Q. In a vibration magnetometer, the time period of a bar magnet oscillating in horizontal component of earth’s magnetic field is 2 s. When a magnet is brought near and parallel to it, the time period reduces to 1 s. The ratio $\frac{F}{H}$ of the fields, $F$ due to magnet and $H$, the horizontal component will be
A) $\sqrt{3}$
B) $\frac{1}{\sqrt{3}}$
C) $\frac{1}{3}$
D) 3
Explanation
(d) As $T = 2\pi\sqrt{\frac{I}{MH}}\ \therefore 2 = 2\pi\sqrt{\frac{I}{MH}}$ …(i)
When an external magnet is brought near and parallel to $H$, and the time period reduces to 1 s, net field must be $(F + H)$.
$\therefore 1 = 2\pi\sqrt{\frac{I}{M(F + H)}}$ ...(ii)
Divided Eq. (i) by Eq. (ii),
$\ \frac{1}{2} = \sqrt{\frac{F + H}{H}} = \sqrt{\frac{F}{H} + 1}$
$\frac{F}{H} + 1 = 4$
$\frac{F}{H} = 4 - 1 = 3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.