Question

The average self-induced emf in a 25mH solenoid when the current in it falls from 0.2A to 0A in 0.01 second, is
A. 0.05V
B. 0.5V
C. 500V
D. 50V

Answer 1

Hint: As we all know that the self-induced emf is the result of the change in the current over the period of the time. And this will also be dependent on the value of the inductance of the material. So here the self-induced emf will be dependent on the inductance value of the solenoid and the change in the current in the particular given period of time.

Complete answer:
There is a basic and very fundamental formula for the induced emf in any conductor,
So, the induced emf= $e = L\dfrac{{di}}{{dt}}$-------equation (1)
Here in the question we have to identify the quantitates given keeping in mind the above formula,
So, we have given the value of the inductance of the solenoid, that is, $L = 25 \times {10^{ - 3}}H$.
Change in the current, that is, $di = (0.2 - 0)A = 0.2A$
Change in the time, that is, $dt = 0.01\sec $
So now putting all the values in the equation (1), we get

$e = 25 \times {10^{ - 3}} \times \left( {\dfrac{{0.2}}{{0.01}}} \right)$
$ \Rightarrow e = 0.5V$
So, the average self-induced emf in a 25mH solenoid when the current in it falls from 0.2A to 0A in 0.01 second, is $e = 0.5V$.

So, the correct answer is “Option B”.

Note:
Here we have discussed the solenoid. Solenoid is basically a generic term used for the coil of wire which acts like an electromagnet. This term solenoid is also used to refer to devices that convert electrical energy to mechanical energy. This device creates a magnetic field with the help of the electric current and then uses this magnetic field to create the mechanical motion. Hence, solenoid is basically a typical and simple form of an electromagnet.