Question

electric current flowing in the coil of self inductance 2mh is increased fro 1A to 2 A IN 2 ms.induced emf in the coil is

Answer 1

Step 1: Given that:

Self-inductance (L) in the coil = $2mH$ = $2\times {10}^{-3}H$

Initial value of current = $1A$

The final value of electric current = $2A$

Change in electric current($di$) = $2A-1A=1A$

Time(t) = $2ms$ = $2\times {10}^{-3}s$

Step 2: Formula used:

The induced emf in a coil of self-inductance L is given by;

${\in }_{induced}=L\frac{di}{dt}$

Step 3: Calculation of the induced emf:

${\in }_{induced}=2\times {10}^{-3}H\times \frac{1A}{2\times {10}^{-3}s}$

${\in }_{induced}=1\times 1\times$

${\in }_{induced}=1V$

Thus,

The induced emf is $1V$.