Question

An electron enters a magnetic field whose direction is perpendicular to the velocity of the electron. Then
A. The speed of the electron will increase
B. The speed of the electron will decrease
C. The speed of the electron will remain the same
D. The velocity of the electron will remain the same

Answer 1

Hint: When a charged particle is moving in a magnetic field then the magnetic force on the charged particle is proportional to the vector product of the magnetic field and the velocity.


Formula used:
\[\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)\], Here \[\overrightarrow F \]is the magnetic force vector acting on a charged particle with charge q and moving with the velocity vector \[\overrightarrow v \]and the magnetic field vector \[\overrightarrow B \]



Complete answer:
It is given that the electron entering a magnetic field is perpendicular to the velocity of the electron.
Using the magnetic force vector formula, the direction of the magnetic force on the electron is perpendicular to the magnetic field as well as the velocity vector.
When a perpendicular force acts on a particle then the path of the motion of the particle is circular motion.
The work done by the force is given as,

\[W = Fs\cos \theta \]
As the magnetic force on the electron is perpendicular to the velocity, so the angle between the force and the displacement is 90°.
\[\theta = 90^\circ \]
So, the work done by the magnetic force is zero.
Using the work-energy theorem, when the net work done is zero then the change in kinetic energy of the body is zero, so the speed of the particle is constant.
Hence, the speed of the electron will remain the same but as the direction of velocity is changing in the circular motion, so the velocity is not constant.
Therefore, the correct option is (C).

Note:The kinetic energy is proportional to the square of the speed. As it is a scalar quantity so it doesn’t depend on the direction of the velocity. So, when the kinetic energy is constant it is not necessary that velocity is constant too.