Electromagnetic Induction Class 12 Physics NCERT Solutions

Step-by-step solution

1. 1Use Lenz's law: the induced current flows so as to oppose the change in flux producing it.
2. 2(a) The magnet approaches, flux through coil pqr increases; the face toward the magnet becomes a like pole (N), so looking from the magnet side the current is anticlockwise — along the path qrpq.
3. 3(b) Two coils: for the coil the magnet is approaching (pqr), flux increases — current along prq; for the coil it is receding from (xyz), flux decreases — current along yzx. Each current opposes its own change of flux.
4. 4(c) Closing the key builds up current, flux through the second coil increases — the induced current opposes it, flowing along yzxy.
5. 5(d) Adjusting the rheostat changes the primary current; for the figure's sense (current increasing) the induced current in the second coil flows along zyxz. If the primary current were being decreased, the direction would simply reverse (along xyzx).
6. 6(e) Releasing the key collapses the field; the coupled loop's flux decreases — the induced current maintains it, flowing along xryx.
7. 7(f) The field of a straight wire forms concentric circles lying in the plane of the coplanar loop, so the flux through the loop is zero at all times — no current is induced.

Step-by-step solution

1. 1(a) Turning the irregular wire into a circle increases its enclosed area, so the (into-the-page) flux through it increases.
2. 2To oppose the increase, the induced current must produce a field out of the page — looking along the field direction, the current is anticlockwise (along adcb as labelled in the figure).
3. 3(b) Deforming the loop into a narrow straight wire shrinks its area to zero, so the flux decreases.
4. 4To oppose the decrease, the induced current must maintain the into-the-page field — the current is clockwise (along abcd in the figure).

Step-by-step solution

1. 1n = 15 turns cm⁻¹ = 1500 turns m⁻¹; loop area A = 2.0 cm² = 2.0 × 10⁻⁴ m².
2. 2dB/dt = μ₀ n dI/dt = 4π × 10⁻⁷ × 1500 × (2.0/0.1).
3. 3dB/dt = 4π × 10⁻⁷ × 1500 × 20 = 3.77 × 10⁻² T s⁻¹.
4. 4ε = A·dB/dt = 2.0 × 10⁻⁴ × 3.77 × 10⁻² = 7.5 × 10⁻⁶ V.

Step-by-step solution

1. 1Motional emf while a side is sweeping out of the field: ε = B l v, with l the length of the side that continues to cut the field.
2. 2(a) Moving out normal to the longer (8 cm) side: l = 8 cm = 0.08 m, v = 10⁻² m s⁻¹.
3. 3ε = 0.3 × 0.08 × 10⁻² = 2.4 × 10⁻⁴ V.
4. 4Duration: the 2 cm width takes (2 × 10⁻²)/(10⁻²) = 2 s to leave the field.
5. 5(b) Moving out normal to the shorter (2 cm) side: l = 2 cm = 0.02 m.
6. 6ε = 0.3 × 0.02 × 10⁻² = 6 × 10⁻⁵ V.
7. 7Duration: the 8 cm length takes (8 × 10⁻²)/(10⁻²) = 8 s to leave the field.

Step-by-step solution

1. 1Every point of the rod sweeps flux with speed v = ωr, so the average speed over the rod is v_avg = ωl/2.
2. 2ε = B l v_avg = (1/2) B ω l².
3. 3ε = 0.5 × 0.5 × 400 × (1.0)² = 100 V.

Step-by-step solution

1. 1(a) ε = B l v = 0.30 × 10⁻⁴ × 10 × 5.0 = 1.5 × 10⁻³ V.
2. 2Direction: the wire falls perpendicular to the horizontal component of the earth's field; by Fleming's right-hand rule the force on a positive charge in the wire is toward the east, so the emf acts from the west end toward the east end.
3. 3(c) The moving charge separation makes the eastern end positive, so the east end is at the higher electrical potential.

Step-by-step solution

1. 1ε = L·|dI/dt|.
2. 2|dI/dt| = 5.0/0.1 = 50 A s⁻¹.
3. 3L = ε/(dI/dt) = 200/50 = 4 H.

Step-by-step solution

1. 1Flux linkage with the other coil is Φ₂ = M I₁, so the change is ΔΦ₂ = M ΔI₁.
2. 2ΔΦ₂ = 1.5 × 20 = 30 Wb.
3. 3(For completeness, the induced emf would be ε = M·dI/dt = 1.5 × 20/0.5 = 60 V — but the exercise asks only for the flux-linkage change.)