A copper rod is placed on two parallel, horizontal conducting rails PQ and SR as

Question

A copper rod is placed on two parallel, horizontal conducting rails PQ and SR as shown below The conducting rails arc connected to a battery and switch X. The rails and the copper rod arc in a region of uniform magnetic field of strength B The magnetic field is normal to the plane of the conducting rods as shown in the diagram below. The length of the copper rod between the rails is I. The mass of the copper rod is M. Friction between the copper rod and the rails is negligible. The switch X is now closed and the current in the copper rod is I and in the direction shown in the diagram. On the diagram, draw an arrow to show the direction of the force F on the copper rod. Derive an expression in terms of B, L, M and I, for the initial acceleration a of the copper rod.

Explanation / Answer

F = I L X B where L X B is the vector cross product L is in the direction of the current I So the force will be directed towards the left. F = M a and a = I L B / M (I is also in the direction that positive charges would move and F = q v X B where v is the speed of the charges in the wire, again giving the force to the left)

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