In 0.2 s, the current in a coil increases from 2.0 A to 3.0 A. If inductance of coil is 60 mH, then induced current in external resistance of 3 Ω will be

1A

0.5 A

0.2 A

0.1 A

Solution

0.1 A

When a current in a coil changes, it induces a back emf back emf in the same coil. The self-induced emf is given by

Induced emf

$|e| = L \frac{di}{dt}$

= $(60 \times 10^{- 3}) \times \frac{(3.0 - 2.0)}{0.2}$

= $\frac{60 \times 10^{- 3} \times 1}{0.2}$

= 0.3 A

A changing magnetic field through a coil of wire therefore must induce an emf in the coil which in turn causes current to flow.

Induced current

i = $\frac{e}{R}$

= $\frac{0.3}{3}$

i = 0.1 A

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