Answer:
[tex]r=\frac{m.v}{q.B}[/tex]
Explanation:
[tex]F_B=q.v.B[/tex]...........................................(1)
where q is the charge on the particle.
Now as we know that the force acting on the particle moving in the circular path is a centripetal force which is given as:
[tex]F_c=m.r.\omega^2[/tex]....................................(2)
where:
m = mass of the particle
r = radius of the particle
[tex]\omega[/tex]= angular velocity of the particle
Also, the relation between the angular and the linear velocity is as :
[tex]\omega=\frac{v}{r}[/tex]........................................(3)
From the eq. (2) & (3)
[tex]F_c=m.\frac{v^2}{r}[/tex]...................................(4)
from eq. (1) & (4)
[tex]F_c=F_B[/tex]
[tex]m.\frac{v^2}{r}=q.v.B[/tex]
[tex]r=\frac{m.v}{q.B}[/tex]