A proton moving with a velocity ` 2 . 5 xx 10^(7) m-s ^(-1)` , enters a magnetic field of intensity 2.5 T at an angle `30^(@)` with the magnetic field. The force on the proton is

Options

A. `3 xx 10^(-12)` N
B. `5 xx 10^(-12) `N
C. `6 xx 10^(-12)` N
D. ` 9 xx 10^(-12)` N

Answer

B

Explanation

Given : Velocity of proton
(V) ` = 2 . 5 xx 10^(7) m - s ^(-1)` .
Intensity of magnetic field (B) = 2 . 5 T and angle between magnetic field and direction of motion of proton acting on the charged particle in magnetic field
`(E) = qvB sin theta`
`= (1 . 6 xx 10^(-19)) xx (2.5 xx 10^(10)) xx 2.5 sin 30^(@)`
` = (10 xx 10^(-12)) xx 0 . 5 `
`= 5 xx 10^(-12)` N