A metal rod with a mass of 010 kg rolls down an inclined pla

A metal rod with a mass of 0.10 kg rolls down an inclined plane (inclined at 10 degrees above the horizontal). The rod makes contact with two metal rails that are connected to a R = 2.0 ohm resistor making a closed circuit. The rails are separated from one another by d = 0.15 m. The entire circuit is in a uniform magnetic field of B = 1.0 T that is directed vertically. What is the terminal velocity of the rod? (Neglect sliding and rolling friction of the rod on the rails.)

Solution

in case of terminal velocity

net force will be zero on rod

=> gravitational force = magnetic force

=> mgsin(thetha) = BIL

I = VBL/R

=> mgsin(thetha) = VB^2L^2/R

=> 0.1*9.8*0.1736 = V*1^2*0.15^2)/(2

=> V = 15.12 m/s