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Question: A particle of charge \[q\;\]and mass\[\;m\;\]is moving with velocity \[\overrightarrow V \]. It is s...

A particle of charge q  q\;and mass  m  \;m\;is moving with velocity V→\overrightarrow V . It is subjected to a uniform magnetic field B→\overrightarrow B directed perpendicular to its velocity. Show that it describes a circular path. Write the expression for its radius.

Explanation

Solution

When a charged particle moves perpendicular to a uniform magnetic field. As the magnetic force is perpendicular to the direction of propagation, a charged particle tracks a curved path in a magnetic field. The particle remains to track this curved path till it forms a complete circle. A new way to look at this is that the magnetic force is constantly perpendicular to the velocity so that it performs no work on the charged particle. The kinetic energy and speed of the particle will remain constant. The direction of motion will be affected whereas the speed will be not.

Complete step by step solution:
Let us consider a particle of charge q  q\; and mass   m  \;m\; is directed perpendicularly towards the uniform magnetic field B→\overrightarrow B with velocity V→\overrightarrow V
The force on the charge will be
F=q(v×B)F = q\left( {v \times B} \right)
The magnetic force will be perpendicular to the velocity of a charged particle at all times. Therefore the magnitude of velocity remains the same but its direction continues to change.
Accordingly, the path of the charged particle in a perpendicular magnetic field will become circular. The magnetic force (qvB)  \left( {qvB} \right)\;delivers the required centripetal force to travel along a circular path. Then,
qvB=mv2rqvB = \dfrac{{m{v^2}}}{r}
This gives us r=mvqBr = \dfrac{{mv}}{{qB}} And Here r=r = radius of the circular path traveled by the charge which is the required solution.

Note:
A negatively charged particle passes in the plane of the paper in an area where the magnetic field is perpendicular to the paper. The magnetic force remains perpendicular to the velocity, as a result, velocity changes in direction but not magnitude. It leads to producing a uniform circular motion. As the charge is negative, the force will be opposite in direction to the proposed right-hand rule.)