The inductor shown in figure has inducance `0.54 H` and carries a current in the direction shown thast is decreasing at a uniform rate `(di)/(dt)=0.03A//s`.

a. Find the self induced emf
b. Which emf of the inductor a or b is at a higher potential?

Explanation

(i) Given, inductance, L= 0.54H
`(di)/(dt)=-0.03`
Self-induced emf is givne as
`e=-L(di)/(dt)=(-0.54)(-0.03)V`
`=1.62xx10^(-2)V`
(ii) We know that, `V_(ba)=L(di)/(dt)=-1.62xx10^(-2)V`
Since, `V_(ba)(=V_(b)-V_(a))` is negative, It implies that `V_(a)gtV_(b)` or a is at higher potential.