The figure below shows a bar of mass m = 0.285 kg that can slide without frictio

Question

The figure below shows a bar of mass m = 0.285 kg that can slide without friction on a pair of rails separated by a distance f = 1.20 m and located on an inclined plane that makes an angle theta = 35.0degree with respect to the ground. The resistance of the resistor is R = 1.80 Ohm, 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?

Explanation / Answer

Since its a constant speed Fg = Fb ===> gravitational force equals magnetic force

Fg = mg sin35

EMF = d/dt = d/dt (Blx cos35) = B*cos35*l*v
I = Blv*cos35/R

Fb = BIL cos35 = B^2 l^2 v (cos35)^2/R
Fb = Fg
B^2 l^2 v cos35/R = mg sin35
v = (R mg sin35)/(B^2 l^2 (cos35)^2)
v = (1.8 * 0.285 * 9.8 * sin 35)/(0.25 * 1.44 (cos35)^2) = 11.94 m/s

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