PHYS 208 Lecture 16

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Biot-Savart Law

Equation for finding a magnetic field at any point $\overrightarrow{r}$ away from a current-carrying wire.

$\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\int \frac{I\mathrm{d}\overrightarrow{\ell }\times \hat{r}}{{\left|\overrightarrow{r}\right|}^2}$

For an infinitely long, straight current carrying wire

$\overrightarrow{B}(r)=\frac{{\mu }_{0}I}{2\pi r}$

Alternate Right-Hand Rule

Thumb along current, fingers point in direction of magnetic field.

For oncoming current, magnetic field is always counter-clockwise

Force Between Parallel Wires

Two infinitely long wires are separated a distance $d$ with the currents $I_1$ and $I_2$ of both flowing in the same direction.

The magnetic field at each point along each wire point in opposing direction.

$\begin{array}{rl}\overrightarrow{F}& =I\overrightarrow{\ell }\times \overrightarrow{B}\\ \overrightarrow{B}& =\frac{m{u}_{0}I}{2\pi d}\\ \frac{\overrightarrow{F}}{\ell }& ={\mu }_0\frac{I_1{I}_2}{2\pi d}\end{array}$

• When the currents are parallel, the wires attract.
• When the currents are anti-parallel (in opposite direction), the wires repel.

Magnetic Field at Center of Current Loop

A circular loop of radius $R$ has a current $I$ flowing CCW. What is the magnetic field at the center of the circle?

(Dr. Webb refused to simplify his equation due to the symmetry, so I have taken the liberty to do it for him)

At every point around the circumference, the magnetic field points out from the loop towards us, so $\overrightarrow{B}$ is in the $\hat{k}$ direction.

$\begin{array}{rl}\overrightarrow{B}& =\frac{{\mu }_0}{4\pi }\int \frac{I\mathrm{d}\ell }{R^2}\hat{k}\\ & =\frac{{\mu }_{0}I}{4\pi R^2}\int \mathrm{d}\ell \hat{k}\\ & =\frac{{\mu }_{0}I}{4\pi R^2}\left(2\pi R\right)\hat{k}\\ & =\frac{{\mu }_{0}I}{2R}\hat{k}\end{array}$

Now that was much easier to understand, involved fewer steps, and still arrived at the same answer.

For any fraction of a circle, $\overrightarrow{B}$ scales accordingly:

• Semicircle (half of a circle): $\frac{1}{2}\overrightarrow{B}=\frac{{\mu }_{0}I}{4R}\hat{k}$
• Quarter circle: $\frac{1}{4}\overrightarrow{B}=\frac{{\mu }_{0}I}{8R}\hat{k}$

Ampere's law

Finding magnetic fields using symmetry:

Suppose we integrate a magnetic field "flowing" through a closed wire loop:

$\oint \overrightarrow{B}\cdot \mathrm{d}\overrightarrow{\ell }={\mu }_0{I}_{en}$