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

\[i = \frac{E}{R}\]

\[ = \frac{{10}^{- 3}}{10} = {10}^{- 4} = 0 . 1 mA\]

(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