Electric Field of a Toroid carrying a changing current I = kt on axis

[PhDeezNutz]

Homework Statement

A toroidal coil has a rectangular cross section, with inner radius $a$, outer radius $a+w$, and height $h$. It carries a total of $N$ tightly wound turns, and the current is increasing at a constant rate ($\frac{dI}{dt} = kt$). If $w$ and $h$ are both much less than $a$, find the electric field at point $z$ above the center of the toroidal. [Hint: Exploit the analogy between Faraday fields and magnetostatic fields, and refer to Example 5.6 (In Griffiths)]

Relevant Equations

$B_{text{toroidal coil}} = \frac{\mu_0 NI}{2 \pi s} \hat{\phi}$ where $s$ is the distance from the center of the toroidal coil (the origin if you will) (also $\hat{\phi}$ because I assumed the toroidal coil lies in the $xy-\text{plane}$.

Since $\nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}$ which I believe implies that

$\vec{E} \left( \vec{r} \right) = \frac{1}{4 \pi} \int {\nabla’ \times \frac{\partial \vec{B}}{\partial t}}{\left| \vec{r} - \vec{r}’ \right|}\, d \tau’$

The relevant equation listed above as

$\vec{E} \left( \vec{r} \right) = \frac{1}{4 \pi} \int {\nabla’ \times \frac{\partial \vec{B}}{\partial t}}\frac{1}{\left| \vec{r} - \vec{r}’ \right|}\, d \tau’$

Seems to me to be a natural consequence of the Helmholtz Theorem so I don’t see why it wouldn’t be valid but strangely I’m finding a curl of $0$. Here’s the calculation.

If $\frac{\partial I}{\partial t} = k \Rightarrow I = kt$. Plugging $I$ into $\vec{B}$ we get

$\vec{B} = \frac{\mu_0 N kt}{2 \pi s’} \hat{\phi}’$
And taking the partial derivative wrt $t$

$\frac{ \partial \vec{B}}{\partial t} = \frac{\mu_0 N k}{2 \pi s’} \hat{\phi}’$

In cylindrical coordinates the primed curl of the above is

$\nabla’ \times \frac{ \partial \vec{B}}{\partial t} = \nabla’ \times \frac{\mu_0 N k}{2 \pi s’} \hat{\phi’} = \frac{1}{s’} \left(\frac{\partial}{\partial s’} \left(s’ \frac{ \partial \vec{B}}{\partial t} \right) \right) \hat{\phi}’$

$= \frac{1}{s’} \left(\frac{\partial}{\partial s’} \left(s’ \frac{\mu_0 N k}{2 \pi s’} \right) \right) \hat{\phi}’ = 0 \hat{\phi’}$

Which makes the following integral zero

$\vec{E} \left( \vec{r} \right) = \frac{1}{4 \pi} \int {\nabla’ \times \frac{\partial \vec{B}}{\partial t}}\frac{1}{\left| \vec{r} - \vec{r}’ \right|}\, d \tau’ = 0$

Where did I go wrong with the curl calculation because the answer is definitely not zero?

Thanks for any help in advanced.

Edit all instances of the integral should be

$\vec{E} \left( \vec{r} \right) = \frac{1}{4 \pi} \int \frac{\nabla’ \times \frac{\partial \vec{B}}{\partial t}}{\left| \vec{r} - \vec{r}’\right|} \, d \tau’$

 

[TSny]

Edit all instances of the integral should be

$\vec{E} \left( \vec{r} \right) = \frac{1}{4 \pi} \int \frac{\nabla’ \times \frac{\partial \vec{B}}{\partial t}}{\left| \vec{r} - \vec{r}’\right|} \, d \tau’$

This doesn't look correct to me.

I believe the Helmholtz theorem leads to $$\vec{E} \left( \vec{r} \right) = -\frac{1}{4 \pi} \nabla \times \int \frac{\frac{\partial \vec B(\vec r \, ', t)}{\partial t}}{\left| \vec{r} - \vec{r}\,’\right|} \, d \tau’.$$ I don't see how to get from this to your expression.

 

[PhDeezNutz]

Thankyou @TSny

I figured it out.

Instead of fussing with the Helmholtz Theorem and its intricacies I just appealed to a formula in Griffiths

$\vec{E} \left(\vec{r}\right) = - \frac{1}{4 \pi} \int \frac{\frac{\partial \vec{B}}{\partial t} \times \vec{R}}{R^3} \, d \tau'$

I made some simplifying assumptions along the way and got the same answer the solution manual got via a slightly different method.

That said the simplifying assumptions regarding smallness parameters were probably favorably made in light of knowing the final answer :D

That said the Helmholtz Theorem is something I should definitely get more familiar with.

Edit: just to be sure I only looked at the final solution. Not the workings. I believe my solution is valid due to the corroboration.