Q. Force between two identical short bar magnets whose centers are $r$ metre apart is 8.1 N, when their axes are along the same line. If separation is increased to 3 $r$ and the axis are rearranged perpendicularly, the force between them would become
A) 2.4 N
B) 1.2 N
C) 0.1 N
D) 1.15 N
Explanation
(c) When axes are in the same line,
$F = \frac{\mu_{0}}{4\pi}\frac{6M_{1}M_{2}}{r^{4}}\ ie,\ F \propto \frac{1}{r^{4}}$
When, $r$ becomes thrice, $F\ becomes\frac{1}{(3)^{4}}$ time
$ie,\frac{1}{81}$ time. Therefore, $F^{'} = \frac{8.1}{81} = 0.1$ N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.