The figure below shows a bar of mass m-0.300 kg that can slide without friction

Question

The figure below shows a bar of mass m-0.300 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 27.0° with respect to the ground. The resistance of the resistor is R 3.30 and a uniform magnetic field of magnitude B 0.50 TIS 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? 10.63 Your response differs from the correct answer by more than 10%. Double check your calculations. m/s Need Help? Read It

Explanation / Answer

component of the magnetic field perpendicular to the plane= B cos27

now induced emf= e=Bcos27 L V

induced current= e/R

i= BLV cos27/R

magnetic force on the bar in upward direction= Fm= i L Bcos27

Fm= (BLVcos27/R)*L Bcos27

Fm= (BLcos27)2 V/R

now Velocity V is constant, so acceleration a=0

mgsin27-Fm= m (0)

mg sin27- (BLcos27)2 V/R=0

0.3*9.8*sin27- (0.5*1.2 cos27)2 V/3.3=0

1.335=0.0866V

V= 15.4 m/s

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