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From what I see in my textbook and several on line resources (such as this, p. 8-8), I am tempted to think that the magnetic force $I\boldsymbol{\ell}\times\mathbf{B}$ acting on a straight conducting wire, flown through by a current whose intensity is $I$, placed in a uniform magnetic field $\mathbf{B}$, is such that the magnetic moment due to it, which I think we can write as $$\boldsymbol{\tau}= \int_0^L \mathbf{x}(t) \times \bigg( \frac{I}{L} \boldsymbol{\ell}\times \mathbf{B}\bigg) dt $$where $L$ is the length of the wire and $\mathbf{x}:[0,L]\to\mathbb{R}^3$ a parametrisation of the wire with respect to the point chosen as the pole of the torque, can be calculated as if the force were all concentred in the midpoint of the wire $\mathbf{x}(L/2)$.

Is that so? If my idea is correct, how can we prove that the torque is the same that we would have if the resultant force were all applied in the midpoint of the wire, i.e. that $ \int_0^L \mathbf{x}(t) \times \left( \frac{I}{L} \boldsymbol{\ell}\times \mathbf{B}\right) dt $ $=\mathbf{x}\left(\frac{L}{2}\right)\times(I\boldsymbol{\ell}\times\mathbf{B})$? I heartily thank any answerer!

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This is a straightforward calculation. Writing the torque as

$$\boldsymbol{\tau}= \int_0^1 \mathbf{x}(s) \times \left( I\boldsymbol{\ell}\times \mathbf{B}\right) \mathrm ds $$

the wire spans from $\mathbf x(0)$ to $\mathbf x(1)=\mathbf x(0)+\boldsymbol{\ell}$, and we can parametrize it linearly with $s$, so that $$ \mathbf x(s)=\mathbf x(0)+s\boldsymbol{\ell}, $$ and therefore \begin{align} \boldsymbol{\tau} &= \int_0^1 \mathbf{x}(s) \times \left( I\boldsymbol{\ell}\times \mathbf{B}\right) \mathrm ds \\&= \int_0^1 (\mathbf x(0)+s\boldsymbol{\ell})\times \left( I\boldsymbol{\ell}\times \mathbf{B}\right) \mathrm ds. \end{align} Here the main cross product is just a linear combination of the components of $\mathbf x(s)$ with constant coefficients, so we can pull in the integration all the way to that factor: \begin{align} \boldsymbol{\tau} &= \left(\int_0^1 \mathbf{x}(s) \mathrm ds \right)\times \left( I\boldsymbol{\ell}\times \mathbf{B}\right) \\&= \left(\int_0^1 (\mathbf x(0)+s\boldsymbol{\ell})\mathrm ds\right)\times \left( I\boldsymbol{\ell}\times \mathbf{B}\right) \\&= \left[ s\mathbf x(0)+\frac{s^2}{2}\boldsymbol{\ell}\right]_0^1\times \left( I\boldsymbol{\ell}\times \mathbf{B}\right) \\&= \left[ \mathbf x(0)+\frac{1}{2}\boldsymbol{\ell}\right]\times \left( I\boldsymbol{\ell}\times \mathbf{B}\right) \\&= \mathbf x\left(\tfrac12\right)\times \left( I\boldsymbol{\ell}\times \mathbf{B}\right), \end{align} as you conjectured. (If the linearity step bothers you, you can just expand the cross products out in terms of components, move the integral, and put the explicit cross products out, but honestly it's not worth it.)

One warning to keep in mind is the fact that finite wires like yours are not physical, because you always need to put in current in one end and take it out of the other - only closed current loops will give meaningful, invariant answers for torques and forces in magnetostatics. However, results like yours are indeed useful, e.g. for calculating the torque on a closed loop composed of a number of straight segments.

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  • $\begingroup$ The linearity step doesn't bother at all: multiplying by a vector is the same as multiplying by a matrix. I heartily thank you for your excellent mathematically detailed answer! $\endgroup$ – Self-teaching worker Dec 19 '15 at 10:42

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