Question

Consider the situation shown in figure. The wire PQ has a negligible resistance and is made to slide on the three rails with a constant speed of 5 cm s⁻¹. Find the current in the 10 Ω resistor when the switch S is thrown to (a) the middle rail (b) the bottom rail.

Answer 1

Given:
Magnetic field, B = 1 T
Velocity of the sliding wire, v = 5 × 10⁻² m/s
Resistance of the connected resistor, R = 10 Ω

(a) When the switch is thrown to the middle rail:
Length of the sliding wire = 2 × 10⁻² m
Induced emf, E = Bvl
= 1 × (5 × 10⁻²) × (2 × 10⁻²) V
= 10 × 10⁻⁴ = 10⁻³ V

Current in the 10 Ω resistor is given by
$$\begin{gathered} i=\frac{E}{R} \\ =\frac{{10}^{-3}}{10}={10}^{-4}=0.1\mathrm{mA} \end{gathered}$$

(b) When the switch is thrown to the bottom rail:
The length of the sliding wire becomes 4 × 10⁻² m.
The induced emf is given by
E = Bvl'
= 1 × (5 × 10⁻²) × (4 × 10⁻²)
= 20 × 10⁻⁴ V
Now,
Current, i = $\frac{20\times {10}^{-4}}{10}$ A
= 2 × 10⁻⁴ A = 0.2 mA