0
\$\begingroup\$

I'm looking at the author's solution for a problem using Ampere's Law, but his steps are very terse, and two steps slip by without a justifying remark. I've added my guesses below with question marks.

Problem

What is the magnetic flux density inside a conductor wrapped into the shape of a torus with radius 2cm, 50 turns, and a constant 1A of current flowing through it?

Solution

\begin{align} \oint H \cdot dl &= I_{enc} + \int\int_S \frac{\delta D}{\delta t} \cdot dS \end{align}

Noting that the current is constant (?), $$\frac{\delta D}{\delta t} = 0$$

$$I_{enc} = (50 \textrm{ turns})(1A) = 50A$$

Assuming the material is homogenous,

\begin{align} \oint \frac{B}{\mu} \cdot dl &= 50A \\ dl &= r d\theta \\ \int^{2\pi}_{0} \frac{B}{\mu} (2cm) d\theta &= 50A \end{align}

With the knowledge that the flux density is rotationally symmetric (?),

\begin{align} \frac{B}{\mu} (2cm) \int^{2\pi}_{0} d\theta &= 50A \\ \implies B &= \frac{50A\mu}{(2cm)(2\pi)} = 500 \mu T \end{align}

Justifications

  1. Is the constant current statement correct?
  2. Is the rotational symmetry the reason for removing the flux density term from the integrand? If the element were something more interesting than a torus, would that step be disallowed?
\$\endgroup\$
  • \$\begingroup\$ Is this an electrostatics problem? \$\endgroup\$ – The Photon Feb 14 '17 at 6:19
  • \$\begingroup\$ @ThePhoton It depends on how you feel about justification 1. If the only current is displacement current, then yes. \$\endgroup\$ – bright-star Feb 14 '17 at 6:20
  • \$\begingroup\$ What I mean is if you're looking for the DC solution, then you can justify taking d/dt to 0 because you're looking for the DC solution. \$\endgroup\$ – The Photon Feb 14 '17 at 6:22
  • \$\begingroup\$ @ThePhoton I'm afraid this is a case of trying to figure out the author's reasoning. I'm assuming that the 1A current implies the solution is DC, which is why I made that guess. \$\endgroup\$ – bright-star Feb 14 '17 at 6:26
  • 1
    \$\begingroup\$ Lots of catchems in this one! \$\endgroup\$ – skvery Feb 14 '17 at 8:30
1
\$\begingroup\$

The starting equation is the famous "Maxwell's own" equation presented as an integral. The constant current here means that it's DC that never changes and never has been changed. It's a starting point and the calculation tries to find, what are the consequences. It can be denied only if one founds that this starting point has a hidden logical contradiction, but nothing like that has been yet found. The assumption is also usable for the practice because all practical waves attenuate quite soon after switchig the DC on due the losses.

The author has assumed that the winding is dense enough and the diameter of a single turn is small enough (when compared to 2 cm) to give the flux that has good enough rotational symmetry (=scalar B does not depend on the angle in the integral)

The second assumption surely is not usable for all practical windings. But this calculation ignores those cases.

\$\endgroup\$
1
\$\begingroup\$

"Is the constant current statement correct?"

Well, the problem states,

and a constant 1A of current flowing through it

so, yes, it is correct.

Is the rotational symmetry the reason for removing the flux density term from the integrand?

Yes. If the flux density changed wrt the angle of integration, it would not be radially symmetric. The two are equivalent.

\$\endgroup\$
1
\$\begingroup\$

First do $$I_{enc} = \int \int_S J_f~dS~.$$

The result of this is the total current through a surface, and if this surface ends totally inside the torus, $$I_{enc} = N~I = 50~A~.$$

(In this case it is DC but it will also hold for AC and this term is the normally called 'the sum of the ampere-turns' and it is equal to the magnetising current.) It can also be written as $$ \Sigma ~N~I~.$$

The second part of the answer is concerned with the integral along the edge of the surface S, chosen for constant H inside the torus $$\oint H \cdot dl = H\cdot \oint dl = H \cdot 2\pi R = H\cdot L(S)$$ with R the radius of the ring of S, inside the torus. If the diameter of the tube of the torus is, say 1 cm , R can be in the range [1,5 cm - 2,5 cm] .

This means that the magnetic field will vary through the thickness according to L, the length of the edge of surface S as discussed.

Also watch out for $$\mu_0$$ lurking somewhere.

If you follow the same principles, the equation is uneversal, but tricky to calculate if the paths are not along constant lines as in the example.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.