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Faraday law state that $$\epsilon_L=\oint \vec{E} \cdot d\vec{l}=-\frac{d\phi}{dt}=-L\frac{dI}{dt}.$$

So by kirchoff's analysis, isn't the equation would be like $$\oint \vec{E}\cdot d\vec{l}=0-IR+\epsilon_c=-L\frac{dI}{dt}$$ as potential drop in resistor and potential gain in cell

But why it is instead $$0+IR- \epsilon_c=-L\frac{dI}{dt}$$


$\epsilon_L=$emf by inductor

$\epsilon_c=$emf of cell

$\oint=$ closed integral of the circuit in clockwise direction

Asumption: inductor are made of super conducting wire (0 ohm resistance)


2 Answers 2


This is a still from Walter Lewin's video 8.02x - Lect 20 - Inductance, RL Circuits, Magnetic Field Energy.

He evaluates $\displaystyle \int \vec{E} \cdot d\vec{l}$ for each of the three circuit elements starting at the dot to the left of the current arrow at the top of the circuit digram, moving in a clockwise direction which is the same as the current direction label.

For the inductor since $\vec E=0$ the integral is zero.
For the resistor the direction of the path taken $d\vec \ell$ is the same as the direction of the electric field so the integral is positive $+IR$.
For the cell the direction of the electric field is opposite to that of the path taken and so the integral is negative, $-\epsilon _{\rm c}$.

This leads to the resulting equation, $\,0+IR-\epsilon_c=-L\frac{dI}{dt}$

Note that in this derivation Walter Lewin is evaluating $\displaystyle \int \vec{E} \cdot d\vec{l}$ which you will remember is equal to $\color {red}{minus}$ the change in potential, which as a differential equation is the electric field is equal to $\color {red}{minus}$ the potential gradient.

  • $\begingroup$ But If i analysis using potential change, isnt that when we follow the direction of current, the potential experience a drop in the resistor, which then give us -IR,while for battery we go from low potential to high potential which we experience a gain in potential, +emf? $\endgroup$
    – Pck Tsp
    Jul 21, 2022 at 10:02
  • $\begingroup$ He is working out $\displaystyle \int \vec{E} \cdot d\vec{l}$ which you will remember is equal to $\color {red}{minus}$ the change in potential, which as a differential equation is the electric field is equal to $\color {red}{minus}$ the potential gradient. $\endgroup$
    – Farcher
    Jul 21, 2022 at 10:51

You are probably confusing two different ways to derive the equation, one not using the concept of potential, and the other one using the concept.

If we are starting with Faraday's law for a closed path, we do not use the concept of potential, instead we need to express contributions to the integral of total electric field $\mathbf E$

$$ \oint \mathbf E \cdot d\mathbf{l} $$ for the oriented path colocated with the conductive path of the circuit, due to each conducting member in the circuit.

For the circuit in the picture, we have

$$ \int \mathbf E \cdot d\mathbf l = 0 ~~~\text{(for the inductor)} $$ We have zero because the inductor is made of perfect conductor, where total electric field vanishes.

$$ \int \mathbf E \cdot d\mathbf l = RI ~~~\text{(for the resistor)} $$ Here we have plus sign because in resistor, electric field is in direction of current.

$$ \int \mathbf E \cdot d\mathbf l = -\epsilon_c ~~~\text{(for the cell)} $$ Here we have minus sign, because electric field inside the cell is opposite to direction of current, which is the direction of integration. Current in a cell flows in direction of electrochemical EMF, which points against electric field there.

So the Faraday law equation becomes

$$ RI - \epsilon_c = -L\frac{dI}{dt}. $$

The other way, used in practice when solving AC circuits, is to write down directly the KVL equation, which states that sum of potential drops in a closed path equals to 0. We have

$$ \text{potential drop on the inductor} = L\frac{dI}{dt}, $$ $$ \text{potential drop on the resistor} = RI, $$ $$ \text{potential drop on the cell} = -\epsilon_c. $$

So the KVL equation is

$$ L\frac{dI}{dt} + RI - \epsilon_c = 0. $$ which is the same as the equation obtained using the first method.


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