Principle and working : A current carrying coil, placed in a uniform magnetic field, can experience a torque.

Consider a rectangular coil for which no. of turns $={\mathit{N}}_{\mathit{i}}$
Area of cross-section $=\mathit{l}\times \mathit{b}=\mathit{A}$,
Intensity of the uniform magnetic field $=\mathit{B}$,
Current through the coil $=\mathit{I}$
$\therefore$ Deflecting torque $=\mathit{B}\mathit{I}\mathit{l}\times \mathit{b}=\mathit{B}\mathit{I}\mathit{A}$
For $\mathrm{N}$ turns, $\mathit{\tau }=\mathit{N}\mathit{B}\mathit{I}\mathit{A}$
Restoring torque in the spring $=\mathit{k}\mathit{\theta }$
( $\mathit{k}=$ restoring torque per unit twist)
$\therefore \mathit{N}\mathit{B}\mathit{L}\mathit{A}=\mathit{k}\mathit{\theta }$
$\therefore \mathit{I}=\left(\frac{\mathit{k}}{\mathit{N}\mathit{B}\mathit{A}}\right)\mathit{\theta }$
$\therefore \mathit{I}\propto \mathit{\theta }$
The deflection of the coil is therefore, proportional to the current flowing through it.

(i) Need for a radial magnetic field:
The relation between the current (i) flowing through the galvanometer coil, and the angular deflection $\left(\mathit{\phi }\right)$ of the coil (from its equilibrium position), is
$\mathit{\phi }=\left(\frac{\mathit{N}\mathit{A}\mathit{B}\mathit{I}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }}{\mathit{k}}\right)$
where $\mathit{\theta }$ is the angle between the magnetic field $\overrightarrow{\mathit{B}}$ and the equivalent magnetic moment $\overrightarrow{{\mathit{\mu }}_{\mathit{m}}}$ of the current carrying coil.
Thus $\mathit{I}$ is not directly proportional to $\mathit{\varphi }$. We can ensure this proportionality by having $\mathit{\theta }={90}^{\circ }$. This is possible only when the magnetic field $\overrightarrow{\mathit{B}}$, is a radial magnetic field. In such a field, the plane of the rotating coil is always parallel to $\overrightarrow{\mathit{B}}$.
To get a radial magnetic field, the pole pieces of the magnet, are made concave in shape. Also a soft iron cylinder is used as the core.
The soft iron core not only makes the field radial but also increases the strength of the magnetic field.
A galvanometer has low resistance and allow only a very small current. When high current is passed the coil will burn hence galvanometer as such is not used for measuring current.
(ii) We have
Current sensitivity $=\frac{\mathit{\theta }}{\mathrm{I}}=\mathit{N}\mathit{B}\mathit{A}∕\mathit{k}$
OR
(a) Definition of self inductance and its SI unit
(b) Derivation of expression for mutual inductance
Self inductance of a coil equals the magnitude of the magnetic flux, linked with it, when a unit current flows through it.
Alternatively
Self inductance of a coil, equals the magnitude of the emf induced in it, when the current in the coil, is changing at a unit rate.
SI unit: henry / (weber/ampere) / (ohm second.)

When current ${\mathit{I}}_2$ is passed through coil , it in turn sets up a magnetic flux through ${\mathrm{S}}_1$ :
$\mathit{\varphi }={\mathit{n}}_1\times {\mathit{\mu }}_0\frac{{\mathit{n}}_2}{\mathit{l}}\times {\mathit{I}}_2\times \mathit{\pi }{\mathit{r}}_1^2$
$=\left({\mathit{\mu }}_0\frac{{\mathit{n}}_1{\mathit{n}}_2}{\mathit{l}}\mathit{\pi }{\mathit{r}}_1^2\right){\mathit{I}}_2={\mathit{M}}_{12}{\mathit{I}}_2$

[Note : If the student derives the correct expression, without giving the diagram of two coaxial coils, full credit can be given]