A ring of mass m and radius R is set into pure rolling on horizontal rough surface, in a uniform magnetic field of strength B – as shown in the figure. A point charge q of negligible mass is attached to rolling ring. Friction is sufficient so that it does not slip at any point of this motion. (θ is measured in clockwise from positive y –axis).

Speed of the point  charge at the position shown = 2v cos $\frac{\theta }{2}$

Magnitude of magnetic force acting on the  point charge F = q ( 2v cos $\frac{\theta }{2})$B = 2vBq $\cos \frac{\theta }{2}.$

Since the line of action of the magnetic force F always passes through the point of contact with the ground , it does not produce any moment about the point of contact and its angular acceleration is always zero and so linear acceleration is also always zero .

The horizontal component of

the magnetic force F 

$=\mathrm{q}\left\{2\mathrm{vcos}\left(\frac{\mathrm{\theta }}{2}\right)\mathrm{Bsin}\frac{\mathrm{\theta }}{2}=\mathrm{Bqvsin}\mathrm{\theta }\right\}$. 

Since resultant horizontal force acting on the ring is zero , f - F = 0 $\Rightarrow f =Bqv \sin \theta$

Vertical component of force F 

$=2{\mathrm{qVBcos}}^2\frac{\mathrm{\theta }}{2},\text{ its maximum value is }2\mathrm{qVB}$

$\therefore \mathrm{N}=\mathrm{mg}-2\mathrm{Bqv}$

$\Rightarrow \text{ To loose the contact, }\mathrm{N}=0$

$\Rightarrow \mathrm{v}=\frac{\mathrm{mg}}{2\mathrm{qB}}$