Magnetism and Matter - Result Question 23
27. 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
[2010]
(a) $1 s$
(b) $2 s$
(c) $3 s$
(d) $4 s$
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Answer:
Correct Answer: 27. (d)
Solution:
- (d) Time period of a vibration magnetometer,
$T \propto \frac{1}{\sqrt{B}} \Rightarrow \frac{T_1}{T_2}=\sqrt{\frac{B_2}{B_1}}$ $\Rightarrow T_2=T_1 \sqrt{\frac{B_1}{B_2}}=2 \sqrt{\frac{24 \times 10^{-6}}{6 \times 10^{-6}}}=4 s$
