Question

b) What rate of change of current in a solenoid having self-inductance 19.4 mH
produces a self-induced emf of 100 mV in it?​

Answer 1

Given :

Self-inductance of the solenoid = 19.4mH = 0.0194H

Self induced emf of the solenoid = 100mV

To Find :

The rate of change of current in the solenoid = ?

Solution :

• For a solenoid the emf is given by the formula

                $E_{s}= -L\frac{dI}{dt}$

• By substituting the given values in the formula, we get

                $\frac{dI}{dt}=\frac{E_{s} }{L}$

                $\frac{dI}{dt}=\frac{0.1}{0.0194}$

                $\frac{dI}{dt}=5.15A/s$

The rate of change of current in the solenoid is 5.15A/s.