he figure below shows a bar of mass m = 0.270 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 ? = 29.5° with respect to the ground. The resistance of the resistor is R = 3.30 Ω, 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

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The figure below shows a bar of mass m = 0.270 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 ? = 29.5° with respect to the ground. The resistance of the resistor is R = 3.30 Ω, 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
7:57
A webassign.net
below shows a bar of ma
a pair of rails separated I
I plane that makes an ang
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B = 0.500 T is directed
ntire region through whic
pes the bar slide along the
B
Transcribed Image Text:7:57 A webassign.net below shows a bar of ma a pair of rails separated I I plane that makes an ang ance of the resistor is R = B = 0.500 T is directed ntire region through whic pes the bar slide along the B
Expert Solution
Step 1

The bar is sliding down with the constant velocity, hence the acceleration of the bar is zero. That means the net force on the bar is zero.

There are two forces acting on the bar, one is gravitational force and the other one is the magnetic force due to the applied magnetic field.

The free-body diagram for the bar is shown in the figure below:

Advanced Physics homework question answer, step 1, image 1

As the bar slides down, there is an induced emf in the loop, and the induced emf is given by

ε=Blv........1B-perpendicular component of the magnetic field in the plane of the loopl-length of the barv-velocity of thebar.ε=Bcosθ lv.......2

Due to this induced emf, there is an induced current I in the loop and hence the loop experiences a magnetic force.

Fm=IBlWhere I=εR=BcosθlvRFm=B2l2vcosθR.......3

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