In the book Basic Laws of Electromagnetism by Igor Irodov, He introduces force on a closed current-carrying loop as;

$$ F = \oint I \vec{dl} \times \vec{B}$$

And then he states the most interesting case of analysis is that of a small planar current loop of sufficiently small size and introduces a quantity $p_m$ to study this where $p_m$ is defined as:

$$ p_m = IS \vec{n}$$

Where $I$ is current, $S$ is the area bounded by the loop, and $ \vec{n}$ is normal to the loop and supposedly the force on it is given by:

$$ F= p_{in} \frac{ \partial B}{\partial n }$$

Where we are taking the directional derivative of $ \vec{B}$ in direction of the normal.

I have a few questions about this :

  1. Why are small current loops interesting?
  2. How did he go from the line integral into the other expression with force? (*)

*: I couldn't find derivations of this formula but I tried to attempt a derivation on my own:

I realized that stokes are not applicable here because the integral is a cross product instead of a dot. Hence, I thought of using what I learned in one form (??) and assumed a planar loop and field perpendicular to x-y plane.

$$ F = I \oint B dx -B dy $$

This looks like a flux one form but I'm not sure what to do after this.

??: I am not completely sure if the expressions I'm using are valid.

The book can be found here, See page-158


2 Answers 2


I think the author's main point is to introduce the concept of a magnetic moment (similar to a dipole moment) which can be thought of as a small current loop. The real magnetic moments (such as spin) are not really current loops... so it is just a pedagogical tool.

Regarding evaluating the moment - the integral here is a contour integral, so you have to choose a shape of your loop, parametrize it appropriately, and integrate along this contour. Square loop might be easier, while a circular loop might be a bit less artificial.

  • $\begingroup$ The way they've written it, it seems that all loop shape should give same integral $\endgroup$
    – Babu
    Commented Oct 19, 2020 at 10:06
  • $\begingroup$ Do you have experience with contour integrals? The notation may be misleading here. $\endgroup$
    – Roger V.
    Commented Oct 19, 2020 at 10:14
  • $\begingroup$ I'm not sure of the precise definition but I think it means a line integral over a loop in this context. The confusion is due to the fact that contour integrals is usually used to describe integrals in the complex plane $\endgroup$
    – Babu
    Commented Oct 19, 2020 at 10:17
  • $\begingroup$ Sorry, I might have used a wrong term: look up line integrals (contour integrals in the complex analysis are a particular case of those): en.wikipedia.org/wiki/Line_integral $\endgroup$
    – Roger V.
    Commented Oct 19, 2020 at 10:20
  • $\begingroup$ Anyhow, the idea is that you split the loop in small elements and add up their contributions. $\endgroup$
    – Roger V.
    Commented Oct 19, 2020 at 10:21

The derivation of the formula:

$$ F = \oint I \vec{dl} \times \vec{B} = - I \oint \vec{B} \times \vec{dl}$$

Now using identity in this MSE post

$$ \oint \vec{B} \times \vec{dl} = \int_S (\nabla \cdot \vec{B}) \hat{n} - (\nabla B_i) n_i dS$$

The first term vanishes due to maxwell's identity that $ \nabla \cdot B=0$ and the second term is the directional derivative along the normal of the contour:

$$ \oint \vec{B} \times dl = -\int_S \frac{ \partial B}{\partial n} dS$$

Now for a 'small enough' loop, the above expression reduces to multiplying the rate of change of B along normal with the surface area. Plugging this back into our original expression:

$$ F = I \frac{ \partial B}{\partial n} dS$$

Which is the identity the book has written.

  • $\begingroup$ Ok, I am a bit puzzled how one would arrive at the direction since the left side is of scalars $\endgroup$
    – Babu
    Commented Oct 19, 2020 at 14:42
  • $\begingroup$ The identity in question is a form of Stokes' theorem - it is a rather essential stuff in line integrals. The output is a vector, so Irodov takes some liberties in this case - he focuses on its magnitude. $\endgroup$
    – Roger V.
    Commented Oct 19, 2020 at 14:47
  • $\begingroup$ So, 1. is my derivation correct, 2. how do I find direction now? $\endgroup$
    – Babu
    Commented Oct 19, 2020 at 14:48
  • $\begingroup$ It is against the rules of Physics StackExchange to (ask to) correct/solve homework problems. It is okay to ask for help, if something blocks you - I think I gave you the key to the problem. $\endgroup$
    – Roger V.
    Commented Oct 19, 2020 at 15:43
  • $\begingroup$ I do get your point, but I came to the realization that the quantity maybe more complex than what was thought initially. I made a post about it on MSE here $\endgroup$
    – Babu
    Commented Oct 19, 2020 at 15:45

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