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 ro b is at a higher potential?

Explanation

a. Self induced emf
`e=-L (di)/(dt)=(-0.54)(-0.03)V`
`=1.62xx10^-2V`
b. `V_(ba)=L(di)/(dt)=-1.62xx10^-2V`
Since `V_(ba)=(=V_b-V_a)` is negative. It implies that `V_a gt V_b` or a is at higher potential.