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Q. 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 $6$ microtesla. When a horizontal field of $18$ microtesla is produced in same direction of field by placing a current carrying wire, the new time period of magnet will be :-

Moving Charges and Magnetism

Solution:

$T =2 \pi \sqrt{\frac{ I }{ MB }} $
$\Rightarrow T \propto \frac{1}{\sqrt{ B }}$
$\frac{ T _{1}}{ T _{2}}=\sqrt{\frac{ B _{2}}{ B _{1}}}=\sqrt{\frac{6+18}{6}}=2$
$\Rightarrow T _{2}=1 sec$