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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

4s

B

1s

C

2s

D

3s

Text Solution

Verified by Experts

The correct Answer is:
A
`T= 2pi sqrt((I)/(B_(H)M))`
`T= 2pi sqrt((I)/(M(B_(H)- B))) rArr T.= T. (1)/(sqrt((1- (B)/(B_(H)))))`
`rArr T.= 2T= 4` sec
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