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.

Answer 1

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