Question

If the rate of change of current is 4 A/s and induces an emf of 20 mV in a solenoid. What is the self-inductance of the solenoid?

Answer 1

The equation of emf in a solenoid is e = L $\dfrac{{di}}{{dt}}$, where L is the self-inductance of the solenoid, e is the emf induced and $\dfrac{{di}}{{dt}}$ is the rate of change of current.

Complete step-by-step solution
The rate of change of current or the change of current with respect to time is given as $\dfrac{{di}}{{dt}}$ = 4 A/s. The emf induced in the solenoid is given as e = 20 mV = $20 \times {10^{ - 3}}$ V.
Self-inductance is the induction of voltage in the circuit due to the changing current in the circuit itself. This changing current leads to a change in the magnetic field of the circuit and this changing magnetic field induces a voltage in the circuit.
We use the equation of emf induced in a solenoid due to self-inductance, and obtain,
e = L $\dfrac{{di}}{{dt}}$ …equation (1)
We now substitute the values of rate of change of current and emf induced in the solenoid due to self-inductance in equation (1). We get,
$20 \times {10^{ - 3}} = L \times 4$
Upon simplifying, we obtain the value of self-inductance in Henry,
L = $\dfrac{{20 \times {{10}^{ - 3}}}}{4} = 5 \times {10^{ - 3}}$ H
This can also be written as L = 5 mH.

Hence, we obtain that the value of the self-inductance of the solenoid is mH or 5 $\times {10^{ - 3}}$ H.

Note The self-inductance of a solenoid depends on a number of factors:
1. Number of turns in the solenoid
2. Radius of the solenoid
3. Length of the solenoid
4. Relative permeability of the material in the solenoid
5. The self-inductance of a solenoid does not depend upon the current passing through the solenoid.