A proton beam enters a magnetic field of `10^(-4) " Wb " m^(-2)` Wb 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 described will be

Options

A. 10 m
B. 1 m
C. 0.1 m
D. 100 m

Answer

D

Explanation

When the proton enters bem enters the magnetic field B normally , it describes radius r given by
` r = (mv)/(eB) = v/((e/m)B)`
where `e/m` is the specific charge of the proton and v I the velocity .
Here ,
` v = 10^(9) " m s"^(-1) , e/m = 10^(11) C kg^(-1)`
`B = 10^(-4) " Wb m"^(-2)` ,
` :. " " r = (10^(9)" m s"^(-1))/((10^(11)" C kg"^(-1))(10^(-4)" Wb m"^(-2))) = 100 m `