Question

A vibration magnetometer placed in magnetic meridian has
a small bar magnet. The magnet executes oscillations with a
time period of 2 sec in earth's horizontal magnetic field of 24
microtesla. When a horizontal field of 18 microtesla is produced
opposite to the earth's field by placing a current carrying wire,
the new time period of magnet will be
(a) 1 s (b) 2 s
(c) 3 s (d) 4 s

Answer 1

$time \: period \: = 2\pi \sqrt{ \frac{moment \: of \: inertia}{hz \: magnetic \: field} }$

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