- #1
robinegberts
- 15
- 4
I am trying to derive the equation ##\vec{\Gamma} = \vec{m} \times \vec{B}##, where ##\vec{m} = I \vec{A}## is the magnetic moment, and ##\vec{A}## is normal to surface ##A##, from the Lorentz force law ##\mathrm{d}\vec{F} = I \vec{dl} \times \vec{B}##. For an arbitrary closed current loop ##C## enclosing the area ##A##, I have tried
$$ \vec{\Gamma} = \oint_C{\mathrm{d}\vec{\Gamma}} = \oint_C{\vec{r} \times \mathrm{d}\vec{F}} = \oint_C{\vec{r} \times I (\vec{dl} \times \vec{B})} = I \oint_C{\vec{r} \times (\vec{dl} \times \vec{B})}.$$
Now I know that ##\vec{A} = \frac{1}{2} \oint{\vec{r} \times \mathrm{d}\vec{l}}##, but I don't know how to complete the proof. I have also tried to consider a square loop giving rise to the magnetic moment ##\vec{m}## to derive the equation, but this looks even more complicated.
$$ \vec{\Gamma} = \oint_C{\mathrm{d}\vec{\Gamma}} = \oint_C{\vec{r} \times \mathrm{d}\vec{F}} = \oint_C{\vec{r} \times I (\vec{dl} \times \vec{B})} = I \oint_C{\vec{r} \times (\vec{dl} \times \vec{B})}.$$
Now I know that ##\vec{A} = \frac{1}{2} \oint{\vec{r} \times \mathrm{d}\vec{l}}##, but I don't know how to complete the proof. I have also tried to consider a square loop giving rise to the magnetic moment ##\vec{m}## to derive the equation, but this looks even more complicated.