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I Magnetic field by ideal toroidal solenoid

  1. Sep 28, 2016 #1
    I am trying to calculate the magnetic field generated by an ideal toroidal solenoid by using the integral of the Biot-Savart law. I do not intend to use Ampère's circuital law.
    Let ##I## be the intensity of the current flowing in each of the ##N## loops of the solenoid, which I will consider an ideal continuous solenoid from this point.
    If ##\mathbf{r}(u,v):[0,2\pi]^2\to\mathbb{R}^3##, ##\mathbf{r}(u,v)=(b+a\cos v)\cos u\mathbf\,{i}+(b+a\cos v)\sin u\,\mathbf{j}+a\sin v\mathbf\,{k}## is a parametrization of the torus, I would say that, in an "infinitesimal spire" of the ideal solenoid, generated by the rotation of ##du## radians of the circumference generating the torus, an "infinitesimal current" ##\frac{IN}{2\pi}du## flows and therefore I would think that the magnetic field at ##\mathbf{x}## could be expressed by $$\frac{\mu_0}{4\pi} \int_{0}^{2\pi}\int_0^{2\pi}\frac{IN \,\partial_v\mathbf{r}(u,v) \times(\mathbf{x}-\mathbf{r}(u,v) )}{2\pi\|\mathbf{x}-\mathbf{r}(u,v)\|^3} dudv.$$
    Am I right?
    I thank anybody for any answer.
     
  2. jcsd
  3. Sep 29, 2016 #2

    vanhees71

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    I think everything looks good, but maybe it's simpler to use the (Coulomb gauge) vector potential, which is given as
    $$\vec{A}(\vec{x})=\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|},$$
    and then the magnetic field by
    $$\vec{B} = \vec{\nabla} \times \vec{A}.$$
    Note, however that in this case, due to the symmetry of the field, the use of Ampere's Law in integral form is much simpler here.
     
  4. Sep 30, 2016 #3
    Thank you so much for your answer! By explicitly writing the formula, I find that, for a point ##\mathbf{x}=x\mathbf{i}+z\mathbf{k}##, the ##x## and ##z## components of the field ##\mathbf{B}## are zero because of reasons of symmetry and $$\mathbf{B}(x,0,z)=\frac{\mu_0 IN}{8\pi^2}\int_0^{2\pi}\int_0^{2\pi}\frac{ax\cos v-ab\cos u\cos v-a^2\cos u+az\cos u\sin v}{(x^2+b^2+a^2+z^2-2x(b+a\cos v)\cos u+2ab\cos v-2az\sin v)^{3/2}}dudv\mathbf{k}$$which, if it agrees with the result given by my books for an ideal solenoid, should be zero at the exterior of the torus and ##-\frac{\mu_0 NI}{2\pi x}\mathbf{j}## for ##z=0##, ##|x|\in (b-a,b+a)## where ##a## is the radius of the circular section of the torus.
    Are my calculation correct until now? If they are, I find the integral quite difficult... Has anybody any idea of how to calculate it?
    I ##\infty##-ly thank you all!
     
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