A vibration magnetometer placed in magnetic meridian has a small bar… | NEET 2010 (AIPMT, Prelims)

Q15NEET 2010 · AIPMT · PRELIMSMedium

Vibration Magnetometer

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

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Answer: (d)

For a vibration magnetometer $T = 2\pi\sqrt{\dfrac{I}{MB_H}}$ , so $T \propto \dfrac{1}{\sqrt{B_H}}$ .
Net horizontal field $= 24 - 18 = 6\ \mu\text{T}$ (wire's field is opposite to earth's field).
$\dfrac{T'}{T} = \sqrt{\dfrac{24}{6}} = 2 \Rightarrow T' = 2 \times 2 = 4\ \text{s}$
So the answer is (d) 4 s.

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Vibration Magnetometer question from NEET 2010 (AIPMT, Prelims)

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