Notes
Magnetic force
- On a charge: $\vec F = q\vec v \times \vec B$. On a current: $\vec F = I\vec\ell \times \vec B$.
- Circular motion in a uniform field: $\displaystyle r = \frac{mv}{qB}$, period $\displaystyle T = \frac{2\pi m}{qB}$ (independent of speed).
- Velocity selector: crossed $\vec E$ and $\vec B$ let through only $\displaystyle v = \frac{E}{B}$.
- A current loop in a field feels a torque: $\tau = NIAB\sin\theta$.
Sources of magnetic fields
- Biot–Savart law: $\displaystyle d\vec B = \frac{\mu_0}{4\pi}\frac{I\,d\vec\ell \times \hat r}{r^2}$. Use it for arcs and loops. At the center of a circular loop: $\displaystyle B = \frac{\mu_0I}{2R}$.
- Ampère's law: $\oint\vec B \cdot d\vec\ell = \mu_0I_{enc}$. Use it with high symmetry:
| Source | Field |
|---|---|
| Long straight wire | $\displaystyle B = \frac{\mu_0I}{2\pi r}$ |
| Inside a long solenoid | $B = \mu_0nI$ ($n$ turns per meter) |
| Inside a thick wire, uniform current, $r < R$ | $\displaystyle B = \frac{\mu_0Ir}{2\pi R^2}$ |
- Parallel currents attract with force per length $\displaystyle \frac{F}{L} = \frac{\mu_0I_1I_2}{2\pi d}$.
Practice
$\hat i \times \hat j = \hat k$, and the negative charge reverses it.
$\displaystyle T = \frac{2\pi m}{qB}$ doesn't depend on speed.
$\displaystyle v = \frac{E}{B}$.
$\displaystyle \frac{\mu_0I}{2R} = \frac{(4\pi \times 10^{-7})(5.0)}{0.20}$.
$\mu_0nI = (4\pi \times 10^{-7})(1000)(2.0)$.
$I_{enc} \propto r^2$, so $B \propto r$.
Field lines are concentrated inside.
$\displaystyle \frac{\mu_0(10)(10)}{2\pi(0.10)} = 2.0 \times 10^{-4}$; opposite currents repel.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A long cylindrical wire of radius $R$ carries a current $I$ spread uniformly over its cross-section.
(a) Use Ampère's law to find $B$ outside the wire ($r > R$).
Show the answer and scoring guide
Circular loop of radius $r$: $\displaystyle B(2\pi r) = \mu_0I \Rightarrow B = \frac{\mu_0I}{2\pi r}$.
Check off the points you earned:
(b) Use Ampère's law to find $B$ inside the wire ($r < R$).
Show the answer and scoring guide
$\displaystyle I_{enc} = I\frac{r^2}{R^2}$. $\displaystyle B(2\pi r) = \mu_0I\frac{r^2}{R^2} \Rightarrow B = \frac{\mu_0Ir}{2\pi R^2}$.
Check off the points you earned:
(c) Describe the graph of $B$ vs $r$.
Show the answer and scoring guide
It rises linearly from 0 at the center to a maximum of $\displaystyle \frac{\mu_0I}{2\pi R}$ at the surface, then falls off as $\displaystyle \frac{1}{r}$.
Check off the points you earned: