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** Edited: I updated the question since the solution is correct.

Problem: A hollow cone(like a party hat) has vertex angle $2\theta$, slant height $L$, and surface charge density $\sigma$. It spins around its symmetry axis with angular frequency $w$. What is the magnetic field at the tip? As the following figure

enter image description here

Solution:

I know that the magnetic field on the axis given by a current $I$ in a ring of radius $r$ and a distance $l$ from anywhere in the ring is

$$\vec{B}_{\text{Ring }z }= \frac{\mu_0 Ir^2}{2l^3}\hat{z} $$

Using this I can consider the magnetic filed of the cone as the sum of a lot of rings with current $dI$ and a variable radius $r$ as :

$$dB_{\text{cone}}= \frac{\mu_0 dIr^2}{2l^3} $$

Using the geometry of the problem we get $r=l\sin(\theta)$, replacing in the previous equation:

$$dB_{\text{cone}}= \frac{\mu_0 l^2 \sin^2(\theta) dI }{2l^3}=\frac{\mu_0 \sin^2(\theta) dI }{2l} $$

Now what is $dI$, we know that the surface current density $\vec{K}$ is defined as :

$$\vec{K}= \sigma\vec{v} \:\:\:\:\text{and}\:\:\:\:\: \vec{K} \equiv \frac{d\vec{I}}{dl_{\perp}} $$

We know that $v=wr$ so $K=\sigma wr$ and with the other equation:

$dI=\sigma w r \cdot dl_{\perp}$

This $dl_{\perp}$ is perpendicular to $K$ and $I$.

Assumption

$dl_{\perp}$ is the same $dl$ that we need to do the integral in the cone from $0$ to $L$

I assume it's correct because of the definition of $K$, but when doing this I'm assuming the following (I think). I'm assuming that the current is steady and that this problem is equivalent to the following:

enter image description here

This is the differential that we need to do the integral, now replacing:

$$dB_{\text{cone}}= \frac{\mu_0 \sin^2(\theta) \sigma w r \cdot dl_{\perp} }{2l}$$

Now integrating :

$$B_{\text{cone}}= \mu_0 \sin^2(\theta) \sigma w \int \frac{ r dl_{\perp} }{2l}$$

Using again $r=l\sin(\theta)$:

$$B_{\text{cone}}= \mu_0 \sin^3(\theta) \sigma w \int \frac{ l dl_{\perp} }{2l}$$

$$B_{\text{cone}}= \frac{\mu_0 \sin^3(\theta) \sigma w}{2} \int_0^L dl_{\perp} $$

$$B_{\text{cone}}= \frac{\mu_0 \sin^3(\theta) \sigma w}{2} L $$

This answer makes sense because of the bigger $w$, the bigger $v$ and consequently $I$. However, if we were to have an infinite hollow cone the field would eventually diverge.

We can also look for an analogy between the electric field and the magnetic field, because is easy to see that the electric field in this tip for the finite hollow cone diverges, however in this case the magnetic field don't. This can be explained by two things:

1) The magnetic field and the electric field for a ring at his axis are the following:

$$\vec{B}_{\text{Ring }z }= \frac{\mu_0 I \sin^2{\theta} }{2l} \hat{z} \:\:\:\:\:\:\:\:\: \vec{E}_{\text{Ring }z }= \frac{1}{4\pi\epsilon_0}\frac{Q \cos(\theta)}{l^2} \hat {z} $$

This difference of $1/l$ makes the difference between the two.

2) Let's suppose that the magnetic field also diverges if I rotate the cone no matter $w$, it would be then a problem to see the limit $w\rightarrow 0$, would there be a magnetic field even if $w\rightarrow 0$.

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closed as off-topic by ACuriousMind, JamalS, Kyle Kanos, Jim, Brandon Enright Nov 3 '14 at 16:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

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  • $\begingroup$ Now this is a homework-related question that shows effort! $\endgroup$ – HDE 226868 Nov 3 '14 at 0:31
  • $\begingroup$ I'm not sure why you are bringing the electric field into it. Obviously as the cone becomes "infinite", so does the amount of charge. Every additional ring of current adds a finite amount to the total B field (ring gets bigger linearly with distance, and current is bigger since the velocity scales with distance. So $Ir^2$ increases as the cube of the distance, and every additional ring contributes the same amount to the integral. Infinite cone = infinite contributions. We don't "check your work" here, but we can agree with your conclusion - the solution should diverge. $\endgroup$ – Floris Nov 3 '14 at 5:04
  • $\begingroup$ I came up with the electric field thing just to make a comparison in order to show why my answer might give reasonable predictions and why another answer don't. The main point of the question is to check if the magnetic field at the tip is right. Is it @Floris ? $\endgroup$ – Keith Nov 3 '14 at 5:15
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    $\begingroup$ This question appears to be off-topic because it is a check-my-work question. $\endgroup$ – ACuriousMind Nov 3 '14 at 14:05
  • $\begingroup$ Edited: I CLARIFIED where and what my question is $\endgroup$ – Keith Nov 3 '14 at 18:01