Question

The deflection of the magnetic needle in a tangent galvanometer is 30$^{0}$ when a current of one ampere flows through it. The deflection of the magnetometer when a current of 4A flows through it is:
$\text{A}\text{. }{{\tan }^{-1}}(8.7)$
$\text{B}\text{. }{{\tan }^{-1}}(2.31)$
$\text{C}\text{. }{{\tan }^{-1}}(1.73)$
$\text{D}\text{. }{{\tan }^{-1}}(\sqrt{2})$

Answer 1

Hint: When the current flows the galvanometer, the coils produces a magnetic field $B=\dfrac{{{\mu }_{0}}Ni}{2r}$. Due to this field and earth’s horizontal magnetic field, the magnet acquires an equilibrium state where, $B={{B}_{H}}\tan \theta$. Use these two formulas to find the deflection of the needle when the current is 4A.

Formula used:
$B={{B}_{H}}\tan \theta$
$B=\dfrac{{{\mu }_{0}}Ni}{2r}$

Complete step by step answer:
A tangent galvanometer is a device, which detects small currents. It is also called a moving magnet galvanometer.
It consists of circular coils of insulated copper wire that are wound on a vertical circular frame. The circular frame is made up of a non-magnetic material. At the centre of this vertical frame a small magnetic compass needle is pivoted.
This needle is placed inside a box made up of a non-magnetic material. The needle rotates freely in a horizontal plane. When a current flows through the coil, it produces a magnetic field (B). Now the magnetic needle is influenced by two perpendicular magnetic fields. One the earth’s horizontal magnetic field (${{B}_{H}}$) and the other is the magnetic field B produced due to the current in the coil.
Due to this the needle comes in an equilibrium state at angle $\theta$, where $B={{B}_{H}}\tan \theta$ …. (i)
The magnetic field produced by the coil is given as $B=\dfrac{{{\mu }_{0}}Ni}{2r}$
Here, N are the number of turns in the coil, i is the current in the coil and r is the radius of the coil.
Substitute the value of B in equation (i).
$\dfrac{{{\mu }_{0}}Ni}{2r}={{B}_{H}}\tan \theta$ ….. (ii)
Now there are two cases given in the question.
In one case, the current is 1A and the angle $\theta ={{30}^{0}}$.
This given us that
$\dfrac{{{\mu }_{0}}N(1)}{2r}={{B}_{H}}\tan ({{30}^{0}})$ ….. (iii).
In the second case, N and r are the same but the current is 4A. Let the angle be $\theta$.
Hence, $\dfrac{{{\mu }_{0}}N(4)}{2r}={{B}_{H}}\tan (\theta )$ …… (iv).
Divide equations (iii) and (iv).
$\Rightarrow \dfrac{\dfrac{{{\mu }_{0}}N(1)}{2r}}{\dfrac{{{\mu }_{0}}N(4)}{2r}}=\dfrac{{{B}_{H}}\tan ({{30}^{0}})}{{{B}_{H}}\tan (\theta )}$
$\Rightarrow \dfrac{1}{4}=\dfrac{\tan ({{30}^{0}})}{\tan (\theta )}$
$\Rightarrow \dfrac{1}{4}=\dfrac{1}{\sqrt{3}\tan (\theta )}$
$\Rightarrow \tan (\theta )=\dfrac{4}{\sqrt{3}}=2.31$
$\Rightarrow \theta ={{\tan }^{-1}}(2.31)$
Hence, the correct option is B.

Note: From equation (ii) we get that the current in the galvanometer is directly proportional to tangent of the angle of deflection, i.e. $i\propto \tan \theta$.
$\tan \theta$ is not linear with the angle of deflection. Hence, the scale on the galvanometer is not uniform.