Question

(20% 2.) we have an infinite line current that generates a magnetic field-y sin(a)in free space 2 rx (e, ), in the vicinity of a square loop located on the xz plane. Each side of the square loop has a length of l cm, and its left bottom, vertex is located at (x = 1 cm, z = 0). The loop carries a resistor with a resistance R 1 Ω, and the magnitude of the induced current flowing through R is 0.87A. Find the angular frequency a 1 cm a A .8 L Cm lc nt (1cm,o) 01广

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Answer 1

This is nothing but the application of Faraday's Law of electromagnetic induction. Follow the steps shown.

The induced e.m.f by Faraday�s Law, E = -N d phi/dt where phi = integral B^vector middot dA^vector Here no of them = N = 1 B^vector = -y^^mu_0/2 pi x sin wt dA^vector is infinitesimal area vector in x direction. i.e dA^vector = 0.01 dx y^^ therefore phi = integral^0.02_0.01 B^vector middot dA^vector = = -0.01 mu_0/2 pi sin(wt) integral^0.02_0.01 dx/x = -0.01 mu_0/2 pi sin wt [ln w]^0.02_0.01 = -0.01 mu_0 ln 2/2 pi sin wt putting mu_0 = 4 pi times 10^-7 H/m phi = -(1.386 times 10^-9)sin wt \r\n therefore E = -N d phi/dt = (1.386 w times 10^-9)cos wt therefore I = E/R = (1.386 w times 10^9)cos wt It is given magnitude of I = 0.87A which is taken as r.m.s value under mentioned. Therefore 1.386 w times 10^-9/squareroot 2 = 0.87 w = 0.8877 times 10^9 rad/s

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