The magnetic force acting on the rod can be described by:

Magnetic force = I(d X B) (*where I=current, d=distance, B=magnetic field, and X=cross product)

Using the work-energy theorem, which is:

(K trans + K rot)initial + magnetic Force * L = (K trans + K rot)final (*where K = kinetic energy, trans=translational, rot=rotational)

Since it starts from rest, the initial Kinetic energy = 0 and hence the equation simplifies to:

magnetic Force = (K trans + K rot)final

Where,

K trans = (1/2) m*v^2 (*where m=mass, v=velocity)

K rot = (1/2)I*w^2 (*where I =moment of inertia, w=angular velocity)

Furthermore, I = (1/2)mR^2 and w = V/r

and so K rot = (1/2)((1/2)mR^2) * (V/r)^2

plugging everything back into the equation, we get:

magnetic Force = (K trans + K rot)final

magnetic Force = ((1/2) m*v^2s) + ( (1/2)((1/2)mR^2) * (V/r)^2)

magnetic force can then be rewritten as simply:

I*d*B*L (*since the cross product when theta = 0, is simply one, we are left with just this)

And so we get:

I*d*B*L = ((1/2) m*v^2s) + ( (1/2)((1/2)mR^2) * (V/r)^2)

and solving for velocity we should get: