Question

A proton enters a uniform magnetic field of strength 2 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’s velocity. What is the magnitude of the force that the proton experiences while it moves through the magnetic field?

Answer 1

Answer:

Magnetic force, $F=9.6\times 10^{-17}\ N$

Explanation:

It is given that,

Magnetic field, B = 2 T

Velocity of the proton, v = 300 m/s

Charge on the proton, $q=1.6\times 10^{-19}\ C$

The magnetic field is oriented perpendicular to the proton’s velocity. The magnetic force on the charged particle is given by :

$F=qvB\ sin\theta$

The magnetic field is oriented perpendicular to the proton’s velocity, $\theta=90^{\circ}$

$F=1.6\times 10^{-19}\times 300\times 2$

$F=9.6\times 10^{-17}\ N$

So, the magnitude of the force that the proton experiences while it moves through the magnetic field is $9.6\times 10^{-17}\ N$. Hence, this is the required solution.