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Question

A conducting rod of length l and mass m is moving down a smooth inclined plane of inclination θ with constant velocity v. A current I is flowing in the conductor in a direction perpendicular to the paper inwards. A vertically upward magnetic field B exists in space. Then, magnitude of magnetic field B is

A
mgIlsin θ
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B
mgIltan θ
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C
mg cosθIl
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D
mgIl sin θ
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Solution

The correct option is B mgIltan θ
Let Fm be the force on the current carrying conductor due to magnetic field.
Let us understand the direction of this force with the help of cartesian coordinate system

mg also acts on the rod in downward direction.
The components of mg and Fm along the inclined plane can be shown as below:
As the rod is moving with constant velocity along the inclined plane.
We can say, mg sinθ=Fmcosθ
mg sinθ=BILcosθ
B=mg tanθIL

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