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Q.A pair of adjacent coils has a mutual inductance of 1.5H. If the current in one coil changes from 0 to 20A in 0.5s, what is the change of flux linkage with the coil? OR Fill in the blanks given below regarding dipole analogy between electrostatics and magnetism (6×1/2=3): Electrostatics column —

(i) 1/ε0,
(ii) [blank],
(iii) -p/(4πε0r),
(iv) [blank],
(v) p×E,
(vi) [blank]. Magnetism column —
(i) [blank],
(ii) m,
(iii) [blank],
(iv) μ0·2m/(4πr^3),
(v) [blank],
(vi) -m.B
Assam AhsecAHSEC Higher Secondary (HS) Final Examination 2024Subjective· 3mImportance★★★★★
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Option 1: flux linkage change =MΔI=30 Wb= M\Delta I = 30\,\mathrm{Wb}. Option 2: the electrostatic and magnetic dipole formulas mirror each other with p↔mp \leftrightarrow m and 1/ε0↔μ01/\varepsilon_0 \leftrightarrow \mu_0.

Option 1 — Flux linkage from mutual inductance

Mutual inductance relates the flux linkage in one coil to the current in the other: N2Φ2=MI1N_2\Phi_2 = MI_1. So the change in flux linkage in the second coil due to a change ΔI\Delta I in the first coil's current is

Δ(NΦ)=M ΔI\Delta(N\Phi) = M\,\Delta I

With M=1.5 HM = 1.5\,\mathrm{H} and ΔI=20−0=20 A\Delta I = 20-0=20\,\mathrm{A}:

Δ(NΦ)=1.5×20=30 Wb\Delta(N\Phi) = 1.5 \times 20 = 30\,\mathrm{Wb}

(The 0.5 s is not needed for the flux-linkage change itself — it would only be needed to find the induced emf, ε=Δ(NΦ)/Δt=30/0.5=60 V\varepsilon = \Delta(N\Phi)/\Delta t = 30/0.5 = 60\,\mathrm{V}.)

Option 2 — Electrostatic / magnetic dipole analogy

Electric and magnetic dipoles obey exactly analogous formulas, with the correspondence p↔mp \leftrightarrow m (dipole moment) and 14πε0↔μ04π\dfrac{1}{4\pi\varepsilon_0} \leftrightarrow \dfrac{\mu_0}{4\pi}:

RowElectrostaticsMagnetism
(i)/(i)1/ε01/\varepsilon_0μ0\mu_0
(ii)/(ii)ppmm

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