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<b>Lecture 15: Magnitism IV: Ampere's Law</b><br>
Mon 09 Mar 2026 11:00
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== Biot-Savat: Complex Loops Continued ==

Continuing from the end of last lecture, we found by symmetry that the <nowiki>$yz$</nowiki> components cancel, leaving only an <nowiki>$x$</nowiki> component for any point <nowiki>$P$</nowiki> along the x-axis.

[[File:EM-Y1S2-lec14-07.png|thumb|center|600px]]

[[File:EM-Y1S2-lec15-01.png|thumb|center|600px]]

Taking magnitudes (and noting that <nowiki>$\hat{r}$</nowiki> is a unit vector): <div class="pn-import-equation"><nowiki>$$\delta B = \frac{\mu_0 I}{4 \pi r^2} \delta l$$</nowiki></div>

Resolving into the x-direction, we get that: <div class="pn-import-equation"><nowiki>$$\delta B_{x} = \frac{\mu_0 I}{4 \pi r^2} \delta l \cos \theta \tag{1}$$</nowiki></div>

And as we have a right angled triangle: <div class="pn-import-equation"><nowiki>$$r^2 + a^2 + x^2 \tag{2}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\cos \theta = \frac{a}{r} \implies \cos \theta = \frac{a}{\left(a^2 + x^2\right)^{(1/2)}} \tag{3}$$</nowiki></div>

Inserting 2 and 3 into 1: <div class="pn-import-equation"><nowiki>$$\delta B_x = \frac{\mu_0 I}{4 \pi} \times \frac{a}{\left(a^2 + x^2\right)^{(3/2)}} \delta l$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$B_x = \frac{\mu_0 I}{4 \pi} \times \frac{a}{\left(a^2 + x^2\right)^{(3/2)}} \oint \delta l$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$B_x = \frac{\mu_0 I}{4 \pi} \times \frac{a}{\left(a^2 + x^2\right)^{(3/2)}} 2 \pi a$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$B_x = \frac{\mu_0 I}{2} \frac{a^2}{\left(a^2 + x^2\right)^{(3/2)}}$$</nowiki></div>

Taking limits, we can observe that at the centre of the loop (where <nowiki>$x=0$</nowiki>): <div class="pn-import-equation"><nowiki>$$B_0 = \frac{\mu_0 I}{2} \frac{a^2}{\left(a^2\right)^{(3/2)}}= \frac{\mu_0 I}{2a}$$</nowiki></div>

And when <nowiki>$x\gg a$</nowiki>, we have: <div class="pn-import-equation"><nowiki>$$B_x = \frac{\mu_0 I a^2 }{2x^3}$$</nowiki></div>

However, we can simplify using the magnetic dipole moment <nowiki>$\mu_m = I \times \text{area} = I \pi a^2$</nowiki>, hence: <div class="pn-import-equation"><nowiki>$$B_x = \frac{\mu_0}{4 \pi} \frac{2 \mu_m}{x^3}$$</nowiki></div>

This is strikingly similar for the expression for an electric dipole: <div class="pn-import-equation"><nowiki>$$E_x = \frac{1}{4 \pi \epsilon_0} \frac{2p}{x^3}$$</nowiki></div>

== E- and B-Fields ==

We have already determined the following, for E- and B-fields respectively: <div class="pn-import-equation"><nowiki>$$\int_S \underline{E} \cdot d \underline{S} = \frac{Q}{\epsilon_0} \tag{E}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\int_S \underline{B} \cdot d \underline{S} = 0 \tag{B}$$</nowiki></div>

We also know that the E-field is conservative, hence no work is done if a charge returns to its original position: <div class="pn-import-equation"><nowiki>$$\oint \underline{E} \cdot d \underline{l} = 0$$</nowiki></div>

Is there an equivalent of this for B-fields?

Consider a circiular path of a B-field created around an infinite line of current at radial distance <nowiki>$r$</nowiki>. By symmetry, <nowiki>$\underline{B}$</nowiki> and <nowiki>$d \underline{l}$</nowiki> are parallel, and are constant for a fixed <nowiki>$r$</nowiki>. Hence: <div class="pn-import-equation"><nowiki>$$\oint \underline{B} \cdot d \underline{l} = \oint B dl = B \oint dL = B 2 \pi r$$</nowiki></div>

We have also determined that: <div class="pn-import-equation"><nowiki>$$B = \frac{\mu_0 I}{2 \pi r} \implies \oint \underline{B} \cdot d \underline{l} = B 2 \pi r = \frac{\mu_0 I}{2 \pi r} \times 2 \pi r = \mu_0 I$$</nowiki></div>

This gives us Ampere’s Law: <div class="pn-import-equation"><nowiki>$$\boxed{\oint \underline{B} \cdot d \underline{l} = \mu_0 I_\text{enclosed}}$$</nowiki></div>

=== Example ===

Consider a long solid cylinder carrying a uniformly distributed current. The cylinder has radius <nowiki>$R$</nowiki> and length

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Ash: Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc