Q. A magnet oscillating in a horizontal plane has a time period of $2\ second$ at a place where the angle of dip is $30{^\circ}$ and $3\ seconds$ at another place where the angle of dip is $60{^\circ}$. The ratio of resultant magnetic fields at the two places is
A) $\frac{4\sqrt{3}}{7}$
B) $\frac{4}{9\sqrt{3}}$
C) $\frac{9}{4\sqrt{3}}$
D) $\frac{9}{\sqrt{3}}$
Explanation
(c) $T \propto \frac{1}{\sqrt{B_{H}}} \propto \frac{1}{\sqrt{B\cos\phi}} \Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{B_{2}\cos\phi_{2}}{B_{1}\cos\phi_{1}}}$
$\Rightarrow \frac{B_{1}}{B_{2}} = \frac{T_{2}^{2}}{T_{1}^{2}} \times \frac{\cos\phi_{2}}{\cos\phi_{1}} = \left( \frac{3}{2} \right)^{2} \times \left( \frac{\cos{60{^\circ}}}{\cos{30{^\circ}}} \right) \Rightarrow \frac{B_{1}}{B_{2}} = \frac{9}{4\sqrt{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.