# Why are small planar current loops of sufficiently small size interesting? (magnetic moments)

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.

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

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.

• The way they've written it, it seems that all loop shape should give same integral
– Babu
Commented Oct 19, 2020 at 10:06
• Do you have experience with contour integrals? The notation may be misleading here. Commented Oct 19, 2020 at 10:14
• 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
– Babu
Commented Oct 19, 2020 at 10:17
• 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 Commented Oct 19, 2020 at 10:20
• Anyhow, the idea is that you split the loop in small elements and add up their contributions. 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.

• Ok, I am a bit puzzled how one would arrive at the direction since the left side is of scalars
– Babu
Commented Oct 19, 2020 at 14:42
• 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. Commented Oct 19, 2020 at 14:47
• So, 1. is my derivation correct, 2. how do I find direction now?
– Babu
Commented Oct 19, 2020 at 14:48
• 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. Commented Oct 19, 2020 at 15:43
• 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
– Babu
Commented Oct 19, 2020 at 15:45