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A helium nucleus makes a full rotation i...

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

A

`(10^(-19))/(mu_(0))`

B

`10^(-19) mu_(0)`

C

`2 xx 10^(-10) mu_(0)`

D

`(2 xx 10^(-10))/(mu_(0))`

Text Solution

Verified by Experts

The correct Answer is:
B
Time period (T) = 2
`i=(q)/(T) = (2 xx 1.6 xx 10^(-19))/(2) = 1.6 xx 10^(-19) A`
Field at centre - `B = (mu_(0) I)/(2R) = (mu_0 xx 1.6 xx 10^(-19))/(2 xx 0.8) = mu_(0) xx 10^(-19) T`
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