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
$i_0=\frac{e}{R}=\frac{4}{20}$ A
The time-constant is given by
$\tau =\frac{L}{R}=\frac{2}{20}=0.1$ s

(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:
$\frac{1}{2}L{i}^2$ = $\frac{1}{2}$ × 2(0.17)²
= 0.0289 = 0.03 J