# Why is the formula written this way?

I am enrolled in AP Physics C, and my class is currently covering electricity and magnetism. When answering free-response questions, we are to initiate all of our calculations with a relationship from the provided formula sheet. The given equation for the force on a wire carrying current through a magnetic field is

$$\vec F = \int I \, d\vec\ell \times \vec B$$

where the quantity $d\vec\ell$ has a magnitude equal to an infinitesimally small portion of wire length and a direction parallel the current.

My instructor noted that when calculating the force on a straight wire in a uniform field, the formulae $\vec F = \ell \vec I \times \vec B$ and $F = BI\ell$ are more intuitive and memorable.

My question is, why not simply write $$d\vec F = \vec I \times \vec B \, d\ell$$ especially when the formula for Biot-Savart law is provided in differential form? From the point of view of a student, this formula is more primed for application to both uniform and variable situations as well as is more representative of the natures of the vectors at play.

I have a few hypotheses of my own:

1. Rigor: It is possible that they provided us with a less clear formula to test if we genuinely know our physics. This could be evidenced by lack of subscripts, while subscripts appear in other formulae.
2. Consistency: The immediately preceding formula is $d\vec B =\frac{\mu_0}{4\pi}\frac{I \, d\vec\ell \times \hat r}{r^2}$, Biot-Savart law. Notice that the same unintuitive cross product appears. I am not sure if this is the most common way to write the formula, but the repetition is surely not coincidental.
3. Mathematical grammar: It occurs to me that some might object to multiplying a vector and a scaler in the order $\vec{v} k$ rather than $k\vec v$. Again, I am not sure if this is widely considered disagreeable or unpalatable notation.
• – ZeroTheHero Apr 16 '17 at 22:52
• There is also the issue that Newton's 3rd law does not hold for the force between the differential current elements but it does hold when integrated for the full loops. – hyportnex Apr 16 '17 at 22:58
• @hyportnex Why is that? Could you elaborate? – gen-z ready to perish Apr 16 '17 at 23:00
• Before I answer I need a clearification: Did you intend to change $I$ to $\vec I$ (from a scalar to a vector), or was that a typo? In other words, is your question why the formula is written as an integral-form giving $\vec F$ instead of as a differential form giving $d\vec F$? Or is the question why $d\vec l$ is defined as a vector and instead of $I$? – Steeven Apr 16 '17 at 23:01
• The differential form of the Biot-Savart law would be $d\textbf{F}_{12}=\frac{\mu_o I_1 I_2}{4\pi r^2} d\textbf{l}_1 \times (d\textbf{l}_2 \times \textbf{r})$. This is the force that element 2 effects on element 1. When you interchange the indices 2 with 1 to find the force element 2 feels you do not get the same pointing in the opposite direction. But when you integrate over the full loop of each current the aggregate forces do satisfy Newton's 3rd law: action= reaction. (The situation can remedied by the introduction of the EM field momentum but that is another story.) – hyportnex Apr 16 '17 at 23:47

The thing is, there is no such thing as a current segment, either finite or infinitesimal. It's a nice fiction to think of $I\mathrm d\boldsymbol\ell$ as a small line segment carrying a current $I$, but that segment by itself would quickly accumulate a huge electrical charge on both ends, which immediately takes you out of the static regime you're meant to be describing. The formula works for the whole circuit because there you don't have open leads that would accumulate charge, and that is the only way to have it make sense.

It sounds a bit nit-picky, but it is a pretty important aspect, particularly since you claim that you can do essentially the same thing with the Biot-Savart law. To put things in perspective, keep in mind that the Biot-Savart law is essentially equivalent to Ampère's law, and that law has the conservation of charge baked into it: if you take the divergence of the differential form $$\nabla\times\mathbf B = \mu_0\mathbf J,$$ you immediately get the continuity equation $\nabla\cdot\mathbf J=0$ for the current. (An analogous form of that argument exists for the integral version of that law, of course.) A finite current segment breaks the continuity equation, which means that its magnetic field as you have defined it does not (can not) obey Ampère's law at all, and that by itself is pretty deadly.

With the Lorentz force law this is not quite as deadly, but it's still something one needs to be careful with. You could write that $$\mathrm d\mathbf F = I\mathrm d\boldsymbol\ell \times\mathbf B$$ is the differential of force that's experienced by a differential segment $d\boldsymbol\ell$ of circuit carrying a current $I$ in a magnetic field $\mathbf B$, so long as you remember that that differential only ever ever ever makes sense when it's part of a full integral. Once that is clear in the subtext, then yes, that's something people do on occasion. On a pedagogical setting, though, the chance for confusion is too great and it's best avoided.

From the point of view of a student, this formula is more primed for application to both uniform and variable situations as well as is more representative of the natures of the vectors at play.

It might be "more representative of the natures of the vectors at play", yes, but I don't agree that it "is more primed for application". Do you really find it easier / more intuitive for application and use in real-life situations to have a differential form like:

$$d \vec F=I \;d\vec l\times \vec B$$

$$\vec F=\int I \;d\vec l\times \vec B$$

The latter expression tells the total force on whatever-we-look-at, while the former tells the force on each tiny, tiny piece of it. For engineering applications you very often just need the entire force pushing on a wire or pulling in a block e.g. - you don't care how much each centimetre is being pushed individually. But they are equal anyways:

$$d \vec F=I \;d\vec l\times \vec B\quad\Leftrightarrow \quad\int d \vec F=\int I \;d\vec l\times \vec B \quad\Leftrightarrow \quad \vec F=\int I \;d\vec l\times \vec B$$

So how the definition is presented is not really important.

To the other part of the question about why the length $d\vec l$ is defined as a vector while the current $I$ isn't, there is being asked another question as we speak, which might attract answers that will help you.