Electromagnetic Induction and Inductance (JEE Advanced)

Question 1

A long solenoid of length ℓ and radius r (ℓ gg r) is wound uniformly with N turns of insulated wire; its self-inductance is L. All of the same wire is unwound and rewound uniformly over the same lengt

Accepted Answer

C. For a long solenoid L = μ₀ N² A/ℓ with A = π r².
The wire length is fixed: N(2π r) = N'(2π r/2), so N' = 2N.
New area A' = π r²/4, same ℓ: L' = μ₀ (2N)² (A/4)/ℓ = L.
Halving the radius doubles the number of turns (N² grows by 4) but quarters the area, so the inductance is unchanged.
Difficulty: Hard

Question 2

A fixed horizontal conducting ring has non-zero resistance and negligible self-inductance. A bar magnet, symmetric about its midpoint, is held vertically on the axis of the ring above it, with its nor

Accepted Answer

B. Outside a bar magnet on its axis the field points along its magnetic moment, i.e. from S to N; here the moment points down, so the flux through the ring is downward at every position of the magnet.
(i) Magnet approaching: downward flux increasing. The induced current must produce an upward field inside the ring, which is anticlockwise when viewed from above.
(ii) Magnet below and receding: flux st

Question 3

Use right-handed axes with +z vertically up. A uniform magnetic field B⃗ = Bẑ (B>0) fills all space. A thin straight conducting rod XY of length L lies in the horizontal xy-plane and rotates in that

Accepted Answer

A. A point of the rod at distance s from P moves with v⃗ = ω⃗×s⃗, tangential. On a positive charge, qv⃗×B⃗: with s⃗ = sx̂, v⃗ = ω sŷ and ŷ×ẑ = x̂, so the force is radially outward from P, of magnitude qω s B.
In the steady state an electrostatic field balances it: E = −ω B s (pointing towards P), so V(s) − V_P = ∫₀^s ω B s' ds' = 1/2 Bω s² on each side of P.
Both ends are above P: V_X − V_P = 1/2

Electromagnetic Induction and Inductance: questions with solutions

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