Question

The coils A and B having turns $300$and $600$respectively are placed near each other, on passing a current of $3.0$ampere on A, the flux linked with A is $1.2\times {10}^{-4}$ weber and with B it is $9.0\times {10}^{-5}$ weber. The mutual induction of the system is?

• A. $2\times {10}^{-5}H$
• B. $3\times {10}^{-5}H$
• C. $4\times {10}^{-5}H$
• D. $6\times {10}^{-5}H$

Answer 1

The correct option is B

$3\times {10}^{-5}H$

Step 1: Given data

For coil A,

Current on it, $i_1=3A$

Flux linked with it, ${\varphi }_1=1.2\times {10}^{-4}weber$

For coil B,

Flux linked with it, ${\varphi }_2=9.0\times {10}^{-5}weber$

Step 2: Formula used

Flux induced (${\varphi }_2$) in the secondary coil B is directly proportional to the current ($i_1$) in the primary coil, and the relation can be given by,

${\varphi }_2=M{i}_1$

where $M$ is the mutual induction of the system.

Step 3: Calculate the mutual inductance of the system.

So,

$\begin{array}{rcl}M& =& \frac{{\varphi }_2}{i_1}\\ & =& \frac{9.0\times {10}^{-5}}{3}\\ & =& 3\times {10}^{-5}H\end{array}$

Hence, option (B) is the correct answer.