Question

# A vibration magnetometer placed in a magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2\mathrm{sec}$ in the earth's horizontal magnetic field of $24$ microteslas. When a horizontal field of $18$ microteslas is produced opposite to the earth's field by placing a current-carrying wire, the new time period of the magnet will be

A

$1\mathrm{s}$

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B

$2\mathrm{s}$

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C

$3\mathrm{s}$

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D

$4\mathrm{s}$

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Solution

## The correct option is D $4\mathrm{s}$Step 1: Given dataThe magnet executes oscillations with a time period of $2\mathrm{sec}$.The earth's horizontal magnetic field of $24$ microteslas.A horizontal field of $18$ microteslas.Step 2: To findWe have to determine the new time period of the magnet.Step 3: CalculationWe know that,The time period T of oscillation of a magnet is:$⇒\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{I}}{\mathrm{MB}}}$Here,l is the moment of inertia of the magnet about the axis of rotation.M is the magnetic moment of the magnet.B is the uniform magnetic field as l, remains the same.Now,$\mathrm{T}\propto \frac{1}{\sqrt{\mathrm{B}}}$ or, $\frac{{\mathrm{T}}_{2}}{{\mathrm{T}}_{1}}=\sqrt{\frac{{\mathrm{B}}_{1}}{{\mathrm{B}}_{2}}}$Then,${\mathrm{B}}_{1}=24\mathrm{ut}\phantom{\rule{0ex}{0ex}}{\mathrm{B}}_{2}=24\mathrm{ut}–18\mathrm{ut}\phantom{\rule{0ex}{0ex}}=6\mathrm{ut}\phantom{\rule{0ex}{0ex}}{\mathrm{T}}_{1}=2\mathrm{seconds}\phantom{\rule{0ex}{0ex}}{\mathrm{T}}_{2}=2\sqrt{\frac{24\mathrm{ut}}{6\mathrm{ut}}}\phantom{\rule{0ex}{0ex}}=2\sqrt{4}\phantom{\rule{0ex}{0ex}}=2×2\phantom{\rule{0ex}{0ex}}=4\mathrm{s}$Therefore, option D is the correct choice.

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