Question

When a magnetic field is applied in a direction perpendicular to the direction of cathode rays, then their
A. Energy decreases
B. Energy increases
C. Momentum increases
D. Momentum and energy remain unchanged

Answer 1

Hint: When a net non-zero work is done on the body by an external agent then using work-energy theorem the body gains energy. When the force is applied on the body then there will be acceleration and hence change in velocity, i.e. the change in momentum.


Formula used:
\[\overrightarrow F = q\left( {\vec v \times \vec B} \right)\] is the magnetic force vector, \[\vec v\]is the velocity of the charged particle and \[\vec B\]is the magnetic field in the region.



Complete answer:
When a magnetic field is applied in a region and charge is moving in the region, then the magnetic force acting on the charge is given as,
\[\overrightarrow F = q\left( {\vec v \times \vec B} \right)\]
It is given that the magnetic field is applied in a direction perpendicular to the direction of the cathode rays. So, the magnetic force acting on the cathode rays is perpendicular to the velocity of the cathode rays.
So the magnetic force on the charge is perpendicular to the velocity of the charge, i.e. displacement of the charge.
So, the work done by the perpendicular force is zero. Using the work-energy theorem, the moving charge will not gain any energy as net zero work is done on it. So, the energy remains unchanged.
The energy remains constant for the moving cathode rays, the speed of the charge is constant too. As the momentum is the product of the velocity and the mass, so the magnitude of the momentum too remains unchanged.
Therefore, the correct option is (D).





Note:We should be careful while considering the momentum. As the momentum is a vector quantity and the charge is in uniform circular motion, the linear momentum direction changes but the magnitude remains constant. So, in this context, we need to consider the magnitude of the momentum and not the direction.