PHYS 208 Lecture 15

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Magnetism and Magnetic Fields

Lorentz force on an electric current:

$\overrightarrow{F}=I\overrightarrow{l}\times \overrightarrow{B}$

Torque on a Current-Carrying Loop

Suppose a rectangular loop of wire (width $w$ and height $h$) has current $I$ running clockwise with the magnetic field going across the loop (from the bottom of our point of view). The axis of rotation is along the $x$-axis.

The magnetic moment [A · m²] is defined to be current × area (in our case, $whI$) and follows the right hand rule of the current direction (in our case, towards us).

$\mu =IA$
where $\hat{\mu }$ follows the right hand rule around the current.

The torque [N · m] experienced by the loop is as follows:

$\tau =\overrightarrow{\mu }\times \overrightarrow{B}$

Magnetic Potential Energy

Still measured in [J]

$U=-\overrightarrow{\mu }\cdot \overrightarrow{B}$

Ch. 28: Sources of Magnetic Fields

Fundamental source is produced by moving charges

Magnetic field $\overrightarrow{B}$ [T] is proportional to:

• charge $q$ [C]
• velocity $\overrightarrow{v}$ [m/s]

$\overrightarrow{B}$ is inversely proportional to:

• distance away from charge $|\overrightarrow{r}|$ [m]

Direction of $\overrightarrow{B}$ is perpendicular to the plane containing $\overrightarrow{v}$ and $\overrightarrow{r}$ (given by right-hand rule). Thus $\overrightarrow{B}$ is a closed loop around $\overrightarrow{v}$.

$\overrightarrow{B}=\left(\frac{{\mu }_0}{4\pi }\right)\frac{q\overrightarrow{v}\times \hat{r}}{r^2}$

where $\frac{{\mu }_0}{4\pi }$ is a constant measured in [T · m / A]

and ${\mu }_0=4\pi \times 10^{-7}\frac{\mathrm{N}{\mathrm{s}}^2}{{\mathrm{C}}^2}$

the constants given above yield

$\frac{1}{c^2}=\frac{1}{{ϵ}_0{\mu }_0}$, where $c$ is the speed of light (from electromagnetic radiation)

Example

Suppose an electron moving with a velocity of 10⁶ m/s in the $x$ direction. Find the magnetic field at y = 1 m and y = 1 mm (when electron is passing through the origin).

Since y is positive, $\overrightarrow{v}\times \hat{r}$ is in the $-\hat{k}$ direction.

The magnitude is given by

$|\overrightarrow{B}|=10^{-7}\frac{(1.6\times 10^{-19}\mathrm{C})\left(10^6\mathrm{m}/\mathrm{s}\right)}{r^2}=\frac{1.6\times 10^{-20}}{r^2}$

At 1 m, $|\overrightarrow{B}|=1.6\times 10^{-20}\mathrm{T}$ (WEAK)
At 1 mm, $|\overrightarrow{B}|=1.6\times 10^{-14}\mathrm{T}$

For Current Carrying Wires (Biot-Savart Law)

Choose a point anywhere around a wire:

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

Alternate Forms:

$\begin{array}{rl}\overrightarrow{B}& =\left(\frac{{\mu }_0}{4\pi }\right)\int \frac{I\mathrm{d}\overrightarrow{\ell }\times \overrightarrow{r}}{r^3}\\ & =\left(\frac{{\mu }_0}{4\pi }\right)\int \frac{I\mathrm{d}\overrightarrow{\ell }\sin \theta }{r^2}\end{array}$