(a) Line integral of magnetic field over a closed loop is equal to the $\mu_0$ times the total current passing through the surface enclosed by the loop.Alternatively $\oint \vec{B} \cdot \vec{dl} = \mu_0 I$ Let the current flowing through each turn of the toroid be I. The total number of turns equals $n$ . $(2 \pi r)$ where $n$ is the number of turns per unit length. Applying Ampere's circuital law, for the Amperian loop, for interior points. $\oint \vec{B} \cdot \vec{dl} = \mu_0 (n 2 \pi r l)$ $B dl \cos 0 = \mu_0 n 2 \pi r l$ $\Rightarrow \; \; B \times 2\pi r \; = \mu_0 n 2\pi r l$ $B \; = \mu_0 n I$(b) The solenoid contains $N$ loops, each carrying a current I. Therefore, each loop acts as a magnetic dipole. The magnetic moment for a current I, flowing in loop of area (vector) $A$ is given by $m = I A$ The magnetic moments of all loops are alignedalong the same direction.Hence, net magnetic moment equals NIAOR(a) $\phi = M I$Mutual inductance of two coils is equal to the magnetic flux linked with one coil when a unit current is passed in the other coil. Alternatively, $e = -M \; \frac{dI}{dt}$Mutual inductance is equal to the induced emf set up in one coil when the rate of change of current flowing through the other coil is unity.SI unit: henry / (Weber ampere ${\ }^{-1}$ ) / (volt second ampere $^{-1}$ )(b) Let a current $I_2$ flow through $S_2$ . This sets up a magnetic flux $\phi_1$ through each turn of the coil $S_1$ . Total flux linked with $S_1$ $N_1 \phi_1 = M_{12} I_2$ ......(i)where $M_{12}$ is the mutual inductance between the two solenoidsMagnetic field due to the current $I_2$ in $S_2$ is $\mu_0 n_2 l_2$ . Therefore, resulting flux linked with $S_1$. $N_1 \phi_1 = [ (n_1 \ell) \pi r_1^2 ] (\mu_0 n_2 I_2)$ ......(ii)Comparing (i) & (ii), we get $M_{12} I_2 \; = (n_1 \ell) \pi r_1^2 (\mu_0 n_2 I_2)$ ∴ $M_{12} \; = \mu_0 n_1 n_2 \pi r_1^2 \ell$(c) Let a magnetic flux be $(\phi_1)$ linked with coil $C_1$ due to current $(I_2)$ in coil $C_2$ :We have: $\phi_1 \; \propto I_2$ ⇒ $\phi_1 \; = M I_2$∴ $\; \frac{d\phi_1}{dt} \; = M \; \frac{dI_2}{dt}$⇒ $e \; = -M \; \frac{dI_2}{dt}$