question_answer

At a place on earth, horizontal component of earth's magnetic field is \[{{B}_{1}}\] and vertical component of earth's magnetic field is \[{{B}_{2}}\]. If a magnetic needle is kept vertical, in a plane making angle \[\alpha \] with the horizontal component of magnetic field, then square of time period of oscillation of needle when slightly distributed is proportional to

A) \[\frac{1}{\sqrt{{{B}_{1}}\,\cos \,\alpha }}\]

B) \[\frac{1}{\sqrt{{{B}_{2}}}}\]

C) \[\frac{1}{\sqrt{{{({{B}_{1\,}}\cos \,\alpha )}^{2}}+B_{2}^{2}}}\]

D) infinite

Correct Answer: C

Solution :

[c] Resultant magnetic field in the plane, \[B=\sqrt{{{({{B}_{1}}\,\cos \,\alpha )}^{2}}+B_{2}^{2}}\] Time period,                 \[T=2\pi \sqrt{\frac{I}{MB}}\]