Consider a disk of radius $R$ rotating with angular velocity $\omega$ and with a surface charge density $\sigma$. I have shown that the magnetic field as a function of the distance along its axis (z) is given by $$\vec{B}=\frac{\mu_0\sigma\omega}{2}\frac{(|z| - \sqrt{R^2+z^2})}{\sqrt{R^2+z^2}}\hat{z}.$$ I am then asked to evaluate the result for $|z|\gg R$ and use this to compute the magnetic moment of the rotating disk. I would have thought that for $|z|\gg R$ the magnetic field would become $$\vec{B}=\frac{\mu_0\sigma\omega}{2}\frac{(|z| - \sqrt{R^2+z^2})}{\sqrt{R^2+z^2}}\hat{z}=\vec{B}=\frac{\mu_0\sigma\omega}{2}\frac{(|z| - \sqrt{z^2})}{\sqrt{z^2}}\hat{z}=\vec{0}$$ but this is wrong. The solution gives $$\vec{B}\simeq\frac{\mu_0\sigma\omega}{2}\frac{R^4}{4|z|^3}\hat{z}=\frac{\mu_0\sigma\omega}{8}\frac{R^4}{|z|^3}\hat{z}=\frac{\mu_0}{4\pi}\frac{3(m\cdot \hat{z})\hat{z}-\vec{m}}{|z|^3}.$$ I don't understand how half of these approximations are made. Am I right in thinking that the end product should be known before the problem is attempted, thus making you able to make approximations that would correctly result in a specific end product, this one being the potential of a magnetic dipole?

Can someone please take me through this a bit slower than the solutions have, and explain each step?

  • 1
    $\begingroup$ Looks like you need to expand $\sqrt{R^2+z^2}$ to second order to get the correct first order solution. $\endgroup$
    – CuriousOne
    Jan 10, 2016 at 17:31

1 Answer 1


@CuriousOne is right on the money. I can provide a general response to your frustration about how these approximations are made. Let's examine this problem more closely.

Let's look at

$$\vec{B}=\frac{\mu_0\sigma\omega}{2}\frac{(|z| - \sqrt{R^2+z^2})}{\sqrt{R^2+z^2}}\hat{z}$$.

We can simplify this as

$$ \vec{B}=\frac{\mu_0\sigma\omega}{2}\left(\frac{|z|}{\sqrt{R^2+z^2}} -\frac{\sqrt{R^2+z^2})}{\sqrt{R^2+z^2}}\right)\hat{z}\\ \vec{B}=\frac{\mu_0\sigma\omega}{2}\left(\frac{|z|}{\sqrt{R^2+z^2}} -1\right)\hat{z} $$

Now you you can probably convince yourself easily that,

$$ R^2+z^2 \sim z^2 $$

if $z\gg R$, but what you really need to approximate is

$$ \frac{1}{\sqrt{R^2+z^2}} $$

for the first term in the parenthesis. The subtlety here is that, as we of course know, the functions $$f(z) = z^2 + R^2 = z^2\left(1+\left(\frac{R}{z}\right)^2\right)$$ and $$g(z) = \frac{1}{\sqrt{z^2+R^2}} = \frac{1}{z\sqrt{1+(\frac{R}{z})^2}}$$ are not the same function; this is extremely important because if we want to approximate their values near $\frac{R}{z} \sim 0$, then they have entirely different expansions. In short, when problems ask you to approximate some expression when a given quantity is "small", the first thing you should try to do is to Taylor expand any functions of that quantity that appear in your solutions about the given quantity being very close to zero, or "small".

A similar example is that of a rigid 2-D planar pendulum of length $L$ moving under the force of gravity. If we just write the summation of the forces on the pendulum before making any approximations, we have

$$ m\vec{a} = \sum \vec{F}\\ $$ In the tangential direction we then have $$ mL\ddot{\theta} = -mg\sin\theta $$ If we now enforce the small angle approximation for $\sin\theta$, which is effectively the Taylor expansion of $\sin\theta$ near $\theta \sim 0$, corresponding to the case when the oscillation of the pendulum are "small", then we obtain

$$ mL\ddot{\theta} \approx -mg\theta\\ \ddot{\theta} \approx -\frac{g}{L}\theta $$

This is now just the equation for simple harmonic motion, which gives the correct result that for small oscillations, the frequency of oscillations of the pendulum are just

$$ \omega = \sqrt{\frac{g}{L}} $$


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