Question

50 cm to mA flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 2.0 T to 6.0 T in 2.0 s. If the coil has a total resistance of 0.40 Ω, what is the magnitude of the induced current?

Answer 1

Answer:

0.025 A

Explanation:

A = 50 cm^2 = 50 x 10^-4 m^2

B2 = 6 T, B1 = 2 T

db = 6 - 2 = 4 T

dt = 2 s

R = 0.4 ohm

Let i be the magnitude of induced current and e be the induced emf.

According to the Faraday's law of electromagnetic induction

e = dФ / dt

e = A dB / dt

e = 50 x 10^-4 x 4 / 2 = 0.01 V

i = e / R = 0.01 / 0.4 = 0.025 A