A straight solenoid has 60 turns per cm in primary and 200 turns in the secondary. The area of cross-section of the solenoide is $4c{m}^2$ Calculate the mutual inductance. Primary is tightly kept inside the secondary.

The magnetic field at any point inside the straight solenoid of primary wi per unit length carrying a current , is given by the relation,

$B={\mu }_0{n}_1{i}_1$

The magnetic flux through the secondary of $N_2$ turns each of area $S$ is given as

$N_2\varphi =N_2\left(BS\right)={\mu }_0{n}_1{N}_2{i}_{1}S$

$M=\frac{N_2{\varphi }_2}{i_1}={\mu }_0{n}_1{N}_{2}S$

Substituting the values, we get

$$\begin{gathered} M=(4\mathrm{\pi }\times {10}^{-7})\left(\frac{50}{{10}^{-2}}\right)\left(200\right)(4\times {10}^{-4}) \\ =5.0\times {10}^{-4}\mathrm{H} \end{gathered}$$