Q. A magnet is suspended in such a way that it oscillates in the horizontal plane. It makes $20\ oscillations\ per\ minute$ at a place where dip angle is $30{^\circ}$ and $15\ oscillations\ per$ $minute$ at a place where dip angle $60{^\circ}$. The ratio of total earth’s magnetic field at the two places is
A) $3\sqrt{3}\ :8$
B) $16\ :9\sqrt{3}$
C) $4\ :9$
D) $2\sqrt{3}\ :9$
Explanation
(b) Given, $v_{1} = \frac{20}{60} = \frac{1}{3}\sec^{- 1}{and}v_{2} = \frac{15}{60} = \frac{1}{4}\sec^{- 1}\ $
Now $v = \frac{1}{2\pi}\sqrt{\frac{MB_{H}}{I}} = \frac{1}{2\pi}\sqrt{\frac{MB\cos\phi}{I}}\ \left\lbrack \because B_{H} = B\cos\phi \right\rbrack$
$\therefore\frac{v_{1}}{v_{2}} = \sqrt{\frac{B_{1}\cos\phi}{B_{2}\cos\phi}} \Rightarrow \frac{B_{1}}{B_{2}} = \left( \frac{v_{1}}{v_{2}} \right)^{2}\left( \frac{\cos\phi_{2}}{\cos\phi_{1}} \right)^{2}$
$\Rightarrow \frac{B_{1}}{B_{2}} = \left( \frac{1/3}{1/4} \right)^{2}\frac{\cos{60{^\circ}}}{\cos{30{^\circ}}} = \frac{16}{9} \times \frac{1/2}{\sqrt{3}/2} = \frac{16}{9\sqrt{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.