Notes
Faraday's and Lenz's laws
- Magnetic flux: $\displaystyle \Phi_B = \int\vec B \cdot d\vec A$.
- Faraday's law: $\displaystyle \varepsilon = -N\frac{d\Phi_B}{dt}$. The flux can change because $B$ changes, the area changes, or the angle changes.
- Lenz's law: the induced current makes a field that opposes the change in flux (the minus sign).
- Motional emf: a rod of length $\ell$ moving at $v$ through $B$: $\varepsilon = B\ell v$. If it's part of a circuit, the induced current feels a force that opposes the motion (magnetic braking).
- Induced electric fields: $\displaystyle \oint\vec E \cdot d\vec\ell = -\frac{d\Phi_B}{dt}$.
Inductance
- An inductor resists changes in current: $\displaystyle \varepsilon_L = -L\frac{dI}{dt}$.
- Solenoid inductance: $L = \mu_0n^2A\ell$. Energy stored: $\displaystyle U = \frac{1}{2}LI^2$.
LR circuits
- Current growing after a switch closes: $\displaystyle I(t) = \frac{\varepsilon}{R}\left(1 - e^{-t/\tau}\right)$, with $\displaystyle \tau = \frac{L}{R}$.
- Right after the switch closes, an inductor acts like a break (no current through it). After a long time, it acts like a wire. That's the opposite of a capacitor.
LC circuits
Energy sloshes between the capacitor and the inductor, like a mass on a spring. $\displaystyle \omega = \frac{1}{\sqrt{LC}}$.
Maxwell's equations
Gauss's law, Gauss's law for magnetism ($\oint\vec B \cdot d\vec A = 0$), Faraday's law, and the Ampère–Maxwell law together describe all of electromagnetism and predict electromagnetic waves.
Practice
$\displaystyle 50\left(\frac{0.060}{0.30}\right) = 10$ V.
$B\ell v = 0.50(0.40)(5.0)$.
If the induced current helped the change, energy would come from nothing.
The inductor opposes the sudden change in current.
$\displaystyle \tau = \frac{L}{R} = \frac{0.20}{40}$.
$\displaystyle \frac{1}{2}(0.50)(16)$.
Induced eddy currents create fields that oppose its motion.
$\displaystyle \omega = \frac{1}{\sqrt{(0.10)(10^{-5})}} = \frac{1}{10^{-3}}$.
Free response
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Free-response question
A metal rod of mass $m$ and length $\ell$ slides without friction on two horizontal rails connected by a resistor $R$. A uniform magnetic field $B$ points straight up. The rod is given an initial speed $v_0$ and then released.
(a) Find the current in the circuit when the rod's speed is $v$.
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$\varepsilon = B\ell v$, so $\displaystyle I = \frac{B\ell v}{R}$.
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(b) Find the magnetic force on the rod and its direction.
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$\displaystyle F = I\ell B = \frac{B^2\ell^2v}{R}$, opposite the rod's velocity (Lenz's law).
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(c) Derive $v(t)$.
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$\displaystyle m\frac{dv}{dt} = -\frac{B^2\ell^2}{R}v \Rightarrow \frac{dv}{v} = -\frac{B^2\ell^2}{mR}dt \Rightarrow v(t) = v_0e^{-B^2\ell^2t/(mR)}$.
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