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:
The value of circular velocity or frequency is [tex]\dfrac{qB}{m}[/tex].
Explanation:
The charged particle is moving in a circular path.
Let the charge of the particle is [tex]q[/tex]. The radius of the circular path is [tex]R[/tex]. The velocity of the particle is [tex]v[/tex] and it is moving in a magnetic field is [tex]B[/tex].
Now, the magnetic force experienced by the charged particle is,
[tex]F=qvB[/tex] (assuming the magnetic field is perpendicular to the direction of motion)
Now, the centripetal force experienced by the particle will be,
[tex]F_c=\dfrac{mv^2}{R}[/tex]
Velocity of the particle can be written as (in the form of angular velocity) [tex]v=R\omega[/tex].
Now, equating the magnetic and centripetal force to get the value of circular velocity,
[tex]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}[/tex]
So, the value of circular velocity or frequency is [tex]\dfrac{qB}{m}[/tex].
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