PHYS 208 Lecture 15

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Magnetism and Magnetic Fields

Lorentz force on an electric current:

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

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:

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

Magnetic Potential Energy

Still measured in [J]

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

Ch. 28: Sources of Magnetic Fields

Fundamental source is produced by moving charges

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

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

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

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

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

${\vec {B}}=\left({\frac {\mu _{0}}{4\pi }}\right){\frac {q{\vec {v}}\times {\hat {r}}}{r^{2}}}$

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

${\frac {1}{c^{2}}}={\frac {1}{\epsilon _{0}\mu _{0}}}$, 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, ${\vec {v}}\times {\hat {r}}$ is in the $-{\hat {k}}$ direction.

The magnitude is given by

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

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

For Current Carrying Wires (Biot-Savart Law)

Choose a point anywhere around a wire:

${\vec {B}}=\left({\frac {\mu _{0}}{4\pi }}\right)\int {\frac {I\mathrm {d} {\vec {\ell }}\times {\hat {r}}}{r^{2}}}$

Alternate Forms:

${\begin{aligned}{\vec {B}}&=\left({\frac {\mu _{0}}{4\pi }}\right)\int {\frac {I\mathrm {d} {\vec {\ell }}\times {\vec {r}}}{r^{3}}}\\&=\left({\frac {\mu _{0}}{4\pi }}\right)\int {\frac {I\mathrm {d} {\vec {\ell }}\sin {\theta }}{r^{2}}}\end{aligned}}$