Question

A tangent galvanometer has a coil with 50 turns and radius equal to 4 cm. A current of 0.1 A is passing through it. The plane of the coil is set parallel to the earth’s magnetic meridian. If the value of the earth’s horizontal component of the magnetic field is \[7\times {{10}^{-5}}Tesla\] and\[{{\mu }_{0}}=4\pi \times {{10}^{-7}}Wb{{A}^{-1}}{{m}^{-1}}\], the deflection in the galvanometer needle will be:
\[\begin{align}
  & \text{A}\text{. }45{}^\circ \\
 & \text{B}\text{. 48}\text{.3 }\!\!{}^\circ\!\!\text{ } \\
 & \text{C}\text{. 50}\text{.7 }\!\!{}^\circ\!\!\text{ } \\
 & \text{D}\text{. }52.7{}^\circ \\
\end{align}\]

Answer 1

Hint: We have to find deflection of galvanometer needle when current is given for the tangent galvanometer. By using formula for the tangent galvanometer we can find the deflection of the galvanometer needle. The permeability in free space is also given. Further, we can discuss briefly about the tangent galvanometer.

Formula used:
\[I=\dfrac{2r{{B}_{H}}\tan \theta }{{{\mu }_{0}}n}\]

Complete answer:
Tangent galvanometer is used to measure electric current with the help of a compass needle. It basically compares the magnetic field produced by an unknown electric current with the earth’s magnetic field. Its name is given on the basis of its operation as works by the tangent law of magnetism. The magnetic field of the coil is given by
\[B=\dfrac{{{\mu }_{0}}nI}{2r}\]
Where n is number of turns, I is the current flowing through the coil, r is the radius of the coil and \[{{\mu }_{0}}\]is the permeability in free space.
From, the tangent law
\[B={{B}_{H}}\tan \theta \]
Where \[{{B}_{H}}\] is the earth’s horizontal component of the magnetic field. By substituting the value of B in the above tangent law we can field the current passing through coil. Hence current for the tangent galvanometer is given as
\[I=\dfrac{2r{{B}_{H}}\tan \theta }{{{\mu }_{0}}n}\]
Here we have to find deflection in the galvanometer needle which is given as
\[\theta ={{\tan }^{-1}}\left( \dfrac{I{{\mu }_{0}}n}{2r{{B}_{H}}} \right)\]
Given values are \[I=0.1A,n=50,r=4cm,{{B}_{H}}=7\times {{10}^{-5}}Tesla,{{\mu }_{0}}=4\pi \times {{10}^{-7}}Wb{{A}^{-1}}{{m}^{-1}}\]
Substituting all value in the above equation we get
\[\theta ={{\tan }^{-1}}\left( \dfrac{(0.1)\times 4\pi \times {{10}^{-7}}\times 50}{2\times 4\times {{10}^{-2}}\times 7\times {{10}^{-5}}} \right)\]
Substituting \[\pi =3.14\]
\[\begin{align}
  & \theta ={{\tan }^{-1}}\left( \dfrac{(0.1)\times 4\times 3.14\times {{10}^{-7}}\times 50}{2\times 4\times {{10}^{-2}}\times 7\times {{10}^{-5}}} \right) \\
 & \theta ={{\tan }^{-1}}\left( \dfrac{62.8\times {{10}^{-7}}}{56\times {{10}^{-7}}} \right) \\
 & \theta ={{\tan }^{-1}}\left( 1.121 \right) \\
 & \theta =48.26{}^\circ \simeq \text{48}\text{.3 }\!\!{}^\circ\!\!\text{ } \\
\end{align}\]
So the angle made by the galvanometer needle is \[\text{48}\text{.3 }\!\!{}^\circ\!\!\text{ }\].

So, the correct answer is “Option B”.

Note:
While calculating, always make sure that all units are the same, if not change it. Like, here the radius was given in centimetre, while substituting we changed its unit in meters. If the unit was not changed we would have got the wrong answer. These are the very small things where we can make mistakes.