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A bar magnet suspended freely in a unifo...

A bar magnet suspended freely in a uniform magnetic field is vibrating with a time period of 3s. If the field strength is increased to 4 times of the earlier field strength, then the time period (in second) will be

A

`12`

B

`6`

C

`1.5`

D

`0.75`

Text Solution

Verified by Experts

The correct Answer is:
C
As, the time period,= T=2pi`sqrt((1)/(MB))`
` " " T alpha (1)/(sqrtB) Rightarrow (T_(2))/(T_(1))=sqrt(B_(1)/(B_(2))`
` " " T_(2)/(3)=sqrt((1)/(4)) or T_(2)=1.5s.`
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