Question

Pl. explain
A proton enters a magnetic field of 1.5 tesla with a velocity of 2.0 X 10 raise to power 7 m/sec at an angle of 30 degree with the field.The magnitude of the force on the proton is :
1) 4.2 * 10 raise to power -12 N
2) 2.4 * 10 raise to power -12 N
3) 4.2 * 10 raise to power -6 N
4) 2.4 * 10 raise to power -6 N

Answer 1

The answer is:     2) 2.4 * 10 raise to power -12 N

The magnitude of the magnetic force on a charg particle is F =  |q| v B sin θ

Here q = charge on a proton = 1.602 x 10⁻¹⁹ C

v = 2 x 10⁷ m/s

θ = 30°

B = 1.5 T

Substituting all these values in the expression for F, we get

F = 1.602 x 10⁻¹⁹ C x 2 x 10⁷ m/s x 1.5 T sin 30°

  = 2.403 x 10⁻¹² N