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\newcommand{\bold}[1]{\mbox{\boldmath $#1$}}
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\noindent
Institutt for fysikk, NTNU

\noindent
TFY4155/FY1003 Elektrisitet og magnetisme

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V{\aa}r 2007

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L{\o}sningsforslag til {\o}ving 13
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\noindent
Veiledning uke 15.

\vspace{0.7cm}
\noindent
{\em Oppgave 2}

\vspace{0.2cm}
\noindent
$a)$ Distance vector (from the origin) to a point in the ring:
$$\bold{r} = s \hat{s} + z \hat{z}$$
Current density is:
$$\bold{j} = \frac{I}{ab} \hat{\phi}$$
Hence:
$$\bold{r} \times \bold{j} = \frac{I}{ab} \left( s \hat{z} - z \hat{s} \right)$$
Magnetic dipole moment becomes:
$$\bold{m} = \frac{I}{2ab} \int \left( s \hat{z} - z \hat{s} \right)
s\,ds\,dz\,d\phi$$
where $\phi$ runs from 0 to $2\pi$, $s$ from $R-a/2$ to $R+a/2$ and
$z$ from $z_0$ to $z_0 + b$. In the first term, i.e., the one which is
proportional with $\hat{z}$, integration over $z$ gives a factor $b$ while
the integration over $\phi$ gives a factor $2\pi$. The integral over $s$ becomes
$$\int_{R-a/2}^{R+a/2} s^2 ds = \frac{1}{3} \left(R+a/2\right)^3 -
\frac{1}{3} \left(R-a/2\right)^3$$ 
The second term, i.e., the one which is proportional with $\hat{s}$, must, by symmetry
arguments, disappear. We may also show this explicitly: The unit vector $\hat{s}$
is {\em not} a constant, but depends on the angle $\phi$ in the following manner:
$$\hat{s} = \left( \cos \phi \right) \, \hat{x} + \left( \sin \phi \right) \, \hat{y}$$
Hence, this term will be proportional with
$$\int_0^{2\pi} \, \hat{s} \, d\phi = \int_0^{2\pi} \, 
\left( \cos \phi \; \hat{x} + \sin \phi \; \hat{y} \right) d\phi = 0$$
So, the magnetic dipole moment of the ring is:
\begin{eqnarray}
\bold{m} &=& \frac{I}{2ab} \cdot b \cdot 2\pi \cdot \left[ \frac{1}{3} \left(R+a/2\right)^3 -
\frac{1}{3} \left(R-a/2\right)^3 \right] \, \hat{z} \nonumber \\
&=& \frac{I \pi R^3}{3a} \left[ \left(1+\frac{a}{2R}\right)^3 -
\left(1-\frac{a}{2R}\right)^3 \right] \, \hat{z} \nonumber
\end{eqnarray}



\vspace{0.2cm}
\noindent
$b)$ With a thin ring, $a \ll R$, this expression for $\bold{m}$ may be expanded
to lowest order in the ratio $a/R$:
$$\left(1+\frac{a}{2R}\right)^3 \simeq 1 + \frac{3a}{2R}$$
$$\left(1-\frac{a}{2R}\right)^3 \simeq 1 - \frac{3a}{2R}$$
Then, one has
$$\bold{m} \simeq I \pi R^2 \hat{z}$$
Alternatively, you may simply multiply out the two parentheses that are raised
to the third power. This becomes no more than 4 terms from each. Then you will find:
$$\left(1+\frac{a}{2R}\right)^3 - \left(1-\frac{a}{2R}\right)^3 =
\frac{3a}{R} + \frac{a^3}{4R^3} \simeq \frac{3a}{R}$$
where the final approximation could be performed because $a \ll R$.



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