Question

The inductor shown in Figure has inductance 0.260 H and carries a current in the direction shown. The current is changing at a constant rate. (a) The potential between points a and b is Vab = 1.04 V, with point a at higher potential. Is the current increasing or decreasing? (b) If the current at t = 0 is 12.0 A, what is the current at t = 2.00 s?

Short answer

a) the current is decreasing whereas the emf is directed to oppose this decrease.

b) the current at t = 2 sec is 4.00 amperes.

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

$\mathit{\epsilon }\mathbf{=}\mathbf{-}\mathit{L}\frac{\mathbf{}\mathbf{d}\mathbf{i}}{\mathbf{d}\mathbf{t}}$

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.

Variation of current

We are given,

The inductor with an inductance, L = 0.260H

The potential between a and b, Vab= 1.04V

The induced emf is from point b to a which is same as the direction of current.

Therefore, the current is decreasing whereas the emf is directed to oppose this decrease.

The current at t = 2.00 s

The induced emf is given by,

$$\begin{gathered} \left|E\right|V_{\alpha b}=V_{\alpha b}=L\left|\frac{di}{dt}\right| \\ \left|\frac{di}{dt}\right|=\frac{Vab}{L} \\ \left|\frac{di}{dt}\right|=\frac{1.04V}{0.26H}=4.00A/s \end{gathered}$$

If the current at localid="1664177006398" $t_i$= 0 is $i_i$= 12.0A,

The current at$t_f$= 2.00s, the rate is,

$\frac{di}{d_t}=\frac{i_f-i_i}{t_f-t_i}=\frac{i_f-i_i}{t_f}$

$i_f=t_f\frac{di}{di}+i_i$

$$\begin{gathered} i_f=\left(2.00s\right)\left(-4.00A\right)+12.0A=4.00A \\ {i}_f=4.00A \end{gathered}$$

Therefore, the current at t = 2 sec is 4.00 amperes.