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I'm wondering why the magnetic forces on a rectangular small current-loop of side $\delta x$ and $\delta y$, lying in the $xy$ plane are the following ones (Eq.(18) of "Small Current-Loops"):

$$ F_x = I \, \delta y \, B_z (x + \delta x) − I \, \delta y \, Bz (x) \\ = I \, \delta y \frac{\partial B_z}{\partial x} \, \delta x = m \, \frac{\partial B_z}{\partial x} $$

I tried to get them by using the Laplace's force (also here) formula ($d\boldsymbol{l} = (\delta x, \, \delta x, \, 0)^T$):

$$ F(x, y, z) = I \, d\boldsymbol{l} \, \times \, \boldsymbol{B} = I \begin{pmatrix} B_z \, \delta y \\ - B_z \, \delta x \\ B_y \, \delta x - B_x \, \delta y \end{pmatrix} $$

and the I tried to evaluate:

$$F_x = F_x (x + \delta x) - F_x (x) = I \, B_z(x + \delta x) \, \delta y - I \, B_z(x) \, \delta y $$

but this is wrong.

EDIT: the $F_y$, according to eq. (18) is:

$$ F_y = I \, \delta x \, B_z (y + \delta y) − I \, \delta x \, Bz (y) \\ = I \, \delta x \frac{\partial B_z}{\partial y} \, \delta x = m \, \frac{\partial B_z}{\partial y} $$

in this case maybe there are two mistakes (1) the last term should be $\delta y$, and (2) there should be a minus sign:

$$ F_y = - I \, \delta x \, \delta y \, \frac{\partial B_z(y)}{\partial y} $$

The Taylor's expansion I applied is:

$$ B_z(y+\delta y)\simeq B_z(y)+\frac{\partial B_z}{\partial y}(y)\,\delta y $$

if I apply it to the eq. 18, "it works" (*); instead, if I apply it to my equation (the component $- B_z(y) \, \delta x$ I reported in the above vector $F(x,y,z)$), it gives me a minus sign.

(*) Actually the term $\delta y$ doesn't appear in the final result and this is an error because it is wrong to write $m = I \, \delta x \, \delta x$; indeed in the equation of the $F_x$ we have $m = I \, \delta x \, \delta y$.

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I don't think there's any mistake in your calculation, you're just missing the final step.

The Taylor expansion of $B_z$ to first order in the neighborhood of $x$ is: $$B_z(x+\delta x)\simeq B_z(x)+\frac{\partial B_z}{\partial x}(x)\,\delta x$$ which yields the result you wanted.

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  • $\begingroup$ You're refering to an equation (18) that doesn't exist in your question. Please make your question as self-contained as possible. Also, remember that the magnetic field is a function of three variables. In my answer I didn't write $y$ and $z$ because they weren't impacted by the Taylor expansion, but for other components of the force, you might need to write them explicitely. $\endgroup$
    – Miyase
    Commented Jul 10, 2022 at 8:24
  • $\begingroup$ Hi @Miyase I reported the $F_y$ from equation 18 in the above link in order to avoid to visit the webpage. I performed the same calculations as before because the $F_x$ and the $F_y$ have the same form. The only difference is that we should replace $x$ with $y$. $\endgroup$
    – Gennaro Arguzzi
    Commented Jul 10, 2022 at 8:27
  • $\begingroup$ Then include the new equation in your post. Making an external link a required part of a question is frowned upon here, because the link could break in the future, making the question impossible to understand. $\endgroup$
    – Miyase
    Commented Jul 10, 2022 at 8:29
  • $\begingroup$ Second line, one of the $\delta x$ should be a $\delta y$ (you didn't apply Taylor expansion correctly). $\endgroup$
    – Miyase
    Commented Jul 10, 2022 at 9:39
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    $\begingroup$ Make a careful drawing with the orientation of the current, and be careful with the orientation of vector $I\,d\vec{l}$ in Laplace force. Each of the four sides of the square must be studied separately. $\endgroup$
    – Miyase
    Commented Jul 10, 2022 at 14:28

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