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A helium nucleus makes full rotation in ...

A helium nucleus makes full rotation in a circle of radius 0.8 m in 2.5 seconds. The value of magnetic field B at the centre of the circle will be

A

`3.1pi xx10^-26 T`

B

`2pi xx10^-26 T`

C

`4pi xx10^-25 T`

D

`2 pi xx10^-25 T`

Text Solution

Verified by Experts

The correct Answer is:
A
`B=(mu_0I)/(2a)(I=(q)/(T)=(2e)/(T))`
Since for helium , `q=2e`
`B=(mu_02e)/(2aT)=(4pi xx 10^-6xx2xx1.6xx10^-19)/(2xx0.8xx2.5)`
`B=3.2pi xx10^-26 T`.
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