Magnetic Properties of Matter: Practice Problem & Solution
14. 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 Pre)
Solution Explained:
To solve this problem, we apply the core principles of Magnetic Properties of Matter. Understanding the underlying formula is key to arriving at the correct answer below:
Initial field $B_1 = 24$ $\mu$T, $T_1 = 2$ s. Net new field $B_2 = 24 - 18 = 6$ $\mu$T.
Since $T \propto \frac{1}{\sqrt{B}}$, we have $\frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}} = \sqrt{\frac{24}{6}} = \sqrt{4} = 2$. Therefore, $T_2 = 2 \times 2 = 4$ s.
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