A single loop of wire with N = 1 is exposed to a uniform magnetic field changing such that the flux Φ = B A; If B is increasing at rate dB/dt = 0.8 T/s and area A = 0.5 m^2, what is the induced emf?

A single loop of wire with N = 1 is exposed to a uniform magnetic field changing such that the flux Φ = B A; If B is increasing at rate dB/dt = 0.8 T/s and area A = 0.5 m^2, what is the induced emf?

• A. 0.4 V
• B. 0.2 V
• C. 0.8 V
• D. 0.6 V

Answer: (A)

The main idea is Faraday’s law: the induced emf in a loop equals the negative rate of change of magnetic flux through the loop. For a single loop with area A in a uniform field, the flux is Φ = B A. If the area stays the same and B changes in time, dΦ/dt = A dB/dt. The induced emf is ε = -N dΦ/dt, where N is the number of turns (here N = 1). Plugging in the values: dΦ/dt = A dB/dt = 0.5 m^2 × 0.8 T/s = 0.4 Wb/s = 0.4 V. Therefore ε = -0.4 V, and the magnitude of the induced emf is 0.4 V. The negative sign (per Lenz’s law) indicates the direction of the induced current opposes the change in flux.

The main idea is Faraday’s law: the induced emf in a loop equals the negative rate of change of magnetic flux through the loop. For a single loop with area A in a uniform field, the flux is Φ = B A. If the area stays the same and B changes in time, dΦ/dt = A dB/dt. The induced emf is ε = -N dΦ/dt, where N is the number of turns (here N = 1).

Plugging in the values: dΦ/dt = A dB/dt = 0.5 m^2 × 0.8 T/s = 0.4 Wb/s = 0.4 V. Therefore ε = -0.4 V, and the magnitude of the induced emf is 0.4 V. The negative sign (per Lenz’s law) indicates the direction of the induced current opposes the change in flux.