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Question
Consider the situation shown in figure. The wires P1Q1 and P2Q2 are made to slide on the rails with the same speed 5 cm s−1. Find the electric current in the 19 Ω resistor if (a) both the wires move towards right and (b) if P1Q1 moves towards left but P2Q2 moves towards right.
Solution
(a) When both wires move in same direction:-
The sliding wires constitute two parallel sources of emf.
The net emf is given by
e = Blv
⇒ e = (1 × 4 × 10−2 ) × 5 × (10−2)
= 20 × 10−4 V
The resistance of the sliding wires is 2 Ω.
∴ Net resistance = \[\frac{2 \times 2}{2 + 2}+19=20\Omega=\frac{2 \times {10}^{- 4}}{20}=0.1\text {mA}\]
(b) When both wires move in opposite directions with the same speed, the direction of the emf induced in both of them is opposite. Thus, the net emf is zero.
∴ Net current through 19 Ω = 0
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