NCERT Solutions for Class 12 Physics Chapter 6: Electromagnetic Induction

Concept and context. According to Faraday’s law an emf ε = −dΦ_B/dt is induced in a circuit whenever the magnetic flux through it changes, and by Lenz’s law (the minus sign) the direction of the induced current is such that it opposes the change in flux that produced it. This is a statement of conservation of energy: if the induced current assisted the change, a small push would grow without limit and create energy from nothing. Practical method: (1) find the direction of the existing flux (field) through the loop and whether it is increasing or decreasing; (2) the induced current must create a field that is opposite to the original if the flux is increasing and in the same direction if it is decreasing; (3) use the right-hand rule (or the rule “a face behaves as a north pole if the current appears anticlockwise when viewed from that face, as a south pole if clockwise”) to convert the required field into a current direction. A useful shortcut: the induced current always makes the face of the coil nearer to the approaching magnet (or current loop) like the approaching pole (repulsion) and the near face like the opposite pole when the magnet recedes (attraction).

Case-by-case solution

1. (a) South pole of a bar magnet moving towards the coil pq (right-hand end of the coil nearest the magnet). The magnet’s field at the coil is increasing in magnitude as the pole approaches. By Lenz’s law the coil must oppose this approach, so its near (right-hand) face must become a south pole (like poles repel). A face acts as a south pole if, viewed from that face, the current is clockwise. Hence the induced current is clockwise as seen from the magnet, i.e. in the lettered figure the current flows through the external wire q → r → p and through the coil from p to q: qrpq.
2. (b) Magnet moving to the left between two coaxial coils (pq on the left, xy on the right). Coil pq: the S-end of the magnet approaches it, the flux rises, the near face (right-hand end) must be S → clockwise as seen from the magnet: the current is prqp (p → r → q in the external wire). Coil xy: the N-end of the magnet is moving away from it, so the flux (towards the right, away from the N pole) decreases; the coil must oppose the decrease by attracting the receding N pole, so its near (left-hand) face must be S as well: clockwise as seen from the magnet (i.e. opposite in sense to that in coil pq when viewed from the same side): yzxy (y → z → x).
3. (c) Two coaxial loops; the tapping key in the left-hand loop is just closed. The current in the left loop rises from zero to a steady value (arrows in the figure show it is clockwise as seen from the right-hand loop), so the field through the loop xyz (directed away from the viewer standing at the right) increases. The induced current in xyz must oppose the increase: it must produce a field in the opposite direction, i.e. the current is anticlockwise as seen from the right (opposite to the current in the left loop, as for two parallel currents that repel), written yzxy (y → z → x). The induced current lasts only while the current is changing (a brief pulse).
4. (d) Rheostat setting being changed in the circuit of the right-hand loop. The current in the right-hand loop changes (the setting is changed so that the current increases, as in the NCERT answer; the arrows drawn on the loop show its sense — anticlockwise as seen from the left-hand loop’s right side); the flux through the left-hand loop xyz changes accordingly. The induced current in xyz must oppose the increase in the flux: its sense is opposite to that in the other loop and it is clockwise as seen from the right: zyxz (z → y → x).
5. (e) Two coaxial solenoids; the tapping key in the circuit of the left-hand solenoid is just released. The current in the first solenoid decreases to zero, so the flux through the second solenoid xy decreases. The second coil must oppose the decrease by producing a field in the same direction as the original: the induced current in the second solenoid flows in the same sense as the original current in the first: xryx (x → r → y in the external wire).
6. (f) A straight wire carrying a decreasing current, perpendicular to the plane of a loop and passing through its centre. The magnetic field lines of the straight wire are circles around the wire lying in planes perpendicular to it — here in the plane of the loop itself. Hence the flux through the loop (B·A, with the area vector along the wire) is zero at all times even though the current changes: no current is induced.

• Verification of method: in every case apply Lenz’s law in the form “the induced current opposes the change of flux — not the flux itself”. When the flux is increasing the induced field opposes the existing field; when it is decreasing the induced field reinforces it.
• Energy interpretation: in (a) and (b) work must be done against the repulsive force to push the magnet towards the coil — that work becomes the electrical energy of the induced current (finally heat).
• Dependence on the winding: the lettered paths depend on the sense of winding drawn in the NCERT figure; the physical statements (clockwise/anticlockwise as seen from a given side) are independent of the lettering.

Conclusion: the induced currents are (a) qrpq; (b) prqp in the left coil and yzxy in the right coil; (c) yzxy; (d) zyxz; (e) xryx; (f) none, each determined by requiring that the induced current oppose the change in the magnetic flux (Lenz’s law).

Concept and context. The flux through a loop in a uniform field is Φ_B = B A cos θ. Even if the field is constant, the flux changes when the area of the loop changes. A wire loop of fixed length encloses the maximum area when it is a circle; deforming an irregular loop into a circle therefore increases the enclosed area (and flux), while squeezing a circular loop into a narrow straight wire decreases the area (and flux) towards zero. Lenz’s law then fixes the direction: the induced current must oppose the change in flux.

(a) Irregular wire turning into a circle (magnetic field directed into the page, ×)

1. Change in flux: the wire has a fixed length; as it turns into a circle the enclosed area increases, so the flux of the field directed into the page increases.
2. Required induced field: to oppose the increase of inward flux, the induced current must produce a field pointing out of the page inside the loop.
3. Direction: a field out of the page inside a loop corresponds to an anticlockwise current (right-hand rule). In terms of the labels of the figure the current flows along a → d → c → b → a (anticlockwise).

(b) Circular loop deformed into a narrow straight wire (field directed out of the page, •)

1. Change in flux: the area enclosed shrinks towards zero, so the flux through the loop (outward) decreases.
2. Required induced field: the induced current must oppose the decrease, i.e. try to maintain the flux: it must produce a field in the same direction as the existing one, out of the page.
3. Direction: again anticlockwise: a′ → d′ → c′ → b′ → a′ as labelled in the figure.

• Interesting point: the two situations look opposite (flux increasing vs. decreasing, field inward vs. outward), but the result is the same sense because the field direction is also reversed — both changes require an anticlockwise induced current.
• Why does the area change create an emf? Elements of the wire move through the field, so the free charges in them experience the Lorentz force qv×B: the induced emf is a motional emf (ε = −dΦ/dt = −B dA/dt).

Conclusion: in both situations the induced current is anticlockwise — along adcba in (a) and along a′d′c′b′a′ in (b) — so that it opposes the increase of inward flux in (a) and the decrease of outward flux in (b).

Concept and context. The magnetic field inside a long solenoid carrying current I is uniform and equal to B = μ₀ n I, where n is the number of turns per unit length. When the current in the solenoid changes, B changes and so does the flux through a small loop placed inside it. By Faraday’s law the emf induced in the loop is ε = −dΦ_B/dt = −A dB/dt (the loop has a single turn and its plane is normal to the axis, so B is normal to the area and cos θ = 1). Because the current changes at a steady rate, the emf is constant.

Given: n = 15 turns/cm = 1500 turns/m; loop area A = 2.0 cm² = 2.0 × 10⁻⁴ m²; I changes from 2.0 A to 4.0 A in Δt = 0.1 s; μ₀ = 4π × 10⁻⁷ T m A⁻¹; the loop is a single turn.

1. Change in the field: ΔB = μ₀ n ΔI = 4π × 10⁻⁷ × 1500 × (4.0 − 2.0) = 4π × 10⁻⁷ × 3000 = 3.77 × 10⁻³ T.
2. Change in the flux through the loop: ΔΦ = A ΔB = 2.0 × 10⁻⁴ × 3.77 × 10⁻³ = 7.54 × 10⁻⁷ Wb.
3. Induced emf (magnitude): ε = ΔΦ/Δt = 7.54 × 10⁻⁷ / 0.1 = 7.54 × 10⁻⁶ V.
4. ε ≈ 7.5 × 10⁻⁶ V (7.5 μV).

ε = A μ₀ n (ΔI/Δt) = 7.5 × 10⁻⁶ V

• Direction: the flux (along the axis) is increasing, so by Lenz’s law the induced current in the loop flows so as to oppose the increase: it produces a field opposite to that of the solenoid, i.e. the induced current in the loop is in the sense opposite to the current in the solenoid.
• Check: dI/dt = 20 A s⁻¹; dB/dt = μ₀ n (dI/dt) = 4π × 10⁻⁷ × 1500 × 20 = 3.77 × 10⁻² T s⁻¹; ε = A dB/dt = 2.0 × 10⁻⁴ × 3.77 × 10⁻² = 7.5 × 10⁻⁶ V ✓.
• Remark: a few microvolts is tiny — the loop is small (2 cm²) and the rate of change is modest; a coil of N turns would give N times the emf.

Conclusion: the emf induced in the loop while the current changes is 7.5 × 10⁻⁶ V, with the induced current opposing the increase of flux (opposite in sense to the solenoid current).

Concept and context. When a conductor of length l moves with velocity v perpendicular to a uniform magnetic field B (and perpendicular to its own length), the Lorentz force on the free charges produces a motional emf ε = B l v. By Faraday’s law the same result comes from the changing flux: if a loop is moving out of the field region, the flux is Φ = B l x (x = length still inside the field) and ε = −dΦ/dt = B l v, where l is the length of the side that is crossing the field boundary (the side perpendicular to the velocity). Because of the cut in the loop no current flows (open circuit), but the emf appears across the cut. The emf exists only while the flux is changing: that is, while the loop is passing out through the boundary of the field region; once it is completely outside (or inside) the flux is constant (zero) and there is no emf.

Given: sides 8 cm = 0.08 m and 2 cm = 0.02 m; B = 0.3 T (normal to the plane of the loop); v = 1 cm s⁻¹ = 0.01 m s⁻¹.

(a) Velocity normal to the longer side

1. The side that cuts across the field boundary is the longer side, so l = 8 cm = 0.08 m.
2. ε = B l v = 0.3 × 0.08 × 0.01 = 2.4 × 10⁻⁴ V.
3. Duration: the loop moves a distance equal to its other dimension (the shorter side, 2 cm) before it is completely out of the field: t = distance/speed = 2 cm ÷ 1 cm s⁻¹ = 2 s.

(b) Velocity normal to the shorter side

1. Now the leading side is the shorter side, so l = 2 cm = 0.02 m.
2. ε = B l v = 0.3 × 0.02 × 0.01 = 6 × 10⁻⁵ V.
3. Duration: the distance to move out is the longer side, 8 cm: t = 8 cm ÷ 1 cm s⁻¹ = 8 s.

• Check with flux: the flux through the loop falls from B × (8 × 2 cm²) = 0.3 × 1.6 × 10⁻³ = 4.8 × 10⁻⁴ Wb to zero; in (a) in 2 s: ε = 4.8 × 10⁻⁴/2 = 2.4 × 10⁻⁴ V ✓; in (b) in 8 s: ε = 4.8 × 10⁻⁴/8 = 6 × 10⁻⁵ V ✓.
• Note: the total flux change ΔΦ is the same in the two cases; a faster change (shorter time) gives a larger emf. The emf is zero before and after the loop crosses the field boundary.
• Why the cut? With a cut there is no closed path, so no current flows and no magnetic braking force acts; the emf appears as a potential difference across the cut.

Conclusion: (a) ε = 2.4 × 10⁻⁴ V lasting 2 s; (b) ε = 6 × 10⁻⁵ V lasting 8 s.

Concept and context. A conductor moving through a magnetic field develops a motional emf because the Lorentz force q(v × B) pushes the free electrons towards one end until the resulting electrostatic field balances the magnetic force. For a rod of length l rotating with angular velocity ω about one end in a uniform field B perpendicular to the plane of rotation (parallel to the axis), different points of the rod move with different speeds, v = ωr (r = distance from the pivot), so the emf is found by adding the contributions of all small elements dr: dε = B v dr = B ω r dr. The sliding contact with the circular ring completes the circuit (NCERT Fig. 6.11 and Example 6.6).

Given: l = 1.0 m; ω = 400 rad s⁻¹; B = 0.5 T.

Derivation of the emf (two methods)

1. Method I — integrating the motional emf of elements: the speed of a point at distance r from the pivot is ωr. The emf of an element dr is dε = B (ωr) dr. Summing from r = 0 to l: ε = Bω ∫₀^l r dr = ½ B ω l².
2. Method II — average speed: the speed varies linearly from 0 at the pivot to ωl at the free end, so the average speed is ωl/2; ε = B l v_avg = B l (ωl/2) = ½ B ω l².
3. Method III — flux method: in time dt the rod sweeps a sector of area dA = ½ l² (ω dt); the flux cut is dΦ = B dA = ½ B l² ω dt; so ε = dΦ/dt = ½ B ω l².

Calculation

ε = ½ B ω l² = ½ × 0.5 × 400 × (1.0)² = 100 V

• Polarity: for a field directed out of the page and an anticlockwise rotation, v × B on a point of the rod points outwards (away from the axle), so positive charges are pushed towards the ring and electrons towards the centre: the ring is at the higher potential and the centre is negative; the polarity reverses if the rotation or the field is reversed.
• Independent of the ring: the ring serves only to complete the circuit through the sliding contact; the emf is generated entirely in the rod (every spoke of a wheel behaves the same way — NCERT Example 6.7 — and emfs of spokes are in parallel so the number of spokes is immaterial).
• Magnitude check: ωl = 400 m s⁻¹ is the speed of the tip; the average speed 200 m s⁻¹; ε = 0.5 × 1.0 × 200 = 100 V ✓.

Conclusion: the emf developed between the centre and the ring is 100 V, given by ε = ½ B ω l².

Concept and context. A conductor moving in a magnetic field develops a motional emf ε = B l v (v, B and l mutually perpendicular) because the magnetic Lorentz force F = q(v × B) separates the charges in the conductor. The direction of the force on a positive charge is given by v × B; the end towards which the positive charges are pushed becomes the positive (higher potential) end, and the emf is directed from the negative to the positive end through the wire. Here the wire is horizontal (east–west), moves vertically downwards (falling) and the horizontal component of the Earth’s field points along the geographic north (from south to north).

Given: l = 10 m; v = 5.0 m s⁻¹ (downwards); B_H = 0.30 × 10⁻⁴ Wb m⁻² (northward); wire along east–west.

(a) Magnitude of the induced emf

ε = B l v = (0.30 × 10⁻⁴)(10)(5.0) = 1.5 × 10⁻³ V (1.5 mV)

• The three quantities are mutually perpendicular — v is vertical, B horizontal (north) and the wire horizontal (east–west) — so sin θ = 1 and the full value is obtained. (Only the horizontal component B_H is effective because the vertical component is parallel to v and gives v × B = 0.)
• Since the wire falls freely, its speed (and hence the emf) increases with time: the value 1.5 mV is the instantaneous value when v = 5.0 m s⁻¹.

(b) Direction of the emf

1. Take x = east, y = north, z = up. Then v = −5.0 ẑ and B = +B_H ŷ.
2. Force on a positive charge: v × B = (−ẑ) × (ŷ) = −(ẑ × ŷ) = −(−x̂) = +x̂ — towards the east.
3. Positive charges are pushed towards the east end, so the emf in the wire is directed from west to east (the emf is directed along the direction of the force on positive charges).

(c) Which end is at higher potential?

• Positive charges accumulate at the east end, leaving a surplus of electrons at the west end, so the east end is at the higher electric potential (the west end is negative). The wire acts like a battery with its positive terminal at the east end.

Conclusion: (a) the induced emf is 1.5 × 10⁻³ V; (b) its direction is from west to east; (c) the east end of the wire is at the higher electrical potential.

Concept and context. When the current in a coil or circuit changes, the magnetic flux linked with it changes and an emf is induced in the same circuit — self-induction. The flux linkage is proportional to the current: NΦ_B = L I, and the induced (back) emf is ε = −L dI/dt, where L is the self-inductance (unit henry, 1 H = 1 V s A⁻¹ = 1 Wb A⁻¹). The negative sign (Lenz’s law) shows that the emf opposes the change in current: it resists the fall of current by tending to keep it flowing. L plays the role of “electrical inertia” in the same way that mass resists a change of velocity. For a long solenoid L = μ_r μ₀ n² A l.

Given: ΔI = 0 − 5.0 = −5.0 A; Δt = 0.1 s; average induced emf |ε| = 200 V.

1. Magnitude of the average rate of change of current: |ΔI/Δt| = 5.0/0.1 = 50 A s⁻¹.
2. From |ε| = L |ΔI/Δt|: L = |ε|/|ΔI/Δt| = 200/50 = 4 H.
3. Sign: the current is falling (dI/dt negative) so ε = −L(dI/dt) is positive — the induced emf drives current in the original direction, opposing the decrease.

L = ε / (ΔI/Δt) = 200 / 50 = 4 H

• Physical meaning: L = 4 H means that a 1 A s⁻¹ change in current induces 4 V — a rather large inductance (iron-core coils or large solenoids; air-core coils are typically μH to mH).
• Energy stored: before the current falls the circuit stored U = ½LI² = ½ × 4 × 25 = 50 J in its magnetic field; this energy is released (as the induced emf drives current, appears as heat/spark) when the current is switched off.
• Estimate: because only the average emf is given, the value of L is an estimate.

Conclusion: the self-inductance of the circuit is estimated to be 4 H.

Concept and context. When two coils are close together, the flux through one is partly due to the current in the other. A changing current I₂ in coil 2 changes the flux linked with coil 1 and induces an emf ε₁ = −M₁₂ dI₂/dt, where M₁₂ is the mutual inductance of coil 1 with respect to coil 2; equivalently the flux linked with coil 1 (flux linkage N₁Φ₁) is N₁Φ₁ = M₁₂ I₂. Mutual inductance depends on the geometry, the number of turns, the separation and orientation of the coils and on the medium between them; it is the same in both directions (M₁₂ = M₂₁ = M). Its unit is the henry. By Faraday’s law, ε₁ = −d(N₁Φ₁)/dt = −M dI₂/dt.

Given: M = 1.5 H; ΔI = 20 − 0 = 20 A; Δt = 0.5 s.

1. Flux linkage: N₁Φ₁ = M I₂ ⇒ the change in flux linkage = M ΔI₂.
2. Substitute: Δ(N₁Φ₁) = 1.5 H × 20 A = 30 H A = 30 Wb (since 1 H A = 1 Wb).
3. Induced emf (additional information): ε = Δ(NΦ)/Δt = 30/0.5 = 60 V (average).

Δ(flux linkage) = M ΔI = 1.5 × 20 = 30 Wb

• Note: the time interval (0.5 s) is not needed for the flux linkage; it is needed only to find the average induced emf (60 V).
• Unit check: 1 H = 1 Wb A⁻¹ so H × A = Wb ✓. NCERT’s Table of physical quantities lists the unit of mutual inductance as henry with dimensions [M L² T⁻² A⁻²].
• Example: in a transformer, the primary and secondary windings are coupled through a large mutual inductance, so a changing primary current induces an emf in the secondary.

Conclusion: the change of flux linkage with the other coil is 30 Wb (corresponding to an average induced emf of 60 V over 0.5 s).