A proton moving with a velocity of `2.5 xx 10^(7) m/s`, enter a magnetic feildof 2T, making an angle of 30◦ with the magnetic field. The force acting on the proton

Options

A. `2 xx 10^(-12)N`
B. `3 xx 10^(-12)N`
C. `4 xx 10^(-12)N`
D. `6 xx 10^(-12)N`

Answer

C

Explanation

`F=qvB sin theta=(1.6 xx 10^(-19))xx (2.5 xx 10^(7))xx 2 xx sin 30=4 xx 10^(-12)N`