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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
4 s
B
1 s
C
2 s
D
3 s
Detailed Solution
Time period of a magnet oscillating in a field B: $T = 2\pi\sqrt{\frac{I}{MB}}$, so $T \propto \frac{1}{\sqrt{B}}$
Initial field: $B_1 = 24\ \mu T$
The applied field is opposite to the earth's field, so the net field is $B_2 = 24 - 18 = 6\ \mu T$
$\frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}} = \sqrt{\frac{24}{6}} = 2$
$T_2 = 2\times2 = 4$ s
Initial field: $B_1 = 24\ \mu T$
The applied field is opposite to the earth's field, so the net field is $B_2 = 24 - 18 = 6\ \mu T$
$\frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}} = \sqrt{\frac{24}{6}} = 2$
$T_2 = 2\times2 = 4$ s