Q. Two short magnets have equal pole strengths but one is twice as long as other. The shorter magnet is placed 20 cm in tan A position from the compass needle. The longer magnet must be placed on the other side of the magnetometer for no deflection at a distance equal to
A) 20 cm
B) $20 \times (2)^{1/3}\ cm$
C) $20 \times (2)^{2/3}\ cm$
D) $20 \times (2)\ cm$
Explanation
(b) For no deflection in tan A position
$\frac{\mu_{0}}{4\pi}\frac{2M_{1}}{d_{1}^{3}} = \frac{\mu_{0}}{4\pi}\frac{2M_{2}}{d_{2}^{3}}$
$\therefore\ \ \ \ \frac{M_{1}}{M_{2}} = \left( \frac{d_{1}}{d^{2}} \right)^{3}$
Or $\frac{1}{2} = \left( \frac{20}{d^{2}} \right)^{3}$
$Or\ \ \ d_{2} = 20 \times (2)^{1/3}cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.