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 $=N_i$ Area of cross-section $=l \times b = A$, Intensity of the uniform magnetic field $=B$, Current through the coil $=I$ $\therefore$ Deflecting torque $=B I l \times b = B I A$ For $N$ turns, $\;\; \tau = N B I A$ Restoring torque in the spring $=k \theta$ ( $k=$ restoring torque per unit twist) $\therefore \; N B L A = k \theta$ $\therefore \; I \; = \left( \frac{k}{N B A} \right) \theta$ $\; \therefore \; I \propto \theta$ 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 $(\phi)$ of the coil (from its equilibrium position), is $\phi = \left( \frac{N A B I \sin \theta}{k} \right)$ where $\theta$ is the angle between the magnetic field $\overset{\rightarrow}{B}$ and the equivalent magnetic moment $\overset{\rightarrow}{\mu_m}$ of the current carrying coil. Thus $I$ is not directly proportional to $\phi$. We can ensure this proportionality by having $\theta = 90^{\circ}$. This is possible only when the magnetic field $\overset{\rightarrow}{B}$, is a radial magnetic field. In such a field, the plane of the rotating coil is always parallel to $\overset{\rightarrow}{B}$. 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 $= \frac{\theta}{I} = \frac{N B A}{k}$ 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 $I_2$ is passed through coil , it in turn sets up a magnetic flux through $S_1$ :$\phi \; = n_1 \times \mu_0 \; \frac{n_2}{l} \times I_2 \times \pi r_1^2$$\; = \left( \mu_0 \; \frac{n_1 n_2}{l} \pi r_1^2 \right) I_2 = M_{12} I_2$$\;\text{where}\; M_{12} \; = \mu_0 \; \frac{n_1 n_2}{l} \pi r_1^2$[Note : If the student derives the correct expression, without giving the diagram of two coaxial coils, full credit can be given]