Q. Two small magnets each of magnetic moment $10\ A$-$m^{2}$ are placed in end-on position $0.1\ m$ apart from their centres. The force acting between them is
A) $0.6 \times 10^{7}N$
B) $0.06 \times 10^{7}N$
C) $0.6\ N$
D) $0.06\ N$
Explanation
(c) $F = \frac{\mu_{0}}{4\pi}\left( \frac{6MM^{'}}{d^{4}} \right)$ in end-on position between two small magnets
$\therefore F = 10^{- 7}\left( \frac{6 \times 10 \times 10}{(0.1)^{4}} \right) = 0.6N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.