- #1

latentcorpse

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[itex]\mathbf{B}=\frac{\mu_0 I a^2}{2(a^2+z^2)^{\frac{3}{2}}} \mathbf{\hat{z}}[/itex]

where I is the current and z is the distance along the x-axis from the centre of the loop.

I've done this part!

{ii} An insulating disc of radius a has uniform surface charge density [itex]\sigma[/itex]. It rotates at angular velocity [itex]\omega[/itex] about a perpindicular axis through its centre. What is the surface current density [itex]\mathbf{K(r)}[/itex] at [itex]\mathbf{r}[/itex] from it's centre?

Find the contribution [itex]d \mathbf{B}(z)[/itex] to the magnetic field on the axis of the disc from a ring with radii between r and r+dr, and thus find the magnetic field on the axis of the spinning disc.

Show that as [itex]z \rightarrow \infty[/itex],

[itex]\mathbf{B}(z) ~ \frac{1}{8} \mu_0 \sigma \omega \frac{a^4}{z^3} \mathbf{\hat{z}}[/itex]

(iii) What would the corresponding results be for a spinning ring of inner radius a and outer radius b. Recover the result in part (i) by taking the limit [itex] b \rightarrow a [/itex]

As I said I've already done (i)

For (ii) I said the surface charge density was given by [itex] \mathbf{K(r)}=\sigma \mathbf{\omega} \wedge \mathbf{r}[/itex]

As for the next part about the contribution, I took the perpendicular axis as the z direction and said that since

[itex]\mathbf{I}=I \mathbf{dr}=\mathbf{K} dA[/itex], we get

[itex]\oint I \mathbf{dr}= \int_S \mathbf{K} dA [/itex] and so

[itex]I 2 \pi r = \sigma \mathbf{\omega} \wedge \mathbf{r} dr[/itex]

Now I can use the formula from (i) with I as above and a=r giving

[itex]B_z = \frac{\mu_0}{2} \int_{0}^{a} \frac{\sigma \omega r^3 dr}{(r^2+z^2)^{\frac{3}{2}}}[/itex] where essentially we integrate in order to add up all the rings and get a disc.

however I can't integrate this though!