question_answer

A magnetic compass needle oscillates 30 times per minute at a place where the dip is \[45{}^\text{o},\]and 40 times per minute where the dip is \[30{}^\text{o}\]. If \[{{B}_{1}}\]and \[{{B}_{2}}\]are respectively the total magnetic field due to the earth at the two places, then the ratio \[{{B}_{1}}/{{B}_{2}}\]is best given by:               [JEE Main Held on 12-4-2019 Morning]

A) 2.2                              

B) 1.8

C) 0.7      

D) 3.6

Correct Answer: C

Solution :

\[{{f}_{1}}=\frac{1}{2\pi }\sqrt{\frac{\mu {{B}_{1}}\cos {{45}^{o}}}{I}}{{f}_{2}}=\frac{1}{2\pi }\sqrt{\frac{\mu {{B}_{2}}\cos {{30}^{o}}}{I}}\] \[\frac{{{f}_{1}}}{{{f}_{2}}}=\sqrt{\frac{{{B}_{1}}\cos {{45}^{o}}}{{{B}_{2}}\cos {{30}^{o}}}}\]\[\therefore \]\[\frac{{{B}_{1}}}{{{B}_{2}}}0.7\]