Question

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

• A. 8 x ${10}^{-5}$ H
• B. 3 x ${10}^{-5}$ H
• C. 4 x ${10}^{-5}$ H
• D. 6 x ${10}^{-5}$ H

Answer 1

The correct option is A 8 x ${10}^{-5}$ H

In passing current $(I=3A)$ through coil $A$, the flux linked $\left({\varphi }_1\right)$ is given by:

${\varphi }_1={\mu }_0{N}_1^{2}I\left(\frac{A}{L}\right)[N_1=300,givenA=Areaofcoil.L=Length]$

and material inductance $\left(M\right)$ between $A$ & $B$ given by:

$M={\mu }_0{N}_1{N}_2\left(\frac{A}{L}\right)$ (Assuming both coils have same $A$ & $L$)

so from above,

$M=\frac{N_2}{N_1}\frac{{\varphi }_1}{I}=\frac{600}{300}\times \frac{1.2\times {10}^{-4}}{3}$

$\Rightarrow M=8\times {10}^{-5}H$ (Ans)