A 8.10cm diameter wire coil is initially oriented so that its plane is perpendicular to a magnetic field of 0.710T pointing up. During the course of 0.120s, the field is changed…

Question

asked Sep 28, 2021 145k views

1 Answer

Chris Fei · Answer 1 · 2021-10-01T13:42:37+0000

Answer:

The average induced emf in the coil is 20.19 mV

Step-by-step explanation:

Given;

diameter of wire, d = 8.10cm = 0.081 m

initial magnetic field strength, B₁ = 0.71 T

final magnetic field strength , B₂= 0.24 T

time = 0.12 s

According to Faraday's law, average induced emf in the coil is directly proportional to the rate of change of flux.

$E.M.F= (\delta \phi)/(dt) = (BA)/(t)$

where;

A is the area of the wire = ¹/₄ (πd²)

= ¹/₄ (π x 0.081²) = 0.005154 m²

$E.M.F = ((0.71-0.24)*0.0053)/(0.12) =0.02019 \ V = \ 20.19 \ mV$

Therefore, the average induced emf in the coil is 20.19 mV