Question

The sensitiveness of tangent galvanometer will be maximum if deflection in it is tending to
(A) $0^\circ$
(B) $30^\circ$
(C) $45^\circ$
(D) $60^\circ$

Answer 1

Hint
A tangent galvanometer will be sensitive if it gives a large deflection for a fractional change in current. Now we can use the formula $I = K\tan \theta$ to find the deflection $d\theta$ and find the condition which gives the maximum deflection.
In this problem, we will be using the following formulas,
$I = K\tan \theta$ where $K$ is given by, $K = \dfrac{{{B_H}}}{{{G_1}}}$ and ${G_1} = \dfrac{{n{\mu _o}}}{{2a}}$
Here, ${B_H} =$ the horizontal component of the magnetic field,
${\mu _o} =$ the permeability of free space = $4\pi \times {10^{ - 7}}N/{A^2}$
$n =$ the number of turns of the coil inside the galvanometer.
$a =$ the radius of the coil.
$\theta =$ the angle of deflection.

Complete Step by Step Solution
Let us consider a magnetic field $B$ produced in the coil due to the current $I$passing through it. So the horizontal component of the magnetic field is ${B_H}$.
Now according to the tangent law,
$B = {B_H}\tan \theta$
And we know that for a current-carrying coil having $n$ number of loops, the magnetic field produced on the plane perpendicular to the coil is given by
$B = \dfrac{{n{\mu _o}I}}{{2a}}$ where $I$ is the current in the coil.
Therefore we can write,
$\dfrac{{n{\mu _o}I}}{{2a}} = {B_H}\tan \theta$
$\Rightarrow I = \dfrac{{2a}}{{n{\mu _o}}}{B_H}\tan \theta$
We can take $K = \dfrac{{{B_H}}}{{{G_1}}}$ where ${G_1} = \dfrac{{n{\mu _o}}}{{2a}}$
Therefore,
$I = K\tan \theta$
Now, differentiating on both sides, we get
$dI = K{\sec ^2}\theta d\theta$
Next, we divide both the sides by $I$ .
$\dfrac{{dI}}{I} = \dfrac{{K{{\sec }^2}\theta d\theta }}{I}$
$\Rightarrow \dfrac{{dI}}{I} = \dfrac{{K{{\sec }^2}\theta d\theta }}{{K\tan \theta }} = \dfrac{{K\cos \theta d\theta }}{{K\sin \theta {{\cos }^2}\theta }}$
Cancelling $K$ and $\cos \theta$ from the numerator and the denominator of the R.H.S of the equation, we get
$\dfrac{{dI}}{I} = \dfrac{{d\theta }}{{\sin \theta \cos \theta }}$
By multiplying 2 on numerator and denominator we get
$\Rightarrow \dfrac{{dI}}{I} = \dfrac{{2d\theta }}{{2\sin \theta \cos \theta }}$
We know that $2\sin \theta \cos \theta = \sin 2\theta$
So,
$\therefore \dfrac{{dI}}{I} = \dfrac{{2d\theta }}{{\sin 2\theta }}$
$\Rightarrow d\theta = \dfrac{{\sin 2\theta }}{2}\dfrac{{dI}}{I}$
From this equation, we can see that the value of $d\theta$ will be maximum when the value of $\sin 2\theta$ will be equal to 1 as $- 1 < \sin 2\theta < 1$ .
$\therefore \sin 2\theta = 1$
$\Rightarrow 2\theta = \dfrac{\pi }{2}$
From here we get,
$\therefore \theta = \dfrac{\pi }{4} = 45^\circ$
So, a tangent galvanometer will be most sensitive when the deflection angle $\theta = 45^\circ$ .
Therefore, the correct option is (C).

Note
We should be careful that we don’t confuse a moving coil galvanometer with a tangent galvanometer. A tangent galvanometer is based on the principle of the tangent law of magnetism.