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:

Answer 1

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

is given by :

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

A = Area of coil. L = length]

and mutual inductance (M) between A & B given by :

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

so from above,

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

$\Rightarrow M=8\times {10}^{-5}H.$