Question

A particle with charge q and kinetic energy KE travels in a uniform magnetic field of magnitude B. If the particle moves in a circular path of radius R, find expressions for its speed v and its mass m. (Use any variable or symbol stated above as necessary.) (a) expression for its speed (Do not use m in your answer.) v = (b) expression for its mass (Do not use v in your answer.) m =

Answer 1

Answer:

a)$v=\dfrac{2.KE}{qBR}$

b)$m=\dfrac{(qBR)^2}{2.KE}$

Explanation:

Given that

Charge = q

Magnetic filed = B

Radius = R

We know that kinetic energy KE

$KE=\dfrac{1}{2}mv^2$                    ----------1

m v² = 2 .KE            

The magnetic force F = q v B

Radial force

$Fr=\dfrac{1}{R}mv^2$

For uniform force these two forces should be equal

$q v B=\dfrac{1}{r}mv^2$

q v B R =m v²

q v B R =  2 .KE

$v=\dfrac{2.KE}{qBR}$

Now put the velocity v in the equation

$KE=\dfrac{1}{2}mv^2$

$m=\dfrac{2 .KE}{v^2}$

$m=\dfrac{2.KE}{\left(\dfrac{2.KE}{qBR}\right)^2}$

$m=\dfrac{(qBR)^2}{2.KE}$