Question

Show all steps I will like the post

A proton moving in the xy-plane is subject to the Lorentz force due to a constant magnetic field that points in the positive direction of the z-axis. Assuming that the electron loses 20% of its energy what will be the ratio of the initial and final radii of the proton orbit?

Answer 1

Let the velocity to be v. Now that it is moving in x-y plane and the applied magnetic field is in z-direction thus the vecloity and magnetic field are always perpendicular to each other and that the produced lorrentz force will be always perpendicular to the velocity and in the x-y plane itself. So this will make the proton to move in a circle.

Now as you can see that the force is always perpendicular to the velocity thus it does no work. However if the particle is losing its energy this means that it has been acted upon by non-conservative forces. This work will only result in change of kinetic energy thus whatever energy is lost is only of the kinetic energy. So initially even it had kinetic energy and finally even it is having kinetic energy.

Now initially the kinetic energy was mv²/2. Now when it has lost 20% kinetic energy thus final kinetic energy

K_f = 0.8mv²/2

=> mv_f²/2 = 0.8mv²/2

=> v_f = (0.8)^1/2v

Now that the lorrentz force is given by F = qvB

when the particle is rotating in a circle this force balances the centifugal force Thus

mv²/r = qvB

=> Now you can see that the mass m, charge q and magnetic field B is always constant

thus r is directly constant to velocity v;

v_i/v_f = r_i/r_f

=> r_i/r_f = v/(0.8)^1/2v

=> r_i/r_f = 1.118