Electromagnetism – Magnetic Forces on a Small Current-Loop in an Inhomogeneous Magnetic Field

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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 $m = I \, \delta x \, \delta y$.

Best Answer

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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