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  4. A proton beam enters a magnetic field of 10-4 Wb m-2 normally. If the specific charge of the proton is 1011 C kg-1 and its velocity is 109 m s-1 then the radius of the ?circle decribed will be

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Q. A proton beam enters a magnetic field of $10^{-4}\,Wb\,m^{-2}$ normally. If the specific charge of the proton is $10^{11}\, C \,kg^{-1}$ and its velocity is $10^9 m\,s^{-1}$ then the radius of the ?circle decribed will be

KCETKCET 2015Moving Charges and Magnetism

Solution:

Given, $\beta=10^{-4} \,Wb \,m ^{-2}$
$\frac{q}{m} =10^{11} C \,kg ^{-1} $
$v =10^{9} \,ms ^{-1}$
We know that, radius of the circle,
$I =\frac{m v}{q B} $
$I =\frac{v}{\left(\frac{q}{m}\right) B} $
$r =\frac{10^{9}}{\left(10^{11}\right)\left(10^{-4}\right)} $
$I =100\, m$