1 points SerCP10 20.P.029. Ask Your Teacher My Notes Question Part Points Submis

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1 points SerCP10 20.P.029.
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The figure below shows a bar of mass m = 0.240 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 = 30.0° with respect to the ground. The resistance of the resistor is R = 1.20 , 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?

m/s

Explanation / Answer

Component of B perpendicular to the plane
= Bcos30°
=> emf induced in the coil
= Bcos30° * 1.2 * v =
=> current in the bar,
I = /R = (v/R) * 1.2 Bcos30°
Force opposing the downward motion
F = I * L * Bcos30° = (v/R) * (1.2 Bcos30°)^2

As the bar starts sliding, v keeps on increasing and so does F till F balances the downward component of the weight of the bar
=> mgsin30° = (v/R) * (1.2 Bcos30°)^2
=> velocity, v
= (mgRsin30°) / (1.2 Bcos30°)^2
= (0.240 * 9.81 * 1.2 * sin30°) / (1.2 * 0.5 * cos30°)^2
= (0.8292) / (0.2957)
= 5.23m/s.

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