A proton enters a magnetic field of 1.5 T with a velocity of `2 xx 10^(7) ms^(-1)` at an angle of `30^@` with the field. The force on the proton will be

Options

A. `2.4 xx 10^(-12) N`
B. `0.24 xx 10^(-12) N`
C. `24 xx 10^(-12)N`
D. `0.024 xx 10^(-12) N`

Answer

A

Explanation

Here `q = +e = 1.6 xx 10^(-19) C, B = 1.5 T , v = 2 xx 10^(7) m s^(-1) and theta = 30^@`
`:.` Force on proton `F = q v B sin theta = 1.6 xx 10^(-19) xx 2 xx 10^(7) xx 1.5 xx sin 30^@ = 2.4 xx 10^(-12) N`.