Question

The inductor in Figure has inductance 0.260 H and carries a current in the direction shown that is decreasing at a uniform rate, di/dt = -0.0180 A/s. (a) Find the self-induced emf. (b) Which end of the inductor, a or b, is at a higher potential?

Short answer

a) the induced emf in the inductor is 4.68 x 10-3 V.

b) the terminal a will be at higher potential than b.

Step-by-step solution

Given Data

Faraday’s law states that a current is induced in a conductor when it is exposed to a time varying magnetic flux. This induced current is driven by a force called electromotive or electromagnetic force. The magnitude of induced emf is given by

$\epsilon =-L\frac{di}{dt}$

Where L is the inductance of the conductor.

An inductor is a passive two-terminal device that stores energy in a magnetic field when current passes through it.

Lenz further explained the direction of this induced current. According to lens, the direction of induced current will be such that the magnetic field created by the induced current opposes the changing magnetic field which caused its induction.

Step 2: The induced emf through the inductor is,

The induced emf through the inductor is,

We are given,

The rate of current, di/dt = -0.0180 A/s

The inductor with an inductance, L = 0.260 H

$$\begin{gathered} \left|E\right|=L\frac{di}{dt}=(0.260H)(0.0180A/s)=4.68{\rm X}{10}^{-3}V \\ E=4.68{\rm X}{10}^{-3}V \end{gathered}$$

Therefore, the induced emf in the inductor is 4.68 x 10-3 V.

The higher potential between a and b

The current in the inductor is decreasing. To oppose the behavior, induced emf will be in the same direction as the current.

Therefore, the terminal a will be at higher potential than b.