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?


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) $.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.