Study the diagram and answer the following questions:
(a) In the given circuit when the key is closed, the galvanometer in the coil B shows momentary deflection. Give a reason for this deflection.
(b) How will the galvanometer deflect if the variable resistance in the closed circuit is decreased? (in the same direction / in the opposite direction)
(c) Give a reason for your answer to (b).
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(a) When the key is closed, the current in the coil A grows from zero to its steady value, so the magnetic field produced by it changes. Since both the coils are wound on the same soft iron core, the magnetic flux linked with the coil B also changes and an e.m.f. is induced in it by electromagnetic induction. As the circuit of B is closed through the galvanometer, an induced current flows in it and the galvanometer shows a deflection.
The deflection is only momentary because once the current in A becomes steady, the flux linked with B no longer changes and the induced e.m.f. falls to zero.
(b) In the same direction.
(c) On decreasing the variable resistance, the current in the coil A increases. This increases the magnetic flux linked with the coil B, i.e., the flux again increases just as it did at the moment of closing the key.
Since the change in flux is of the same kind as before, by Lenz's law the induced current in B flows in the same direction as earlier, and hence the galvanometer deflects in the same direction.