This post consists of two questions both relating to the directional aspects of emf, in some way. I could not include both the questions in the title, so I chose the most troubling one.

Introduction: To understand the questions we need to understand the meaning of induced emf for a loop at rest. For such loop induced emf is defined as the work done by net NON-conservative electric field on a charge divided by the charge. Hence it is a scalar quantity. Mathematically: $\epsilon =\oint_C\vec{E}\cdot d\vec{l}.$ But what is $d\vec{l}?$ It is a vector representing an element on the loop of length $dl$ and the direction of its direction is tangential to the curve. But there exist two such directions, both opposite to each other (see the fig). Hence we can calculate induced emf $\epsilon$ in two ways; one in an anticlockwise way and the other in an anticlockwise way. enter image description here

$1^{st}$ question: Faraday's law states that $\epsilon =\oint_C\vec{E}\cdot d\vec{l}=-\frac{d\phi_m}{dt}$ where $C$ represents a loop. Which emf out of two is used in Faraday's law?

$2^{nd}$ question: In many books and articles "direction of induced emf" is used. What does the direction of induced emf mean? It's a scalar quantity and hence doesn't have direction in the sense that vectors have it. What does direction mean here? What does it mean, physically, when we say that induced emf is in a clockwise direction as seen from a point in a loop? Does induced emf here mean current?

Note: This question originally consisted of this question as well, but I found that it made my question too lengthy and confusing.


3 Answers 3


You can get an induced emf with no resulting induced current and hence no "opposition" but then no work is done. Thus any mention of induced current should be thought of as considering what might happen if an induced current was allowed to flow.

Your question about direction is resolved if you follow the right-hand convention as shown in the diagram below.

enter image description here

Decide on the direction of the normal unit vector to the area $\hat n$ and then the positive direction for the line integral around loop $C$ is decided by the right-hand rule, eg thumb of the right hand in direction of the normal unit vector and curled fingers of right hand give positive direction for the line integral.

Suppose that the magnetic field is in the direction of $\hat n$ and increasing then $\vec B\cdot \hat n$ is positive and so the right-hand side of Faraday's law, $-\frac {d\phi}{dt}$, will be negative.
The line integral on the left-hand side, the emf $(\displaystyle \oint_{\rm C} \vec E\cdot d \vec l)$ will thus be negative ie in a clockwise direction looking from the top in the diagram.
That will also be the direction of the induced current which is consistent with Lenz.

$\displaystyle \oint_{\rm C} \vec E\cdot d \vec l$ is the work done by the electric field on a unit positive charge when the charge goes around a complete loop which is at rest.
In my example, the electric field direction is clockwise which means the induced current (movement of positive charges) is clockwise.
This is produced by the induced emf which is also in a clockwise direction in that it drives positive charges that way ie in the opposite direction to the arrow with the $C$ by it.

  • $\begingroup$ But still, what does the direction of induced emf mean? What physical significance does it have? $\endgroup$
    – Osmium
    Feb 6 at 9:08
  • $\begingroup$ @Osmium I have added a paragraph to try and answer your question. $\endgroup$
    – Farcher
    Feb 6 at 11:31
  1. Both methods/laws are valid for determining the direction. According to me, there is only one difference. The first method using left hand rule can be called a 'trick' as it provides no reasoning for why it works. On the other hand, Lenz's law provides a reasoning for why induced current should oppose changing magnetic field (conservation of energy).

  2. Direction of emf is physically related to the direction of the induced current. I don't understand what your question is, so I don't know if this answers your question.

  3. You can assume the direction of $d \vec l$ in any of the two directions. If you pick a certain direction, then you will know the direction of $d \vec A$ in calculating the magnetic flux using the right hand rule. It will be along the thumb if you curl your fingers along the loop in the direction of $d \vec l$. Now, you can calculate the dot products in $\oint_C\vec{E}\cdot d\vec{l}=-\frac{d\phi_m}{dt}$ and both ways would provide the correct direction.

  • $\begingroup$ I edited the post slightly, hope it made clear question 2. $\endgroup$
    – Osmium
    Feb 6 at 10:45
  • $\begingroup$ Note: points in this answer correspond to a previous version of this question which has been edited. $\endgroup$
    – Dodo
    Feb 6 at 11:32
  • $\begingroup$ What do you mean by "Direction of emf is physically related to the direction of the induced current."? $\endgroup$
    – Osmium
    Feb 7 at 0:25
  1. Both laws are correct. The -ve sign in the Faraday's law gives the direction of the emf. While the lenz's states that the direction of emf induced is such that it opposes the cause which has induced the emf. So lenz's law gives us the direction of the emf induced.

For example suppose a conducting loop has a current which is increasing with time. So the magnetic flux through the loop will increase and therefore an EMF will induce in the loop. Then according to the lenz's law the direction of EMF induced will oppose the increase in current. So the current produced (induced current) due to the EMF induced will be opposite to the actual current direction. Which in result produce an magnetic field opposite to the increasing Magnetic field produced due to actual increasing current. So both laws tells the same direction, but Faraday's law gives magnitude of the Emf induced also.

  1. By the direction, we simply mean the polarity of the EMF induced. If we assumes the induced emf like a battery then its higer polarity will be in the direction of the current induced to oppose the change in flux. But always remember that the EMF will only try to oppose, it will not oppose the cause completely. But Emf induced in a super conducter will be able to do that.

  2. It is the question of Mathematics. We always take the small element along the change. As here we are integrating along the loop so the change in length we always take along the Electric field of along the component of the electric field.


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