# Magnetic force on a wire: how to calculate it

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?

## 1 Answer

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

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