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In this book it is written on pg $313$ in the last paragraph that Ohms law i.e. $R (constant)=\frac{\epsilon_{ind}}{I}$ is valid for induced current in a circuit. They define $R$ to be the sum of the resistance of all the resistive elements part of the circuit, $I$ to be the current and $\epsilon_{ind}$ to be the induced EMF. I have two doubts related to the meaning of the terms $I$ and $\epsilon_{ind}$.

  1. What current does $I$ represents? Does it represent induced or net current through the circuit? Suppose in a circuit there is a battery connected as well, the circuit is kept in a region where its magnetic flux changes. The battery and changing magnetic flux will both produce current. Does $I$ represent net current through the circuit i.e. the net sum of currently produced by battery and flux or just induced current?
  2. Around what loop is $\epsilon_{ind}$ calculated? There can be infinitely many closed lines along which we can calculate $\epsilon_{ind}$, then for which loop does $\epsilon_{ind}$ corresponds? See the diagram. The two black lines represent curves passing on the surface of a wire, the green and blue lines represent a loop inside the wire, and the red lines represent an uniform magnetic field that is increasing. I can calculate $\epsilon_{ind}$ along blue, green and also along the two black loops. But whose $\epsilon_{ind}$ is to be used in the formula? enter image description here
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3 Answers 3

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Given the loop $\partial S$ that is the boundary curve of a simple otherwise arbitrary surface $\mathcal S$ and calculate the magnetic flux $\Phi$ through this surface defined by $\Phi = \int_{\mathcal {S}} \mathbf B \cdot d\mathbf{S}$. According to Faraday's induction there is an emf $\mathcal V$ induced along the perimeter$\partial \mathcal {S}$ : $\mathcal V = \oint_{\partial S} \mathbf E \cdot d\mathbf{\ell} = -\frac{\partial \Phi}{dt}$. This $\mathcal V$ exists as a contour integral irrespective whether there is any current anywhere but if $\partial \mathcal S$ is along an honest to goodness conductor, a metal wire, then this emf will make the charges move inside the wire, so if the wire's resistance is $R$ then there will be a current $I$ flowing such that $I=\frac{\mathcal V}{R}$. So the current will depend not only on its resistance and by the time varying magnetic field but also on the shape of the wire and its disposition relative to the field. Important to note that since the $\mathbf {B}$ field is solenoidal, $\nabla \cdot\mathbf{B}=0$, the flux $\Phi$ itself is independent of the details of the spanning surface, instead geometrically the flux $\Phi$ depends only on the boundary curve $\partial \mathcal {S}$. With different loops you get different flux and emf.

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According to the book, there is an AC current in the circuit, induced in the coil II by a coil I. The differential equation for the circuit is: $$L\frac{\partial I}{\partial t} + RI = emf$$

If L (the inductance of the coil) is too low compared to R, the circuit can be considered as basically resistive, and $emf = RI$. The current is always all current of the circuit.

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  • $\begingroup$ But along which curve is emf measured? $\endgroup$
    – Osmium
    Feb 4 at 1:07
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The induced current is due to the permeability of the wire not the resistance. So the reaction of the wire to induced current measured in ohms . depends on the self inductance of the wire and the frequency of the "current "

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