Q. In an experiment with vibration magnetometer, the value of $4\pi^{2}I/T^{2}$ for a short bar magnet is observed as $36 \times 10^{- 4}.$ In the experiment with deflection magnetometer with the same magnet, the value of $4\pi d^{3}/2\mu_{0}$ is observed as $10^{8}/36$. The magnetic moment of the magnet used is
A) 50 A-m
B) 100 A-m
C) 200 A-m
D) 1000 A-m
Explanation
(b) In a vibration magnetometer,
$T = 2\pi\sqrt{\frac{I}{MH}.}$
$\therefore 4\pi^{2}\frac{I}{T^{2}} = MH = 36 \times 10^{- 4}$ ….(i)
In a deflection magnetometer,
$H = \frac{\mu_{0}}{4\pi}\frac{2M}{d^{3}}$
$\frac{4\pi d^{3}}{2\mu_{0}} = \frac{M}{H} = \frac{10^{8}}{36}$ ….(ii)
Multiplying Eq. (i) and Eq. (ii), we get
$M^{2} = 36 \times 10^{- 4} \times \frac{10^{8}}{36} = 10^{4}$
$M = 10^{2} = 100$ A-m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.