Question

A proton moving with a velocity of 2.5x10^7 m/s enters magnetic field of 1Wbm making an angle 30 with magnetic field. the force on proton is?

Answer 1

Answer:

the force is 2.0625 x10^26 newton

Explanation:

the formula for force is F=qvbsintheta

here v = 2.5x 10^7 m/s

theta = 30°

and B = 1Wbm

and we know that the charge on the proton is 1.65 X 10^19 coulomb

now ,

F = 1.65x10^19 X 2.5 X 10^7x 1 x sin30°

=2.0625 x10^26 Newton.