Question

If the coefficient of mutual induction of the primary and secondary coils of an induction coil is $5H$ and a current of $10A$ is cut-off in $5 \times {10^{ - 4}}s$ , the emf inducted (in volt) in the secondary coil is
(A) $5 \times {10^4}$
(B) $1 \times {10^5}$
(C) $25 \times {10^5}$
(D) $5 \times {10^6}$

Answer 1

Hint: In order to solve this question, we will use the general relation between mutual induction between the coils, rate of change of current, and the induced emf to calculate the induced emf in the coil. When the change in current in one coil induces a voltage in its nearby coil this phenomenon is called mutual induction.

Formula used:
If M is the mutual induction between the coils and di is the change in current and dt is the change in time then the relation is $e=M\left(\dfrac {di}{dt} \right)$ where ‘e’ is the emf of the coil.

Complete answer:
According to the question, we have given that mutual induction between the primary and secondary coil is $M = 5H$ and the change in current is given by $di = 10A$ and the change in time is given by $dt = 5 \times {10^{ - 4}}s$ and let us assume ‘e’ be the emf induced in the coil then using the formula $e=M\left(\dfrac {di}{dt} \right)$ and putting the values of parameters we get,
$
  e = 5(\dfrac{{10}}{{5 \times {{10}^{ - 4}}}}) \\
  e = 1 \times {10^5}V \\
 $
So, The value of emf induced in the coil is $1 \times {10^5}V$

Hence, the correct answer is option (B)

Note: It should be remembered that the induced emf has the negative sign to represent that the induced emf will flow in a direction opposite to its origin here we have calculated the magnitude of the emf so, the negative sign is ignored and the units of emf is voltage represented by V.