Question

A particle of charge q and rest mass m is moving with velocity v where the magnetic field is B. Here B is perpendicular to v, and there is no electric field. Show that the path of the particle is a curve with radius of curvature R given by R = p/qB, where p is the momentum of the particle, ? mv. (Hint: Note that the force qv × B can only change the direction of the momentum, not the magnitude. By what angle delta ? is the direction of p changed in a short tim

Answer 1

Answer:

$\Delta \theta = \frac{qB}{m} t$

Explanation:

As we know that charge is moving in magnetic field then we will have

$F = qvB$

since this force is perpendicular to the velocity

so this force is considered as centripetal force and hence we will have

$F = F_c$

$qvB = \frac{mv^2}{R}$

so here we have

$R = \frac{mv}{qB}$

$R = \frac{P}{qB}$

here P = mv (momentum)

So angle changed by the charge is given as

$\Delta \theta = \omega t$

$\Delta \theta = \frac{v}{R} t$

$\Delta \theta = \frac{qB}{m} t$