Q. A circular coil of radius $4\ cm$ and of 20 turns carries a current of $3\ amperes$. It is placed in a magnetic field of intensity of $0.5\ weber/m^{2}$. The magnetic dipole moment of the coil is
A) $0.15\ ampere - m^{2}$
B) $0.3\ ampere - m^{2}$
C) $0.45\ ampere - m^{2}$
D) $0.6\ ampere - m^{2}$
Explanation
(b) $M = NiA = 20 \times \frac{22}{7}\left( 4 \times 10^{- 2} \right)^{2}3 = 0.3A - m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.