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Q.Define self-inductance and mutual inductance. Find an expression for mutual inductance of two coaxial solenoids.

Bihar BsebBihar Board Intermediate 2025Subjective· 5mImportance★★★★★
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Self-inductance opposes current change in the same coil (φ = Li); mutual inductance links two coils (φ₂ = Mi₁). For coaxial solenoids M = μ₀n₁n₂Al.

Self-inductance (L). When the current i in a coil changes, the magnetic flux linked with the coil itself changes and induces an EMF in it that opposes the change (Lenz's law). The flux linkage is proportional to the current:

Nϕ=Li,ε=−Ldidt.N\phi = L i, \qquad \varepsilon = -L\frac{di}{dt}.

Here L is the self-inductance; it is the flux linkage per unit current, or the EMF induced per unit rate of change of current. SI unit: henry (H).

Mutual inductance (M). When two coils are placed near each other, a change of current in one coil (the primary) changes the flux linked with the second coil (the secondary) and induces an EMF in it. If i₁ is the primary current and φ₂ the flux linked with the secondary,

N2ϕ2=Mi1,ε2=−Mdi1dt.N_2\phi_2 = M i_1, \qquad \varepsilon_2 = -M\frac{di_1}{dt}.

M is the mutual inductance of the pair; SI unit: henry (H).

Mutual inductance of two coaxial solenoids.

Consider a long solenoid S₁ of length l, cross-sectional area A, with N₁ turns (n₁ = N₁/l turns per unit length), and a second solenoid S₂ of N₂ turns (n₂ = N₂/l) wound closely over the same axis (coaxial).

Let current i₁ flow in S₁. The magnetic field inside a long solenoid is uniform:

B1=μ0n1i1=μ0N1li1.B_1 = \mu_0 n_1 i_1 = \mu_0 \frac{N_1}{l} i_1.

This field also threads all N₂ turns of S₂ (area A each), so the flux linked with S₂ is …

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