Q. A small bar magnet has a magnetic moment $1.2\ A$-$m^{2}.$ The magnetic field at a distance $0.1\ m$ on it axis will be: $(\mu_{0} = 4\pi \times 10^{- 7}$-$m/A)$
A) $1.2 \times 10^{- 4}T$
B) $2.4 \times 10^{- 4}T$
C) $2.4 \times 10^{4}T$
D) $1.2 \times 10^{4}T$
Explanation
(b) $B = \frac{\mu_{0}}{4\pi}.\frac{2M}{d^{3}} \Rightarrow B = 10^{- 7} \times \frac{2 \times 1.2}{(0.1)^{3}} = 2.4 \times 10^{- 4}T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.