A proton enters a magnetic field of flux density 1.5 Wb/`m^(2)` with a spped of `2 xx 10^(7)` m/s at angle of `30^(@)` with the field. The force on the proton will be

Options

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

Answer

A, B, D

Explanation

F = Bqv sin `theta = 1.5 xx 1.6 xx 10^(-19) xx 2 xx 10^(7) xx sin 30^(@)` N