A 10.0cm long solenoid of diameter 0.400 cm is wound uniformly with 800 turns. A second coil with 50 turns is wound around the solenoid at its center. What is the mutual inductance of the combination of the two coils?

When changing current in one circuit changes the magnetic flux in other circuit, an emf is induced in the latter circuit, this phenomena is called mutual induction. Thee mf is directly proportional to the time rate of current in former circuit. The constant so introduced removing the proportionality is called mutual inductance.

Mutual inductance is a measure of the mutual induction between two magnetically linked circuits, given as the ratio of the induced emf to the rate of change of current producing it.

It is expressed as,

$M=\frac{{\varphi }_{B2}}{I_1}{N}_2$

Whereis magnetic flux linked with second coil having turns when current is flowing in first coil.

Mutual inductance between two coils depend on area of cross-section, number of turns in each coil, Space between the two coils, Permeability of medium between the two coils and Length (in case of the solenoid).

Consider two coaxial solenoids of which the outer solenoid has radius and turns whereas the inner solenoid has radius andturns. Both the solenoids are of equal length.

When there is a current in the solenoid, the magnetic induction (field) due to is given by,

$$\begin{gathered} B_2={\mu }_o{n}_2{I}_2 \\ {B}_2={\mu }_o\frac{N_2}{L}{I}_2 \end{gathered}$$

The corresponding flux linkage with solenoid is,

$$\begin{gathered} N_1{\varphi }_{B_1}=N_1{B}_2{A}_1 \\ {N}_1{\varphi }_{B_1}=N_1{\mu }_0\frac{N_2}{L}{I}_2\pi {r_1}^2 \end{gathered}$$

As, $N_1{\varphi }_{B_1}=M{I}_2$compare it with above equation to get,

$M=\frac{{\mu }_0{N}_1{N}_2\pi {r_1}^2}{L}$

A long solenoid of diameteris wound uniformly with turns. A second coil with is wound around the solenoid at its center, mutual inductance of the combination is calculated as

$M=\frac{{\mu }_0{N}_1{N}_2\pi {r_1}^2}{L}$

$$\begin{gathered} M=\frac{\left(4\pi \times {10}^{-7}\right)\times 800\times 50\times \pi \times {\left(0.200\times {10}^{-2}\right)}^2}{0.100} \\ M=6.32\times {10}^{-6}H \end{gathered}$$

Therefore,mutual inductance of the combination of two coils is$6.32\times {10}^{-6}H$.