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

The Magnetic Field Energy is 0.03 J

Explanation:

Given:

Self-inductance of the coil, L = 2.0 H

Resistance in the coil, R = $20\ \Omega$

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 = $i_0(1 - e^{-t/\tau})$

= $\frac {4}{20} (1 - e^{-0.2/0.1})$

= $\frac {1}{5}(1 - e^-^2)$

= 0.17 A

b.) Magnetic field energy at the given instant:

$\frac {1}{2}Li^2 = \frac {1}{2}\times 2(0.17)^2$  

0.0289 = 0.03 J