Question

A flat loop of wire consisting of a single turn of cross-sectional area 8 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.5 T to 2.5 T in a time of 1.19 s. The total resistance is 0.719 Ω. What is the resulting induced current? Answer in units of mA.

Answer 1

Given Data

No. of turns N = 1

change in magnetic field dB = B₂ - B₁

                                         = 2.50 T - 0.500 T

                                         = 2.00 T

Corresponding time    dt = 1.19s

Area of loop   A = 8.00 cm²   = 8.00 * 10^-4 m²

Solution :-

Induced EMF    E = - dφ/dt = - N * A * dB/dt

E = - N * A * dB/dt

E = - 1 * 8.00 * 10^-4 * 2.00 / 1.19

   = - 1.344 * 10^-3     V

   = - 1.344 mV

- ve sign can be discarded as it indicates direction only.

Current I = E / R

                     = 1.344 mV / 0.719 Ω

                     = 1.87 mA    --> Answer