Question

The sensitivity of tangent galvanometer is maximum when deflection is
(A) ${45^o}$
(B) ${90^o}$
(C) ${30^o}$
(D) ${60^o}$

Answer 1

Hint The angle of deflection per unit current flowing through the galvanometer is known as the sensitivity of the galvanometer. Therefore consider an elementary current $dI$ through a deflection of $\theta $ and apply the tangent law.

Compete Step by step solution
According to the tangent law,
$B = {B_H}\tan \theta $
where, $B$ is the magnetic field produced inside the coil due to the current $I$ passing through it.
${B_H}$ is the horizontal component of the magnetic field.
For a current- carrying conductor,
$B = \dfrac{{n{\mu _o}I}}{{2a}}$
where, $n$ is the number of loops.
Therefore equating the above two equations we get
$\dfrac{{n{\mu _o}I}}{{2a}} = {B_H}\tan \theta $
$ \Rightarrow I = \dfrac{{2a{B_H}\tan \theta }}{{n{\mu _o}}}$
Let us consider $K = \dfrac{{{B_H}}}{G}$ where $G = \dfrac{{n{\mu _o}}}{{2a}}$
$\therefore I = K\tan \theta $
Let us differentiate both sides.
$\therefore dI = K{\sec ^2}\theta d\theta $
Here $dI$ stands for an elementary change in the current flowing through the galvanometer, and $d\theta $ stands for an elementary change in the angle of deflection, or the angle through which the needle will turn inside the galvanometer.
Therefore we can say that $\dfrac{{d\theta }}{{dI}}$ is the sensitivity of the galvanometer.
Dividing both sides by $I$we get,
$\dfrac{{dI}}{I} = \dfrac{{K{{\sec }^2}\theta d\theta }}{I}$
$ \Rightarrow \dfrac{{dI}}{I} = \dfrac{{K{{\sec }^2}\theta d\theta }}{{K\tan \theta }}$
$ \Rightarrow \dfrac{{dI}}{I} = \dfrac{{d\theta }}{{\sin \theta \cos \theta }} = \dfrac{{2d\theta }}{{2\sin \theta \cos \theta }}$
We know that $\sin 2\theta = 2\sin \theta \cos \theta $
Substituting this value within the above equation we get,
$\dfrac{{dI}}{I} = \dfrac{{2d\theta }}{{\sin 2\theta }}$
$ \Rightarrow d\theta = \dfrac{{\sin 2\theta }}{2}\dfrac{{dI}}{I}$
The maximum sensitivity of the tangent galvanometer means the maximum possible value of $d\theta $.
From the above equation, we can see that $d\theta $ is maximum when $\sin 2\theta $ is maximum.
$\therefore \sin 2\theta = 1$
$ \Rightarrow 2\theta = {90^o}$
$ \Rightarrow \theta = {45^o}$

Therefore, option A. is the correct answer.

Additional Information
For the measurement of small electric currents, tangent galvanometers are used. Hence a tangent galvanometer is very sensitive and accurate. For a fractional change in the value of current, there is a large deflection in the galvanometer. It consists of a coil made from an insulated copper wire wound on a circular non-magnetic frame.

Note A tangent galvanometer is dependent on the law of tangent magnetism, hence it should not be confused with a moving coil galvanometer. A moving coil galvanometer works when it is acted upon by a torque.