91Ó°ÊÓ

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,

$M=\frac{{\mathrm{Ï•}}_{B2}}{I_1}{N}_2$where ${\mathrm{Ï•}}_{B2}$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,

$B_1=\frac{{\mathrm{μ}}_o{N}_1{I}_1}{2\mathrm{π}r}$

$N_2{\mathrm{Ï•}}_{B2}=N_2{B}_{1}A$

$N_2{\mathrm{ϕ}}_{B2}=\frac{N_2{\mathrm{μ}}_0{N}_1{I}_{1}A}{2\mathrm{π}r}$

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

$M=\frac{{\mathrm{μ}}_o{N}_1{N}_{2}A}{2\mathrm{π}r}$

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$M=\frac{{\mathrm{μ}}_o{N}_1{N}_{2}A}{2\mathrm{π}r}$;(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$M=2.40×{10}^{-5}H$