Question

In a tangent galvanometer, the magnetic induction produced by the coil of wire situated in the magnetic meridian is found to be equal to the horizontal component of the earth’s magnetic field. The deflection produced in it will be:
A. ${{30}^{0}}$
B. ${{60}^{0}}$
C. ${{45}^{0}}$
D. ${{90}^{0}}$

Answer 1

Hint: This problem can be solved by applying the relation between the magnetic induction produced by the coil of wire in the magnetic meridian, the horizontal component of the earth’s magnetic field and the deflection of the tangent galvanometer.
The tangent galvanometer produces a deflection depending upon the strength of the magnetic field (magnetic induction) produced by it in comparison to the horizontal component of the earth’s magnetic field in that meridian.
Formula used:
$B={{B}_{H}}\tan \theta $
Where B is the magnetic induction produced by the tangent galvanometer in that meridian, ${{B}_{H}}$ is the horizontal component of the earth’s magnetic field in that meridian, $\theta $ is the deflection of the tangent galvanometer.

Complete Step-by-Step solution:
The tangent galvanometer produces a deflection depending upon the strength of the magnetic field (magnetic induction) produced by it in comparison to the horizontal component of the earth’s magnetic field in that meridian.
First we will analyze the information given to us.
Magnetic induction produced by the tangent galvanometer is B.
Horizontal component of the earth’s magnetic field in that meridian is ${{B}_{H}}$.
We are required to find out the deflection of the tangent galvanometer $\left( \theta \right)$.
Now,
$B={{B}_{H}}\tan \theta $
$\tan \theta =\dfrac{B}{{{B}_{H}}}$
$\theta ={{\tan }^{-1}}\left( \dfrac{B}{{{B}_{H}}} \right)$ ------------------(1)
Now, it is given to us that the magnetic induction produced by the coil of wire situated in the magnetic meridian is found to be equal to the horizontal component of the earth’s magnetic field.
$\therefore B={{B}_{H}}$
$\therefore \dfrac{B}{{{B}_{H}}}=1$ ---------------(2)
Now, putting (2) in (1), we get,
$\theta ={{\tan }^{-1}}\left( 1 \right)={{45}^{0}}$ ------------------$\left( \because {{\tan }^{-1}}1={{45}^{0}} \right)$
Therefore, the deflection produced in the tangent galvanometer is ${{45}^{0}}$.
Hence, the correct option is C) ${{45}^{0}}$.

Note: The tangent galvanometer gives a good feel for the measure of the horizontal component of the earth’s magnetic field at the point. At the magnetic poles, the horizontal component is zero and thus, the tangent galvanometer remains perfectly vertical. The more towards the horizontal the deflection of the tangent galvanometer, the more is the horizontal component of the magnetic field at that point.