At the instant when the current in an inductor is increasing at a rate of 0.0640 A/s, the magnitude of the self-induced emf is 0.0160 V. (a) What is the inductance of the inductor? (b) If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is 0.720 A?

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

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

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.

We are given,

The rate of current, di/dt = 0.0640 A/s

The self-induced emf, E = 0.0160 V

The induced EMF in the circuit is given by

$E=L\left|\frac{\Delta i}{\Delta t}\right|$

Where, L is the self-inductance,

$L=\frac{E}{\left(\frac{\Delta i}{\Delta t}\right)}$

Plug the values for E and di/dt to get L

$L=\frac{E}{\left|\right(\Delta i/\Delta t\left)\right|}==\frac{0.0160V}{0.0640A/s}=0.250H=250mH$

Therefore, the inductance of the inductor is 250 milli-henry.

The number of turns, N = 400 turns,

The current, i = 0.720 A.

This inductance is given by,

$L=\frac{N{\phi }_B}{i}$

Where,${\phi }_B$= average flux for N turns of the coil.

${\phi }_B=\frac{Li}{N}$

Now, put the values for L, i, and N to get

$$\begin{gathered} {\phi }_B=\frac{Li}{N} \\ =\frac{\left(0.250H\right)\left(0.720A\right)}{400} \\ =4.50X{10}^{-4}WB \end{gathered}$$

Therefore, the average magnetic flux through each turn is 4.5 x 10-4 webster.