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A vibration magnetometer placed in magne...

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

3s

B

4s

C

1s

D

2s

Text Solution

Verified by Experts

The correct Answer is:
B
Time period of bar magnet in vibration magnetometer is `T = 2pi sqrt( (I)/(MB) )`
` T = (1)/(sqrtB)`
Given, T=2 s in the Earth.s horizontal magnetic field `B_H = 24 muT`, and now another horizontal magnetic field `B_h = 18mu T` is acting just opposite to the Earth.s field let new time period be `T_2`
` T_1/T_2 = sqrt( (B_H - B_h)/(B_H) )`
` (2)/(T_2) = sqrt( (24-18)/(24) ) = sqrt(6/24) = 1/2 `
` T_2 = 4s `
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