Q15 of 805 Page 1

The figure shows a metal rod PQ of length l, resting on the smooth horizontal rails AB positioned between the poles of a permanent magnet. The rails, rod and the magnetic field B are in three mutually perpendicular directions. A galvanometer G connects the rails through a key ‘k’. Assume the magnetic field to be uniform. Given the resistance of the closed loop containing the rod is R.

(i) Suppose K is open and the rod is moved with a speed v in the direction shown. Find the polarity and the magnitude of induced emf.


(ii) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience a magnetic force due to the motion of the rod. Explain.


(iii) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?


1) When the rod is moved with a velocity v, there is a force acting on the rod due to a magnetic field which induces emf in the rod which is equal to e = Blv


From right-hand thumb rule, we can find out that the direction of emf is A to B as A is positive and B is negative.


2) There is no net force when K is open. This is because the magnetic force Fm = -e(v + B) is cancelled by the electric force


Fe = eE which is set up due to excess of charge of opposite sign at the ends of the rod( when the circuit was closed).


3) Zero. In this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.


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