Question

In a circular conducting coil, when current increases
from 2 A to 18 A in 0.05 s, the induced emf is 20 V.
The self inductance of the coil is
(1) 62.5 mH
(2) 6.25 mH
(3) 50 mH
(4) Zero​

Answer 1

Answer:62.5mH

Explanation:

We know L=n¢/I

Differentiating wrt time,

N d¢/dt = L dI/dt

=> induced emf, €= L dI/dt

According to the question,

20=L (16) /0.05

=> L =62.5

Thank you...

Answer 2

The correct answer is (1) 62.5mH

In case of self inductance the equation we get is

NФ = LI

where N is number of turns

L denotes self inductance

I is current

differentitaing it with respect to time

N $\frac{d\phi}{dt}$ = L $\frac{dI}{dt}$       ...(i)

We know emf induced is given by

Emf = N \frac{d\phi}{dt}     ...(ii)

Putting (ii) in LHS of (i) we get

Emf = L $\frac{dI}{dt}$

Putting values

20 = L $\frac{18-2}{0.05}$

L = 62.5mH