Question

A proton travels with a speed of 1.8×106 m/s at an angle 53◦ with a magnetic field of 0.49 T pointed in the +y direction. The mass of the proton is 1.672 × 10−27 kg. What is the magnitude of the magnetic force on the proton?

Answer 1

Answer:

Magnetic force, $F=1.12\times 10^{-13}\ N$

Explanation:

It is given that,

Velocity of proton, $v=1.8\times 10^6\ m/s$

Angle between velocity and the magnetic field, θ = 53°

Magnetic field, B = 0.49 T

The mass of proton, $m=1.672\times 10^{-27}\ kg$

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

The magnitude of magnetic force is given by :

$F=qvB\ sin\theta$

$F=1.6\times 10^{-19}\ C\times 1.8\times 10^6\ m/s\times 0.49\ Tsin(53)$

$F=1.12\times 10^{-13}\ N$

So, the magnitude of the magnetic force on the proton is $1.12\times 10^{-13}\ N$. Hence, this is the required solution.