{Δ t} = \frac{1 0 ^{- 3}}{2 \times 0.7} = 1.0 mV$

And the magnitude of the current is

$I = \frac{ε}{R} = \frac{1 0 ^{- 3} V}{0.5Ω} = 2 mA$

Note that the earth's magnetic field also produces a flux through the loop. But it is a steady field (which does not change within the time span of the experiment) and hence does not induce any emf.

A circular coil of radius 10 cm, 500 turns and resistance $2Ω$ is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through $18 0^{\circ}$ in 0.25 s. Estimate the magnitudes of the emf and current induced in the coil. Horizontal component of the earth's magnetic field at the place is $3.0 \times 1 0^{- 5} T$ . Sol. Initial flux through the coil, $Φ_{B (initial)} = BA cos θ$

$= 3.0 \times 1 0^{- 5} \times (π \times 1 0^{- 2}) \times cos 0^{\circ} = 3 π \times 1 0^{- 7} Wb$

Final flux after the rotation,

$Φ_{B (final)} = 3.0 \times 1 0^{- 5} \times (π \times 1 0^{- 2}) \times cos 18 0^{\circ} = - 3 π \times 1 0^{- 7} Wb$

Therefore, estimated value of the induced emf is,

$ε = N \frac{ΔΦ}{Δ t} = \frac{500 \times ( 6 π \times 1 0 ^{- 7} )}{0.25} = 3.8 \times 1 0^{- 3} V I = \frac{ε}{R} = 1.9 \times 1 0^{- 3} A$

Note that the magnitudes of $ε$ and I are the estimated values. Their instantaneous values are different and depend upon the speed of rotation at the particular instant.

Figure shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader. Determine the direction of induced current in each loop using Lenz's law.

Sol.

Note that there are no induced current as long as the loops are completely inside or outside the region of the magnetic field.

(a) A closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets? (b) A closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop (i) When it is wholly inside the region between the capacitor plates. (ii) When it is partially outside the plates of the capacitor? The electric field is normal to the plane of the loop. (c) A rectangular loop and a circular loop are moving out of a uniform magnetic field region (figure) to a field-free region with a constant velocity $v$ . In which loop do you expect the induced emf to be constant during the passage out of the field region? The field is normal to the loops.

(d) Predict the polarity of the capacitor in the situation described by figure.

Sol.

A metallic rod of 1 m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular to the plane of the ring (figure). A constant and uniform magnetic field of 1 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?

$ε = \int d ε = \int_{0}^{R} Bvdr = \int_{0}^{R} B ω rdr = \frac{B ω R ^{2}}{2}$

Note that we have used $v = ω r$ . This gives

$ε = \frac{1}{2} \times 1.0 \times 2 π \times 50 \times (1^{2}) = 157 V$

A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth's magnetic field $H_{E}$ at a place. If $H_{E} = 0.4 G$ at the place, what is the induced emf between the axle and the rim of the wheel? Note that $1 G = 1 0^{- 4} T$ . Sol. Induce $emf = (1/2) ω BR^{2}$

$= (1/2) \times 4 π \times 0.4 \times 1 0^{- 4} \times (0.5)^{2} = 6.28 \times 1 0^{- 5} V$

The number of spokes is immaterial because the spokes are in parallel.

Two concentric circular coils, one of small radius $r_{1}$ and the other of large radius $r_{2}$ , such that $r_{1} ≪ r_{2}$ , are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. Sol. Let a current $I_{2}$ flow through the outer circular coil. The field at the centre of the coil is $B_{2} = \frac{μ _{0} I _{2}}{2 r _{2}}$ . Since the other co-axially placed coil has a very small radius. $B_{2}$ may be considered constant over its cross-sectional area. Hence.

$Φ_{1} = π r_{1}^{2} B_{2} = \frac{μ _{0} π r _{1}^{2}}{2 r _{2}} I_{2} = M_{12} I_{2}$

Thus, $M_{12} = \frac{μ _{0} π r _{1}^{2}}{2 r _{2}}$ From Eq.

$M_{12} = M_{21} = \frac{μ _{0} π r _{1}^{2}}{2 r _{2}}$

(a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B , area A and length $l$ of the solenoid. (b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor?

Sol.

(a) The magnetic energy is

$U_{B} = \frac{1}{2} L I^{2} = \frac{1}{2} L (\frac{B}{μ _{0} n})^{2}$

(since $B = μ_{0} nI$ , for a solenoid)

$= \frac{1}{2} (μ_{0} n^{2} A l) (\frac{B}{μ _{0} n})^{2} = \frac{1}{2 μ _{0}} B^{2} A l$

(b) The magnetic energy per unit volume is,

$u_{B} = \frac{U _{B}}{V}$

(where V is volume that contains flux)

$u_{B} = \frac{U _{B}}{A l} = \frac{B ^{2}}{2 μ _{0}}$

We have already obtained the relation for the electrostatic energy stored per unit volume in a parallel plate capacitor

$u_{E} = \frac{1}{2} ε_{0} E^{2}$

In both the cases energy is proportional to the square of the field strength.

Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a 100 turn coil of area $0.10 m^{2}$ . The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil? Sol. Here, $v = 0.5 Hz; N = 100$ ,

$A = 0.1 m^{2} and B = 0.01 T .$

Maximum voltage

$ε_{0} = NBA (2 π v) = 100 \times 0.01 \times 0.1 \times 2 \times 3.14 \times 0.5 = 0.314 V$

The maximum voltage is 0.314 V We urge you to explore such alternative possibilities for power generation.

EXERCISE QUESTIONS WITH SOLUTIONS

Predict the direction of induced current in the situations described by the following figures (a) to (f).

Use Lenz's law to determine the direction of induced current in the situations described by figure (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire.

A long solenoid with 15 turns per cm has a small loop of area $2.0 cm^{2}$ 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 ? Sol. Here; $n = 15 turns / cm = 1500 turns / m$ ,

$A = 2.0 cm^{2} = 2.0 \times 1 0^{- 4} m^{2} \frac{dI}{dt} = \frac{4.0 - 2.0}{0.1} = 20 A / s$

Induced emf, $ε = \frac{d ϕ _{B}}{dt} = \frac{d}{dt} (ϕ_{B}) = \frac{d}{dt} (BA)$

$= \frac{d}{dt} (μ_{0} nIA) = μ_{0} nA \frac{dI}{dt} = 4 π \times 1 0^{- 7} \times 1500 \times 2.0 \times 1 0^{- 4} \times 20 = 7.5 \times 1 0^{- 6} V$

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? Sol. Here; length of the loop, $ℓ = 8 cm = 0.08 m$ , breadth of the loop, $b = 2 cm = 0.02 m$ , Magnetic field, $B = 0.3 T$ , velocity of the loop, $v = 1 cm / s = 1 0^{- 2} m / s$ (a) If the loop moves perpendicular to its longer side then induced emf.

$ε = v B ℓ = \frac{1}{100} \times 0.3 \times \frac{8}{100} = 2.4 \times 1 0^{- 4} V$

It will last till the edge PS comes out of field. Time after which emf disappear = time taken by loop to pass the field

$t_{1} = \frac{distance}{velocity} = \frac{2}{1} = 2 s$

(b) If the loop moves perpendicular to its shorter side then emf is induced only in side RS which also appears across the cut. Induced em,

$ε = vB ℓ = \frac{1}{100} \times 0.3 \times \frac{2}{100} = 0.6 \times 1 0^{- 4} V$

It will last till the edge PQ comes out of field

$t_{2} = \frac{distance}{velocity} = \frac{8}{1} = 8 s$

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.

Sol. Length of the rod, $ℓ = 1.0 m$ Angular frequency,

$ω = 400 rad / s$

Magnetic field strength,

$B ε = 0.5 T = \frac{1}{2} B ω ℓ^{2} = \frac{1}{2} \times 0.5 \times 400 \times (1)^{2} = 100 V$

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

A horizontal straight wire 10 m long extending from east to west is falling with a speed of $5.0 ms^{- 1}$ , at right angles to the horizontal component of the earth's magnetic field, $0.30 \times 1 0^{- 4} Wb m^{- 2}$ . (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of emf ? (c) Which end of the wire is at the higher electrical potential? Sol. (a) Induced emf,

$ε = B_{H} v ℓ (B ⊥ v ⊥ ℓ) = 0.3 \times 1 0^{- 4} \times 5 \times 10 = 1.5 mV$

(b) Direction of induced emf ⇒ west to east. (c) Eastern end is at higher electric potential.

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. Sol. Initial current,

$I_{1} = 5.0 A; Final current, I_{2} = 0.0 A$

Change in current, $dI = I_{1} - I_{2} = 5 A$ Time taken for the change, $dt = 0.1 s$ Average emf, $ε = 200 V$ Self-inductance (L) of the coil,

$L = \frac{ε}{( \frac{dI}{dt} )} = \frac{200}{( \frac{5}{0.1} )} = 4 H$

Hence, the self-inductance of the coil is 4 H .

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? Sol. Here;

$M = 1.5 H, dI = 20 - 0 = 20 A ε = - \frac{d ϕ _{B}}{dt} = - M \frac{dI}{dt} \frac{d ϕ _{B}}{dt} = M (\frac{dI}{dt}) \Rightarrow d ϕ_{B} = MdI = 1.5 \times 20 = 30 Wb$

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Get NCERT Solutions for Class 12 Physics covering all chapters, with exercise answers, useful formulas, and clear explanations to help students learn concepts and solve questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 6

Clear Use of Lenz’s Law: The solutions explain how to find the direction of induced current using Lenz’s law and cover common types of questions based on it.
Step-by-Step Derivations: Important derivations, such as motional emf in a moving or rotating conductor and mutual inductance of two long coaxial solenoids, are explained step by step.
Work and Energy: The solutions explain the relationship between mechanical work and electrical energy during electromagnetic induction, such as when a magnet is moved near a coil.
Self-Inductance Calculations: The solutions explain the factors that affect the self-inductance of a solenoid, including the number of turns, length, cross-sectional area, and nature of the core.
Eddy Currents and Their Uses: The solutions explain how eddy currents produce heat and cover their practical applications, such as magnetic braking and induction heating.

Frequently Asked Questions

Which parts of Electromagnetic Induction are covered in NCERT Solutions Class 12 Physics Chapter 6?

What is the role of magnetic flux in NCERT Solutions Class 12 Physics Chapter 6?

Which law helps determine the direction of induced current in NCERT Solutions Class 12 Physics Chapter 6?

A conductor moves through a magnetic field. What does NCERT Solutions Class 12 Physics Chapter 6 explain about the induced emf?

How are self-inductance and mutual inductance treated in NCERT Solutions Class 12 Physics Chapter 6?

What practical applications of eddy currents can I learn from NCERT Solutions Class 12 Physics Chapter 6?

Can NCERT Solutions Class 12 Physics Chapter 6 help with JEE and NEET preparation?

Join ALLEN!

NCERT Solutions Class 12 Physics Chapter 6 - Electromagnetic Induction

1.0Class 12 Physics Chapter 6 Electromagnetic Induction: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 6 Electromagnetic Induction : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 6

Table of Contents

Frequently Asked Questions

Which parts of Electromagnetic Induction are covered in NCERT Solutions Class 12 Physics Chapter 6?

What is the role of magnetic flux in NCERT Solutions Class 12 Physics Chapter 6?

Which law helps determine the direction of induced current in NCERT Solutions Class 12 Physics Chapter 6?

A conductor moves through a magnetic field. What does NCERT Solutions Class 12 Physics Chapter 6 explain about the induced emf?

How are self-inductance and mutual inductance treated in NCERT Solutions Class 12 Physics Chapter 6?

What practical applications of eddy currents can I learn from NCERT Solutions Class 12 Physics Chapter 6?

Can NCERT Solutions Class 12 Physics Chapter 6 help with JEE and NEET preparation?

Join ALLEN!

NCERT Solutions Class 12 Physics Chapter 6 - Electromagnetic Induction

1.0Class 12 Physics Chapter 6 Electromagnetic Induction: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 6 Electromagnetic Induction : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 6

Table of Contents