The figure below shows a bar of mass m = 0.260 kg that can slide without frictio

Question

The figure below shows a bar of mass m = 0.260 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 ? = 33.0

The figure below shows a bar of mass m = 0.260 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 ? = 33.0° with respect to the ground. The resistance of the resistor is R = 3.30 ? 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? in m/s

Explanation / Answer

The magnetic force:

F = iLXB

By Newton's 2nd law:Fnet= ma

Fnet = F - mg sin 26 = 0,....=>constant velocity: a = 0

iLxB = mg sin 26 = 0.28*9.81*0.438 = 1.2 N

The angle between the vector B and length L is: 90 - 26 = 64 deg.
Therefore, the cross-product of the two vectors is:

LxB = |L| |B| sin (90 - 26) = 1.2*0.5 sin 64 = 0.6*0.899 = 0.539

The loop current is:

i = mg sin 26 / LxB = 1.2 / 0.539 = 2.23 A

The emf is:

emf = BLv = 0.5*1.2 v = iR

The velocity of the bar is:

v = iR / BLv = 2.23*3.9 / 0.5*1.2 = 14.5 m/s

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