A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N1 turns. A second toroidal solenoid with N2 turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius. (a) What is the mutual inductance of the two solenoids? Assume that the magnetic field of the first solenoid is uniform across the cross section of the two solenoids. (b) if${\mathit{N}}_{\mathbf{1}}\mathbf{=}\mathbf{500}$,${\mathit{N}}_{\mathbf{2}}\mathbf{=}\mathbf{300}$,$\mathit{r}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$and$\mathit{A}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{800}\mathit{c}{\mathit{m}}^{\mathbf{2}}$, what is the value of the mutual inductance?

Mutual inductance is a measure of the mutual inductionbetween 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,

$M=\frac{{\varphi }_{B2}}{I_1}{N}_2$where ${\varphi }_{B2}$is magnetic flux linked with second coil having turns $N_2$ when current $I_1$is flowing in first coil.

A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with $N_1$turns. A second toroidal solenoid with $N_2$turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius. Assume that the magnetic field of the first solenoid is uniform across the cross section of the two solenoids.

When there is current $I_1$in toroidal solenoid $S_1$ with mean radius r and cross-sectional area A is wound uniformly with $N_1$turns, then magnetic induction is given by,

$B_1=\frac{{\mu }_o{N}_1{I}_1}{2\pi r}$

$N_2{\varphi }_{B2}=N_2{B}_{1}A$

$N_2{\varphi }_{B2}=\frac{N_2{\mu }_0{N}_1{I}_{1}A}{2\pi r}$

As $N_2{\varphi }_{B2}=M{I}_1$, compare it with above equation to get,

$M=\frac{{\mu }_o{N}_1{N}_{2}A}{2\pi r}$

If $N_1=500$, $N_2=300$, $r=10.0cm$and$A=0.800c{m}^2$then mutual inductance is calculated as,

Therefore,(a) the mutual inductance of the two solenoids is$M=\frac{{\mu }_o{N}_1{N}_{2}A}{2\pi r}$;(b) If N1 = 500 turns, N2 = 300 turns, r = 10.0 cm, and A = 0.800 cm2 then the value of the mutual inductance is$M=2.40\times {10}^{-5}H$