(a) Line integral of magnetic field over a closed loop is equal to the ${\mathit{\mu }}_0$ times the total current passing through the surface enclosed by the loop.

Alternatively
$\oint \overrightarrow{\mathit{B}}\cdot \overrightarrow{\mathit{d}\mathit{l}}={\mathit{\mu }}_0\mathit{I}$

Let the current flowing through each turn of the toroid be I. The total number of turns equals $\mathit{n}$ . $\left(2\mathit{\pi }\mathit{r}\right)$ where $\mathit{n}$ is the number of turns per unit length. Applying Ampere's circuital law, for the Amperian loop, for interior points.
$\oint \overrightarrow{\mathit{B}}\cdot \overrightarrow{\mathit{d}\mathit{l}}={\mathit{\mu }}_0\left(\mathit{n}2\mathit{\pi }\mathit{r}\mathit{l}\right)$
$\mathit{B}\mathrm{dl}\cos 0={\mathit{\mu }}_0\mathit{n}2\mathit{\pi }\mathit{r}\mathit{l}$
$\Rightarrow \mathit{B}\times 2\mathit{\pi }\mathit{r}={\mathit{\mu }}_0\mathit{n}2\mathit{\pi }\mathit{r}\mathit{l}$
$\mathit{B}={\mathit{\mu }}_0\mathit{n}\mathit{I}$
(b)
The solenoid contains $\mathrm{N}$ 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) $\mathit{A}$ is given by $\mathit{m}=\mathit{I}\mathit{A}$ The magnetic moments of all loops are alignedalong the same direction.
Hence, net magnetic moment equals NIA
OR
(a) $\mathit{\varphi }=\mathit{M}\mathit{I}$
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,
$\mathit{e}=-\mathit{M}\frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{t}}$
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 ${\mathrm{I}}_2$ flow through ${\mathrm{S}}_2$ . This sets up a magnetic flux ${\mathit{\phi }}_1$ through each turn of the coil ${\mathit{S}}_1$ . Total flux linked with ${\mathrm{S}}_1$
${\mathit{N}}_1{\mathit{\varphi }}_1={\mathit{M}}_{12}{\mathit{I}}_2$ ......(i)
where ${\mathit{M}}_{12}$ is the mutual inductance between the two solenoids
Magnetic field due to the current ${\mathrm{I}}_2$ in ${\mathrm{S}}_2$ is ${\mathit{\mu }}_0{\mathit{n}}_2{\mathit{l}}_2$ . Therefore, resulting flux linked with ${\mathrm{S}}_1$.
${\mathrm{N}}_1{\mathit{\varphi }}_1=\left[\left({\mathit{n}}_1\mathit{\ell }\right)\mathit{\pi }{\mathit{r}}_1^2\right]\left({\mathit{\mu }}_0{\mathit{n}}_2{\mathit{I}}_2\right)$ ......(ii)
Comparing (i) & (ii), we get
${\mathrm{M}}_{12}{\mathit{I}}_2=\left({\mathit{n}}_1\mathit{\ell }\right)\mathit{\pi }{\mathit{r}}_1^2\left({\mathit{\mu }}_0{\mathit{n}}_2{\mathit{I}}_2\right)$
∴ ${\mathit{M}}_{12}={\mathit{\mu }}_0{\mathit{n}}_1{\mathit{n}}_2\mathit{\pi }{\mathit{r}}_1^2\mathit{\ell }$
(c) Let a magnetic flux be $\left({\mathit{\varphi }}_1\right)$ linked with coil ${\mathit{C}}_1$ due to current $\left({\mathit{I}}_2\right)$ in coil ${\mathit{C}}_2$ :
We have: ${\mathit{\varphi }}_1\propto {\mathit{I}}_2$
⇒ ${\mathit{\varphi }}_1=\mathit{M}{\mathit{I}}_2$
∴ $\frac{\mathit{d}{\mathit{\varphi }}_1}{\mathit{d}\mathit{t}}=\mathit{M}\frac{\mathit{d}{\mathit{I}}_2}{\mathit{d}\mathit{t}}$
⇒ $\mathit{e}=-\mathit{M}\frac{\mathit{d}{\mathit{I}}_2}{\mathit{d}\mathit{t}}$