The figure below shows a bar of mass m 0225 kg that can sli

The figure below shows a bar of mass m = 0.225 kg that can slide without friction on a pair of rails separated by a distance = 1.20 m and located on an inclined plane that makes an angle = 29.0° with respect to the ground. The resistance of the resistor is R = 3.90 , and a uniform magnetic field of magnitude B = 0.500 T is directed downward, perpendicular to the ground, over the entire region through which the bar moves. With what constant speed v does the bar slide along the rails?
m/s

Solution

since its a constant speed Fg = Fb
gravitational force equals magnetic force

Fg = mg sin29

EMF = d/dt = d/dt (Blx cos29) = B cos29 lv
I = Blv cos29/R

Fb = BIL cos29 = B^2 l^2 v (cos29)^2/R
Fb = Fg

B^2 l^2 v cos29/R = mg sin29
v = (R mg sin29)/(B^2 l^2 (cos29)^2)
v = (3.9 * 0.225 * 9.8 * sin 29)/(0.25 * 1.44 (cos29)^2) = 15.13 m/s