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 l = 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

here,

Magnatic field, B = 0.5 T
length of bar, L = 1.20 m

theta, A = 35 degrees

mass of bar, m = 0.285 kg
Resistance of rod, R = 1.8 Ohms

From newtonSecond law, SUM(F) = 0
F = mgSinA
I*(L B)*Sin(90 - 35) = mgSin35

Solving for Current, I = mgSin35 /(L*B*Sin(90-35))
I = (0.285*9.81*Sin35) /(1.20*0.5*Sin(90-35))
I = 3.263 A

Also Emf , E = R*I = 1.8 * 3.263 = 5.873 V

Since EmF, E = B*L*V

Equating and solvignf or Velocity, V

V = 5.873/(0.5 * 1.20)

V = 9.788 m/s

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