1.

When a helium nucleus revolves in a circular orbit of radius 0.8 m,if it takes 2 sec to comlete one revolution.find magnetic field produced at the centre would be _____ T.

Answer»

`10^(-18)mu_(0)`
`5XX10^(-19)mu_(0)`
`2xx10^(-19)mu_(0)`
`10^(-19)mu_(0)`

Solution :Here Q = 2e = ne, f = 1 HZ, a = 0.8 m
`thereforeB=(mu_(0)I)/(2a)=(mu_(0)Qf)/(2a)""(becauseI=Qf)`
`thereforeB=(mu_(0)xx2xx1.6xx10^(-19)XX1)/(2xx0.8)`
`thereforeB=2xxmu_(0)xx10^(-19)T`


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