Question

A proton travels at a speed of 5.0 × 107 m/s through a 1.0-T magnetic field. What is the magnitude of the magnetic force on the proton if the angle between the proton's velocity and the magnetic field vector is 30°?

Answer 1

The force experienced by proton is  is 4.0 x 10⁻¹² N

Explanation:

When a proton moves in magnetic field , it experiences a force .

The force can be calculated with the following relation

F = q v B sin θ

here q is the charge of proton and v is its velocity

B is the magnetic field strength and θ is the angle between field and velocity vector .

Thus F = 1.6 x 10⁻¹⁹ x 5.0 x 10⁷ x 1.0 x sin 30

= 4.0 x 10⁻¹² N

Answer 2

Answer:

$F_m=4.0\times 10^{-12} N$

Explanation:

The magnitude of the magnetic force is given by

$F_m=evB\sin\theta$

where e, is the proton charge, v, is the velocity and B, is the magnetic field

the proton charge $e=1.6\times 10^{-19}$

$\therefore F_m=1.6\times 10^{-19}\times 5.0\times 10^{7}\times 1.0\times \sin 30^0$

$\implies F_m=4.0\times 10^{-12} N$