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I read in my book of physics, E.W. Gettys' Physics 2, that the magnetic force acting on an "infinitesimal segment" $d\boldsymbol{\ell}$ (I apologise for such language, often used in elementary physics books, but I am using the words that my textbook uses) of conducting wire, flown through by a current whose intensity is $I$, is $d\mathbf{F}=I\,d\boldsymbol{\ell}\times \mathbf{B}$, where $\mathbf{B}$ is the magnetic field, and therefore the magnetic force acting on any conducting wire is $$\mathbf{F}=\int I\,d\boldsymbol{\ell}\times \mathbf{B}$$

I must admit I have never seen an integral written this way. By analogy with the writing $\int_\gamma \mathbf{F}\cdot d\boldsymbol{\ell}$ $:= \int_a^b\mathbf{F}(t)\cdot\boldsymbol{\ell}'(t)\,dt $ where $\boldsymbol{\ell}:[a,b]\to\mathbb{R}^3$ parametrises the pievewise smooth curve $\gamma$, I would suppose that $\int I\,d\boldsymbol{\ell}\times \mathbf{B}$ might be defined as $\int_a^b I(\boldsymbol{\ell}(t))\boldsymbol{\ell}'(t)\times \mathbf{B}(\boldsymbol{\ell}(t))\,dt$ (which belongs to $\mathbb{R}^3$, of course)... Am I right and, if I am not, what does $\int I\,d\boldsymbol{\ell}\times \mathbf{B}$ mathematically means?

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Yes you are correct. If the current $I$ flows into a wire described by the piecewise smooth curve $ \gamma $ and $ \mathbf{l} = \mathbf{l}(t)$ for $ t \in [a,b] $ parametrises $\gamma$, then the magnetic force acting on the wire due to the magnetic field $ \mathbf{B}$ is:

$$ \mathbf{F} = \int_{\gamma}Id\mathbf{l} \times \mathbf{B} = \int_a^bI(\mathbf{l}(t))\mathbf{l}'(t) \times \mathbf{B}(\mathbf{l}(t))dt$$

where $ \mathbf{l}'(t) = \frac{d}{dt} \mathbf{l}(t) $.

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  • $\begingroup$ You are welcome, but you had already found the answer by yourself basically :) $\endgroup$ – NNec Dec 14 '15 at 13:55

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