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This problem is in reference to Q3.239 in Problems in General Physics by I.E. Irodov.

Question: N = $2.5*10^3 $wire turns are uniformly wound on a wooden toroidal core of very small cross-section. A current I flows through the wire. Find the ratio 1 of the magnetic induction inside the core to that at the centre of the toroid.

Assuming an Amperian loop in the central space of toroid directly gives us B = 0 at .the center, similarly inside the toroid gives us $$ B = \mu NI / (2 \pi r) $$ and this is also what I have found in many reference books too. But the answer given is $$ \eta = N / \pi$$

Where have i missed? Thanks.

CodeBlooded
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2 Answers2

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You are correct in your calculation of magnetic field inside the toroid, it is indeed $$ B_\text{inside} = \mu NI / (2 \pi r).$$

However, in the center the magnetic field is non-zero. Indeed, each loop is not a closed circle but a turn of a helix. Each such turn contain a shift of $2\pi r /N$ in toroidal direction. And so while poloidal components of the current do no contribute to the magnetic field at the center, the toroidal components do. And their contributions (for thin toroid) equivalent to that of single circular coil in the toroidal direction. (For definition of directions on torus look here).

Thus, by Biot-Savart law, at the center magnetic field would be $$ \mathbf{B}_\text{center} = \frac{\mu}{4 \pi} \int_C \frac{I d{\boldsymbol\ell} \times \hat{\mathbf{r}}}{r^2}=\hat{\mathbf{n}}\, \frac{\mu}{4\pi} \,\frac{ 2\pi r\, I }{r^2}=\hat{ \mathbf{n}}\,\frac{\mu I}{2 r}, $$ where we have taken into account that the direction of vector product is the same at each point on the circle.

And so the ratio is $$ \eta =\frac{B_\text{inside}}{B_\text{center}}=\frac N \pi.$$

A.V.S.
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  • I couldn't understand why did you ignore the winding structure, isn't it the main thing which makes it a toroid. The direction of current in every small loop in the windings passes through the centre (ideally), then it should qualitatively also be 0. – CodeBlooded Jan 21 '18 at 14:27
  • Each small loop is not closed. It has nonzero toroidal component. See my edits where I tried to clarify this point. – A.V.S. Jan 21 '18 at 15:46
  • By " thin toroid " you mean very small cross section ? As in when we compute magnetic field taking $\dl$ components along radial and saying as cross section is very small so all the little current element paths along a loop can be treated as being along a same distance from centre ? Thats why only we were able to say that equivalent thing right ? If the area was large or say the helical turns were having large deviation that is the titledness , then its not same as that equivalent right ? We are assuming here the tiltedness as very small and cross section very small right ? @A.V.S. – ProblemDestroyer Oct 01 '22 at 12:23
  • @ProblemDestroyer: Yes, small cross-section. In Biot-Savart law if $\hat{\mathbf{r}}/r^2$ is approximately constant on a single turn of the coil then the poloidal components of the current are integrated out (do not contribute to the field at the center). – A.V.S. Oct 01 '22 at 13:26
  • It will be automatically be constant as such all loops are same isnt ? As in dimensions , do you mean something else by approximately constant ?
    2) so this techinique of considering superposition by assuming ciruclar loops and a circular wire will not work for helical wire having large tiltedness ? As in then we cannot take those radial components to be coming from a circle right since each small element is part of a different circle . We cannot approx with a unique circle right ? @A.V.S. (for a turn)
    • – ProblemDestroyer Oct 01 '22 at 16:52