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Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc

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<b>Lecture 14: Magnetism III: Magnetic Fields from Currents</b><br>
Thu 05 Mar 2026 11:00
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[[Year 1 Sem 2/LC Electromagnetism I/Lecture 13|← Previous lecture]] · [[Year 1 Sem 2/LC Electromagnetism I|Module]] · [[Year 1 Sem 2/LC Electromagnetism I/Lecture 15|Next lecture →]]

Recap of last lecture:

<ul>
<li>

Special cases of the magnetic force.

</li>
<li>

Force on a current carrying conductor: <div class="pn-import-equation"><nowiki>$$\underline{F} = I \underline{L} \wedge \underline{B}$$</nowiki></div>

</li>
<li>

Current loops and magnetic dipoles: <div class="pn-import-equation"><nowiki>$$\underline{\mu} = I \underline{A}$$</nowiki></div>

</li>
<li>

Torque on a magnetic dipole in a B-field <div class="pn-import-equation"><nowiki>$$\underline{\tau} = \underline{\mu} \wedge \underline{B}$$</nowiki></div>

</li>
<li>

Potential energy of a magnetic dipole in a B-field: <div class="pn-import-equation"><nowiki>$$U = - \underline{\mu} \cdot \underline{B}$$</nowiki></div>

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</ul>

== Magnetic Fields from a Moving Charge (Current) ==

Danish physicist Hans Christian Oersted in 1820 demonstrated that a magnetic field exists around a current-carrying wire.

This was the first connection between electric and magnetic phenomena.

Consider a B-field at point <nowiki>$\underline{r}$</nowiki> from a charge <nowiki>$q$</nowiki>, moving with velocity <nowiki>$\underline{v}$</nowiki>. The angle between <nowiki>$\underline{r}$</nowiki> and <nowiki>$\underline{v}$</nowiki> is given by <nowiki>$\phi$</nowiki>.

The induced B-field is perpendicular to both <nowiki>$\underline{u}$</nowiki> and <nowiki>$\underline{r}$</nowiki> and is given by: <div class="pn-import-equation"><nowiki>$$B \propto \frac{qv \sin \phi}{r^2}$$</nowiki></div>

In vector form, we have: <div class="pn-import-equation"><nowiki>$$\underline{B} \propto \frac{q}{r^2} \underline{u} \wedge \hat{r}$$</nowiki></div>

In SI units, the constant of proportionality (involving “permeability of free space”, <nowiki>$\mu_0$</nowiki>) is such: <div class="pn-import-equation"><nowiki>$$\underline{B} = \frac{\mu_0}{4 \pi} \frac{q}{r^2} \underline{u} \wedge \hat{r} \qquad \mu_0 = 4 \pi \times 10^{-7} \mathrm{T m A^{-1}}$$</nowiki></div>

[[File:EM-Y1S2-lec13-01.png|thumb|center|600px]]

In a more complex 3D case, this looks as such:

[[File:EM-Y1S2-lec14-02.png|thumb|center|600px]]

[[File:EM-Y1S2-lec14-03.png|thumb|center|320px]]

The B-field sweeps out a screw shape around the motion of direction. The direction is given by the right hand rule, where a “thumbs up” with the thumb pointing in the direction of the motion of the charge indicates the direction with the curl of the fingers.

=== Currents ===

Consider a wire with a current flow through it. The total charge flowing through some infinitesimal element in a unit time is: <div class="pn-import-equation"><nowiki>$$I = \frac{qv}{\delta L}$$</nowiki></div> Hence: <div class="pn-import-equation"><nowiki>$$q \underline{v} = \frac{qv}{\delta L} \delta \underline{L} = I \delta \underline{L} \qquad \text{where $\delta L$ is in the direction of current.}$$</nowiki></div>

Plugging this in, we get the Biot-Savat Law: <div class="pn-import-equation"><nowiki>$$\delta \underline{B} = \frac{\mu_0}{4 \pi} \frac{I \delta L \wedge \widehat{r}}{r^2}$$</nowiki></div>

We can split our object into infinitely many infinitesimal elements and integrate across them all to find the total B-field. We want to:

<ul>
<li>

Define <nowiki>$d \underline{B}$</nowiki> in terms of a single variable.

</li>
<li>

Integrate between the relevant limits.

</li>
<li>

Be careful about the directions of the included vector quantities.

</li>
<li>

Use symmetry where possible.

</li>
</ul>

== Current Loops ==

Consider a current loop with a clockwise current flow. The centre of the loop’s circle is <nowiki>$P$</nowiki>, with a radius from <nowiki>$P$</nowiki> of <nowiki>$R$</nowiki>.

[[File:EM-Y1S2-lec14-04.png|thumb|center|600px]]

The Biot-Savat law gives: <div class="pn-import-equation"><nowiki>$$\delta \underline{B} = \frac{\mu_0}{4 \pi} I \frac{\delta l \wedge \hat{r}}{r^2}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\delta B = \frac{\mu_0}{4 \pi} I \frac{\delta \underline{L}}{r^2}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$B = \frac{\mu_0}{4 \pi} \frac{I}{r^2} \oint dL = \frac{\mu_0}{4 \pi} \frac{I}{4 \pi} \times 2 \pi r$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$B = \frac{\mu_0 I}{2r}$$</nowiki></div>

== Finite Wire ==

[[File:EM-Y1S2-lec14-05.png|thumb|center|600px]]

We want to find the B-field from the lime of current. The B-field is going into the page at <nowiki>$P$</nowiki>, given by the cross. The Biot Savat Law again gives: <div class="pn-import-equation"><nowiki>$$\delta \underline{B} = \frac{\mu_0 I}{4 \pi r^2} \delta \underline{L} \wedge \hat{r}$$</nowiki></div>

As <nowiki>$\hat{r}$</nowiki> is a unit vector, the magnitude is given with the magnitude of L and the sine of phi: <div class="pn-import-equation"><nowiki>$$\delta B = \frac{\mu_0 I}{4 \pi r^2} \delta L \sin \phi \tag{1}$$</nowiki></div>

We want this in terms of <nowiki>$\theta$</nowiki> instead, and given these two angles form a right angled triangle we have: <div class="pn-import-equation"><nowiki>$$\sin \phi = \sin \left(\frac{\pi}{2} - \theta\right) = \cos \theta \tag{2}$$</nowiki></div>

Hence: <div class="pn-import-equation"><nowiki>$$\delta B = \frac{\mu_0}{4 \pi r^2} \delta L \cos \theta$$</nowiki></div>

By trig we simply have: <div class="pn-import-equation"><nowiki>$$r \cos \theta = R \implies r = \frac{R}{\cos \theta} \tag{3}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\tan \theta = \frac{L}{R} \implies L = R \tan \theta \implies \frac{dL}{d \theta} = \frac{R}{\cos^2 \theta}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\implies \delta L = \frac{R}{\cos^2 \theta} \delta \theta \tag{4}$$</nowiki></div>

Inserting (2), (3), (4) back into (1) gives: <div class="pn-import-equation"><nowiki>$$\delta B = \frac{\mu_0 I}{4 \pi} \times \frac{\cos^2 \theta}{R^2} \times \cos \theta \times \frac{R}{\cos^2 \theta} \delta \theta$$</nowiki></div>

Hence: <div class="pn-import-equation"><nowiki>$$\delta B = \frac{\mu_0 I}{4 \pi} \frac{1}{R} \cos \theta \delta \theta$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$B = \frac{\mu_0 I}{4 \pi R} \oint^{\theta_2}_{\theta_1} \cos \theta d \theta$$</nowiki></div>

Where <nowiki>$\theta_1, \theta_2$</nowiki> depend the angles given by the length of the wire.

<div class="pn-import-equation"><nowiki>$$\boxed{B = \frac{\mu_0 I}{4 \pi R} \left(\sin \theta_2 - \sin \theta_1\right)}$$</nowiki></div>

If the wire was infinitely long, <nowiki>$\theta_2 = \pi/2$</nowiki> and <nowiki>$\theta_1 = -\pi/2$</nowiki> (technically a limit as <nowiki>$L \to \infty$</nowiki>). This gives us: <div class="pn-import-equation"><nowiki>$$B = \frac{\mu_0 I}{2 \pi R}$$</nowiki></div>

== More Complex Loops ==

Consider a current loop, with a set of axes centred at the middle of the loop. The loop sits in the <nowiki>$yz$</nowiki> plane, and we step out some perpendicular distance <nowiki>$x$</nowiki> in the <nowiki>$x$</nowiki> plane.

[[File:EM-Y1S2-lec14-06.png|thumb|center|600px]]

We step some distance <nowiki>$\delta L$</nowiki> along this loop, and the vector from the point we’re investigating to B-field at from this <nowiki>$\delta L$</nowiki> is <nowiki>$\underline{r}$</nowiki>, at angle <nowiki>$\theta$</nowiki>.

We can consider symmetry to see that transverse components to the axis will cancel each other out and therefore do not need considering:

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

Therefore all of the components of the B-field in the <nowiki>$yz$</nowiki> planes will cancel out, and the only resultant B-field is in the x-direction. We will fully evaluate this on Monday.

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