This document summarizes the calculation of induced emf using Faraday's law of induction for two cases:
1) A rectangular loop of wire moving with constant velocity v perpendicular to a magnetic field B produced by a current-carrying wire. The induced emf is calculated to be (μ0/4π)*(2Ia)*(v/r).
2) The induced electric field E within a rectangular loop of wire a distance c from a current-carrying wire is calculated to be (μ0/4π)*(2Ia)*(1/c+1/c+a).
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whore A SinG ssed as u LA of we loop constant velocity v- Calculate th.docx
1. whore A SinG ssed as u LA of we loop constant velocity v. Calculate the induced emf using (a)
The Faraday's law of inductionse the connhe between induced emf and the induced electric field
E. (5-points) uot
Solution
a) Consider a small area of the rectangle having length a and width b, which is placed at a
distance r from the current carrying wire .
Magnitude of the magnetic field due to a current carrying l ,B=( u 0 /4?)2I/R
The small amout of magnetic flux asociated with rectangular loop,d? B = ( u 0 /4?)*(2I a dr/r)
Therefore,amount of magnetic flux linked with the rectangular loop is
d? B = ( u 0 /4?)*(2Iadr/r)
intregrating distance dr taking limit r and r+L
now,the total amount of magnetic flux linked with the square loop can be calculated as
d? B = ( u 0 /4?)*(2Ia)*(log e (r+b/r))
according to faraday law of induction,
induced emf =d? B = ( u 0 /4?)*(2Ia)*(log e (r+b/r))
b)speed of motion of the rectangular loop, v=v
distance between the wire and loop, c having length a, so the induced emf on account of its
motion in the magnetic field of the current carrying straight wire is given by
induced emf =Blv=Bav
induced emf = ( u 0 /4?)*(2Ia)*(v/r)
2. For finding induced electric field E, the induced emf is given by the line integral,
induced emf =line integral of electric field *small length(dl) cos (theta)
Induced electric field =induced emf /distance along electric field
= ( u 0 /4?)*(2Ia)*(1/c+1/c+a)