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TSny
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I'm not following this. Suppose you let ##\varepsilon## equal the rate of change of magnetic flux through the circular loop, which is assumed to be given. Then Faraday's law for the circular loop is ##\oint Edl = \varepsilon##.Amplitude said:Assume iL flows from A to B , iR flows from B to A , iv flows from B to A via voltmeter .
Moving anticlockwise in the circular loop and applying KVL ,
-iL r + ∫ELdl -iR r + ∫ERdl = 0
This may be written as ##\int_{ALB} Edl + \int_{BRA} Edl = \varepsilon##. Then substitute ##\int_{ALB} Edl = i_{_L} r## and ##\int_{BRA} Edl =i_{_R} r##, where ##r## is the resistance of half the circular loop. This will give you one equation involving the unknowns ##i_{_L}## and ##i_{_R}##.
Similarly, get a second equation using loop BVAB.
Finally, use the junction rule for currents to get a third equation.