Question

A proton is moving with velocity ${10^4}m/s$ parallel to the magnetic field of intensity 5 tesla. The force on the proton is
A. $8 \times {10^{ - 15}}N$
B. ${10^{14}}N$
C. $1.6 \times {10^{ - 19}}N$
D. Zero

Answer 1

Hint: You can start by mentioning the equation of the magnetic force that a moving charged particle experiences in a magnetic field, i.e. $F = qvB\sin \theta $ . Then just simply substitute the given values in this equation to reach the solution.

Complete answer:
In the problem, we are required to calculate the magnetic force on the proton due to the magnetic field (the source of the magnetic field is not important in this case). Since a proton is a charge that is moving with a velocity ${10^4}m/s$ , so it will be affected by the magnetic field.
The equation of magnetic force exerted on a moving charge in a magnetic field is
$F = qvB\sin \theta $
$F = $ The magnetic force on the moving charge
$q = $ The magnitude of the moving charge
$v = $ The velocity with which the charge is moving
$B = $ The intensity of the magnetic field
$\theta = $ The angle that the direction of the moving charge makes with the direction of the magnetic field.
In the solution, we are given
$q = + 1C$ (The magnitude of the proton is $ + 1$ )
$v = {10^4}m/s$
$B = 5T$
$\theta = 0^\circ $ (The proton is moving parallel to the direction of the magnetic field)
So, for the given problem
$F = 1 \times {10^4} \times 5 \times \sin \left( {0^\circ } \right)$
$F = 1 \times {10^4} \times 5 \times 0$
$F = 0$
So, for a proton moving in a magnetic field, if the direction of the moving proton is parallel to the direction of the magnetic field then no magnetic force acts on the moving charged particle.

So, the correct answer is “Option D”.

Note:
In the solution above we saw that the moving particle experiences no magnetic force when the particle moves in a direction parallel to the direction of the magnetic field. The magnetic force exerted on the moving particle is maximum when the direction of the moving particle is in a direction perpendicular to the direction of the magnetic field ( $F = qvB$ ).