The current in self -inductance `L=40` mH is to be be increased uniformly from 1 A to 11 A is 4 millisecond . The emf induce in inductor during the process is

Options

A. 100 V
B. 0.4 V
C. 40 V
D. 440 V

Answer

A

Explanation

Consider the inductor of inductance L.
The current flowing through the inductor is i.
Now, we can write `phi=Li`
where, `phi` is magnetic flux linked with the inductor
`rArr" "(dphi)/(dt)=L(di)/(dt)`
Given, L = 40 mH
dt = change in time `=t_(2)-t_(1)=4ms =Deltat`
`di=i_(2)-i_(1)=11-1=10A=Deltai`
So, `(dphi)/(dt)=(Deltaphi)/(Deltat)=L(Deltai)/(Deltat)=(40mH)((10))/(4ms)`
`-10xx10=100` ....(i)
According to Faraday's law of electromagnetic induction,
Emf induced, `e=-(dphi)/(dt)`
`rArr" "|e|=(Deltaphi)/(Deltat)=100V" [From Eq. (i)]"`