Question

A rod, PQ, of length l , slides with a constant velocity on two conducting rails. The system of the rod and conducting rails, is present in a region where a uniform (normal) magnetic field, of strength B, is present. The two ends of the rod PQ, (rod’s resistance = R) are joined to the wheat stone bridge, abcd, shown here. If the current in the arms and and dc, of the wheat stone bridge, equals i₀ , the velocity, v, of the sliding rod, equals

(1) \(\frac{16Ri_0}{3Bl}\)

(2) \(\frac{16Ri_0}{Bl}\)

(3) \(\frac{Bl}{(16Ri_0)}\)

(4) \(\frac{3Bl}{(16Ri_0)}\)

Answer 1

 (1) \(\frac{16Ri_0}{3Bl}\)

The emf, induced in the rod, equal Blv where, v, is the (constant) velocity of the sliding rod. 

The wheat stone bridge, abcd, as shown is a balanced bridge. Its equivalent resistance, r, is given by

This gives r = 3R 

The total resistance of the circuit is, therefore (R+3R) = 4R

∴ The induced current, in rod PQ, equals Blv/4R

In the wheat stone bridge, the part, i₀ , going into the arms (ad and (dc), is 

 

Of this current.