Faraday's law of electromagnetic induction

Q. Faraday's law of electromagnetic induction

Question                                                      

Figure shows a bar of mass m = 0.2 kg

that can slide without friction on a pair

of rails separated by a distance l =1.2 m

and located on an incline plane that

makes an angle = 25.0° with respect to

the ground. The resistance of the resistor

is R = 1.0 and a uniform magnetic field

of magnitude of B = 0.5 T is directed

downward, perpendicular to the ground,

over the entire region through which the

bar moves. With what speed v

does the bar slide along the rails?

Answer

Component of B perpendicular to the plane
= Bcos25°

=> emf induced in the coil 
ε = Bcos25° * 1.2 * v = 
=> current in the bar,
I = ε/R = (v/R) * 1.2 Bcos25°
Force opposing the downward motion
F = I * L * Bcos25° = (v/R) * (1.2 Bcos25°)^2

As the bar starts sliding, v keeps on increasing and so does F till F balances the downward component of the weight of the bar
=> mgsin25° = (v/R) * (1.2 Bcos25°)^2
=> velocity, v
= (mgRsin25°) / (1.2 Bcos25°)^2
= (0.2 * 9.81 * 1.0 * sin25°) / (1.2 * 0.5 * cos25°)^2
= (0.8292) / (0.2957)
= 2.80 m/s.