A horizontal rod PQ of mass 10g and length 0.10m is placed on a smooth plane inclined at 60° to the horizontal. A uniform vertical magnetic field of value B is applied in the region PQ. Calculate B if the rod remains stationary on the plane when a current of 1.73A flows in the rod. What is the direction of the current in the rod?
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Answer:
To balance the forces, the force on the rod due to the B-field, FB, is in the direction as shown:
FB = BIL
If the rod remains stationary, resolving forces,
FB cos 60° = W sin 60°
B (1.73) (0.1) (0.5) = (0.01kg * 9.81) (√3 / 2)
B = 0.982 T
For FB to be in the direction as shown, by Fleming's Left Hand rule, the direction of current in the rod should be out of the paper.
