Question

(a) Obtain the expression for the torque on a loop placed in a uniform magnetic field.

(b) Describe the construction and working of a moving coil galvanometer.

Answer 1

a) Plane of the loop i at an angle with the direction of the magnetic field. Let $\theta$ be the angle between normal and the field. The forces on BC and DA are equal and opposite and they cancel each other.

Force on AB is $F_1$ and the force on CD is $F_2$

$F_1=F_2=IbB$

Magnitude of the torque is shown in the figure

$\tau =F_1\frac{a}{2}sin\theta +F_2\frac{a}{2}sin\theta$

$=IbB\frac{a}{2}sin\theta +IbB\frac{a}{2}sin\theta$

$=I\left(ab\right)Bsin\theta$

$=IABsin\theta$ ; where $A=ab$

If there are 'n' such turns then there torque will be $nIABsin\theta$

Magnetic moment of the current, $m=IA$

$\overrightarrow{\tau }$= $\overrightarrow{m}\times \overrightarrow{B}$

b) Principle: MCG is based on the principle that a current carrying coil placed in uniform magnetic field experiences a torque.

Construction:

The galvanometer consists of a coil, with many turns, free to rotate about a fixed axis (Fig. b), in a uniform radial magnetic field. There is a cylindrical soft iron core which not only makes the field radial but also increases the strength of the magnetic field. When a current flows through the coil, a torque acts on it.

Working: torque acting on the coil is given by:

$\tau =NIAB$

Since the field is radial by design, we have taken $sin\theta =1$ in the above expression for the torque.

The magnetic torque $NIAB$ tends to rotate the coil. A spring $Sp$ provides a counter torque $k\phi$ that balances the magnetic torque $NIAB$; resulting in a steady angular deflection $\phi$. In equilibrium,

$k\phi =NIAB$

where $k$ is the torsional constant of the spring; i.e. the restoring torque per unit twist. The deflection $\phi$ is indicated on the scale by a pointer attached to the spring.

$\varphi =\frac{NABI}{k}$