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Question
A coil having inductance 2.0 H and resistance 20 Ω is connected to a battery of emf 4.0 V. Find (a) the current at the instant 0.20 s after the connection is made and (b) the magnetic field energy at this instant.
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Solution
Given:-
Self-inductance of the coil, L = 2.0 H
Resistance in the coil, R = 20 Ω
Emf of the battery, e = 4.0 V
The steady-state current is given by
\[i_0 = \frac{e}{R} = \frac{4}{20}\]
The time-constant is given by
\[\tau = \frac{L}{R} = \frac{2}{20} = 0 . 1\]
(a) Current at an instant 0.20 s after the connection is made:-
i = i0(1 − e−t/τ)
\[=\frac{4}{20}(1 − e^{−0.2/0.1})\]
\[=\frac{1}{5}(1-e^{-2})\]
= 0.17 A
(b) Magnetic field energy at the given instant:-
\[\frac{1}{2}L i^2=\frac{1}{2}\times 2(0.17)^2\]
= 0.0289 = 0.03 J
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