I am trying to derive the magnetic flux through a toroid of inner radius $\rho_0$ and outer radius $\rho_0 + 2a$, having $N$ turns
I started off, by saying $$\phi = \int \vec{B} \cdot \vec{dS}$$
We know that $$\vec{B} = \mu_0\frac{n}{2\pi\rho_0}I \vec{a_\phi} $$
What should I choose $dS$ as ?
dS should be the area element between the loops of the wire wound around the toroid ? (Am I right in saying this ?)
The answer is $$ \phi = \mu_0 NI[\rho_0 – (\rho_0^2 -a^2 )^{1/2}]$$
Can you please guide me through this problem
Thank YOu very much
Anupam
Best Answer
There are detailed discussions of this problem in most E&M texts, including Griffith and Jackson.
The easiest way to approach the problem is using Ampere's Law, with an Amperian loop through the center of the toroid, as seen here.
We know that $$\int B \cdot dS = \mu I_{inc}$$
and the current enclosed by that particular loop is just the number of loops $N$ times the current through the wire. So the equality becomes $$\int B \cdot dS = \mu I N$$
and from the Biot Savart law, we know that the magnetic field through the center of the toroid is constant and circumferential. So this in turn becomes $$2\pi R B = \mu I N$$
which simplifies to $$B = \frac{\mu I N}{2\pi R}$$
There are a lot of assumptions made in this calculation, so the Biot Savart approach can be more satisfying. Try with a straight solenoid first, and then curve it. Parameterize the solenoid $$\langle{rcos(\theta),rsin(\theta),k\theta}\rangle$$
and then take the cross product of the derivative with the vector to get the vertical component, and then take a further dot product with $(0,0,1)$ to get the vertical component.
Hope this helps.