Question

Two protons enter a region of the transverse magnetic field. What will be the ratio of the time period of revolution if the ratio of energy is $2 \sqrt{2}: \sqrt{3}$?

Answer 1

Hint: The magnetic force is pointed towards the centre of circular motion experienced by the object and performs as a centripetal force. Thus, if v and B are right angles to each other, the particle represents a circle. The period of revolution does not depend on the velocity, so for both protons, the period of revolution will continue the same.

Complete step-by-step solution:
Protons revolve in a magnetic field so; centripetal force acts and magnetic force also acts in it.
$F_{c} = F_{m}$
$\dfrac{mv^{2}}{r} = qvB$
$\implies r \propto v$
The time period of revolution is given by:
$T = \dfrac{2 \pi r}{v}$
where, v is the proton’ s speed.
r is the radius of the circular path.
 We know that $r \propto v$,
So, Time period is independent of velocity and is constant. It is independent of the energy.
Time period is the same whenever energy is changed.
$\dfrac{ T_{1}}{ T_{2}} = \dfrac{1}{1} $
The ratio of the time period of revolution is $1:1$.

Note:A force operating on a particle is supposed to perform work when there is a force component in the particle's direction of motion. In the case under reflection where we have a charged particle giving a charge q moving in a steady magnetic field B, the magnetic force works perpendicular to the particle's velocity. Here we assume that the magnetic force makes no work on the particle, and hence, no difference in the particle's velocity can be viewed.