Question

If the resulting trajectory of the charged particle is a circle, what is ω, the angular frequency of the circular motion?

Answer 1

For this answer you need to have in mind the following formulas:

Fmag = qvB

Also have in mind that Centripetal acceleration is: ar = v2/R

centripetal force will be :Fc = mv2/R

And velocity is: v = ωR

Remember that the magnetic force is the centripetal force then: 

Fmag = Fc
qvB = mv2/R
qB = mv/R

Next you need to substitute for velocity:

qB = mωR/R
qB = mω

ω = qB/m

Answer 2

Answer:

The value of circular velocity or frequency is $\dfrac{qB}{m}$.

Explanation:

The charged particle is moving in a circular path.

Let the charge of the particle is $q$. The radius of the circular path is $R$. The velocity of the particle is $v$ and it is moving in a magnetic field is $B$.

Now, the magnetic force experienced by the charged particle is,

$F=qvB$ (assuming the magnetic field is perpendicular to the direction of motion)

Now, the centripetal force experienced by the particle will be,

$F_c=\dfrac{mv^2}{R}$

Velocity of the particle can be written as (in the form of angular velocity) $v=R\omega$.

Now, equating the magnetic and centripetal force to get the value of circular velocity,

$F=F_c\\qvB=\dfrac{mv^2}{R}\\qR\omega B=\dfrac{m(R\omega)^2}{R}\\qR\omega B=\dfrac{mR^2\omega^2}{R}\\\omega=\dfrac{qB}{m}$

So, the value of circular velocity or frequency is $\dfrac{qB}{m}$.

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