A square frame of side 10 cm and a long straight wire carrying current 1A are in the plate of the paper. Starting from close to the wire, the frame moves towards the right with a constant speed of 10ms⁻¹ (see figure).

The e.m. finduced at the time the left arm of the frame is at x = 10 cm from the wire is

Concept:

The e.m.f. induced in a moving conductor is given by the formula:

e.m.f. = - (dΦ / dt)

Where Φ is the magnetic flux, and the flux through a moving conductor is expressed as:

Φ = (μ₀ * i * a) / (2π) [ ln(x + a) - ln(x) ]

Where:

• μ₀ = Permeability of free space
• i = Current through the wire
• a = Length of the side of the square loop
• x = Distance of the left arm of the frame from the wire

The e.m.f. induced is calculated by differentiating the flux with respect to time.

Calculation:

Given: Current i = 1 A, Distance x = 10 cm, Speed of movement v = 10 m/s, and Side of the square a = 10 cm.

e.m.f. = - (dΦ / dt) = (μ₀ * i * a) / (2π) * [(1 / (x + a)) * dx/dt - (1 / x) * dx/dt]

Substitute the values:

e.m.f. = (μ₀ * i * a) / (2π) * [(1 / (x + a)) * v - (1 / x) * v]

e.m.f. = (μ₀ * i * a * v) / (2π) * [(1 / (x + a)) - (1 / x)]

Substituting μ₀ = 4π × 10⁻⁷ T·m/A, i = 1 A, a = 0.1 m, and x = 0.1 m:

e.m.f. = 2 × 10⁻⁷ × ((0.1)² × 10) / (0.1 × 0.2) = 1 μV

∴ The induced e.m.f. is 1 μV.

Hence, the correct answer is option 2.