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.