Question

A certain baryon (a subatomic particle) has a charge q > 0. The baryon moves with a kinetic energy K in a circular path of radius R in the presence of a uniform magnetic field of magnitude B. (Use any variable or symbol stated above as necessary.) (a) What is the speed of the baryon? Assume the speed is low compared to the speed of light. (b) What is the mass of the baryon?

Answer 1

Answer:

$v=\frac{2K}{RqB}$

$m=\frac{(RqB)^2}{2K}$

Explanation:

The formula for kinetic energy K of a particle of mass m moving at velocity v is $K=\frac{mv^2}{2}$ and the formula for the Lorentz force F experimented by a particle of charge q and velocity v under a magnetic field B is (asuming v and B are perpendicular) $F=qvB$.

Since the particle would be moving in circles, this force would be a centripetal force given by $F_{cp}=ma_{cp}=\frac{mv^2}{R}$, where R is the radius of the trajectory.

Then we have (q, K, R and B would be what we know):

$F=F_{cp}$

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

$v=\frac{2K}{RqB}$

And:

$m=\frac{2K}{v^2}=\frac{2K(RqB)^2}{(2K)^2}=\frac{(RqB)^2}{2K}$