Question

A particle carrying a charge of 200 $\mu C$ moves at an angle of 30 degrees to a uniform magnetic field of induction $5 \times {10^{ - 5}}Wb/{m^2}$ with a speed of $2 \times {10^5}m/s$. Calculate the force acting on the particle.

Answer 1

The presence of a charge in a magnetic field exerts a force on it. This force is dependent on the intensity of the magnetic field as well as the motion of the charge.

Formula used:
$F = qvB\sin \theta$, where F is the force exerted on the particle with charge q, moving in a magnetic field with intensity B and at an angle $\theta$. The SI unit of this force is Newton (N).

Complete step by step answer
The magnetic field exerts a force only on a moving charged particle. This is because the motion of a charge is associated with current flow, which gives rise to a magnetic field around it. The direction of the force acting on a charged particle in a magnetic field is determined using the right hand thumb rule.
In this question, we are provided with the following information:
Charge of the particle $q = 200\mu C = 200 \times {10^{ - 6}}C$ [As $1\mu C = {10^{ - 6}}C$]
Angle with the magnetic field $\theta = 30^\circ$
Magnetic field intensity $B = 5 \times {10^{ - 5}}Wb/{m^2}$
Speed of the particle $v = 2 \times {10^5}m/s$
We know that the force exerted on the particle is given as:
$F = qvB\sin \theta$
Substituting the known values in this equation, we get:
$\Rightarrow F = 200 \times {10^{ - 6}} \times 2 \times {10^5} \times 5 \times {10^{ - 5}} \times \sin 30$
$\Rightarrow F = 2 \times {10^{ - 3}} \times \dfrac{1}{2} = {10^{ - 3}}N$

Hence, this is the force experienced by the particle in the given conditions.

Note: When a charged particle moves at some angle with the uniform magnetic field exerted on it, a helical motion is generated. Due to this helical motion of particles trapped in a magnetic field, a very peculiar phenomenon of the Aurora is observed both in the Northern and the Southern hemispheres.