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.

Answer 1

(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