Magnetic field above a thin charged disc

[russdot]

Homework Statement

A thin disc of radius R carries a surface charge $$\sigma$$. It rotates with angular frequency $$\omega$$ about the z axis, which is perpendicular to the disc and through its center. What is B along the z axis?

Homework Equations

General Biot-Savart law:
B(x) = $$\frac{\mu_{0}}{4\pi}\int\frac{J(x') x (x-x')}{|x-x'|}d^{3}x'$$

K $$\equiv \frac{dI}{dl_{perpendicular}}$$
K = $$\sigma$$v

The Attempt at a Solution

I'm wondering if the general form Biot-Savart law can be 'generalized' to a 2-D surface current density K instead, and if the form would be the same?
Giving:
B(x) = $$\frac{\mu_{0}}{4\pi}\int\frac{K(x') x (x-x')}{|x-x'|}d^{2}x'$$

 

[marcusl]

I don't follow your equations, you aren't using a standard notation. Regardless, you don't need to use a surface current. You can treat every infinitesimal interval of radius dr as a loop of current. Integrate the equation for field from a current loop over r from 0 to R

 

[gabbagabbahey]

Hi russdot, try using \vec{B} instead of B in your LaTeX equations, and put the whole equation in LaTeX to make it easier to read.

The Biot-Savart Law for a surface current is:

$$\vec{B}(\vec{r})=\frac{\mu _0}{4 \pi} \int \frac{\vec{K}(\vec{r'}) \times \widehat{\vec{r}-\vec{r'}}}{|\vec{r}-\vec{r'}|^2} d^3r'$$

Such integrals are tedious to evaluate, so if you are allowed to use Ampere's law or calculate the vector potential $\vec{A}$ first, I would do that.

 

[russdot]

Hi gabbagabbahey,
Thanks for the tips, I'm still getting used to the LaTeX notation.

marcusl,
Ok, and then each infinitesimal loop will have a current I = $$\sigma 2 \pi R dr$$.

 

[gabbagabbahey]

Hi gabbagabbahey,
Thanks for the tips, I'm still getting used to the LaTeX notation.

marcusl,
Ok, and then each infinitesimal loop will have a current I = $$\sigma 2 \pi R dr$$.

You're welcome

Shouldn't each loop have a current of $2 \pi r \sigma \omega dr$?

 

[russdot]

Ah yes, I really should eat some food...
but since $$\vec{I} = \lambda \vec{v}$$
and $$\lambda = \sigma 2 \pi dr$$
and $$\vec{v} = r \vec{\omega}$$
then $$\vec{I} = \sigma 2 \pi \vec{\omega} r dr$$, correct?

 

[gabbagabbahey]

Yep, looks good to me (my sigma didn't show up right the first time)...what then is the magnetic field of each loop? What do you get for the total magnetic field?

 

[russdot]

what then is the magnetic field of each loop?

For the magnetic field of each loop (along Z axis), I get:
$$\vec{B}(z) = \frac{\mu_{0}}{4 \pi} \int \frac{\sigma 2 \pi \omega x'^{2} (\hat{\phi} \times \hat{r})dx'd\phi}{x'^{2} + z^{2}}$$
where $$\hat{\phi} \times \hat{\r} = \hat{k}cos\psi + \hat{R}sin\psi$$
($$\psi$$ is the angle between$$\vec{r} = \vec{x} - \vec{x}'$$ and $$\hat{R}$$ where $$\hat{R}$$ is the cylindrical radial unit vector)

the $$sin\psi$$ term integrates out to zero, because $$\hat{R} = \hat{i}cos\phi + \hat{j}sin\phi$$ and I used the trig substitution $$cos\psi = \frac{x'}{\sqrt{x'^{2} + z^{2}}}$$

$$\vec{B}(z) = \mu_{0} \pi \sigma \omega \int_{0}^{R} \frac{ x'^{3} dx'}{(x'^{2} + z^{2})^{3/2}} \hat{k}$$

What do you get for the total magnetic field?

Which gives a total magnetic field:
$$\vec{B}(z) = \mu_{0} \pi \sigma \omega (\frac{2z^{2} + R^2}{\sqrt{z^{2} + R^{2}}} - 2z) \hat{k}$$

 

[Merlino91]

Uhm... wouldn't the current flowing through an infinitesimal loop be the charge divided by a period T=$$\frac{2\pi}{\omega}$$ ?
so
$$i=\sigma\omega rdr$$

P.s: how did you resolve the integral? Substituction with sinh(t) ? Are there any faster method?