What is Ohm's law for induced current?

Question

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?

Answer 1

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.

Answer 2

For the first question: there is only one current and it is generally not possible to distinguish the induced current from the current associated with the generator. In general, the induced electric field modifies the distribution of charges on the surface of the conductor and these charges contribute to the electric field and thus to the current.

For the second question: Faraday's law $e=-d\phi/dt $ assumes that we can define the magnetic flux $\phi$ through the circuit. Therefore, we must use the thin wire approximation. In this case, the flux can be considered as a the same through all the loops. (Note that for the calculation of the inductance, it is not possible to assume that the circuit is infinitely thin!)

If this approximation is not valid, it is much more complicated. The skin effect must be taken into account and the current is no longer distributed uniformly over the conductor cross-section. Except in special cases, Maxwell's equations must be solved in the (magneto)quasistatic approximation.

Hope it can help and sorry for my poor english.

Answer 3

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.

Answer 4

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 "