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. ${10}^{4}N$
• C. $1.6\times {10}^{-19}N$
• D. Zero

Answer 1

The correct option is D Zero

Answer is D.

The magnitude and direction of F depend on the velocity of the particle and on the magnitude and direction of the magnetic field B.
When a charged particle moves parallel to the magnetic field vector, the magnetic force acting on the particle is zero.
When the particles velocity vector makes any angle with the magnetic field, the magnetic force acts in a direction perpendicular to both v and B; that is, F is perpendicular to the plane formed by v and B.
The magnitude of the magnetic force is
$F=qvBsin\theta$
where $\theta$ is the smaller angle between v and B. From this expression, we see that F is zero when v is parallel or anti-parallel to B or 180) and maximum when v is perpendicular to B.
In this case, as the proton moves parallel to the magnetic field, the force is zero.