Electrodynamics

Ohm's Law

Force pushes charges to move to create current, so current density is proportional to force per unit charge $$\textbf{J} = \sigma \textbf{f}$$ $\sigma$ is called the conductivity. The resistivity is defined as $\rho = \frac{1}{\sigma}$. If the force is due to electromagnetic force, which is usually, then $$\textbf{J} = \sigma (\textbf{E} + \textbf{v} \times \textbf{B})$$ And usually the speed is small and negligible, so $$\bbox[5px,border:2px solid #666] { \textbf{J} = \sigma \textbf{E} }$$ This is called the Ohm's law. It is found that $I$ is proportional to $V$, so Ohm's law can be written as $$\bbox[5px,border:2px solid #666] { V = IR }$$ where $R$ is called the resistance.

Example (Griffiths Third Edition Ex 7.2)

Two long cylinders (radii $a$ and $b$ are separated by material of conductivity $\sigma$. If they are maintained at a potential difference $V$, what current flows from one to the other, in a length $L$?

Solution: The field between the cylinders is $$\vec{E}=\frac{\lambda}{2\pi\epsilon_0s}\hat{s},$$ where $\lambda$ is the charge per unit length on the inner cylinder. The current is therefore $$I=\int\vec{J}\cdot d\vec{a}=\sigma\int\vec{E}\cdot d\vec{a}=\frac{\sigma}{\epsilon_0}\lambda L$$ The potential difference between the cylinders is $$V=-\int^a_b\vec{E}\cdot d\vec{l}=\frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{b}{a}\right).$$ So, $$I=\frac{2\pi\sigma L}{\ln(b/a)}V$$

Checkpoint (Griffiths Third Edition Q7.1)

Two concentric metal spherical shells, of radius $a$ and $b$, respectively, are separated by weakly conducting material of conductivity $\sigma$. If they are maintained at a potential difference $V$, what current flows from one to the other?

Let $Q$ be the charge on the inner shell. The field between the spherical shells is $$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{r}$$ The current is therefore $$I=\int\vec{J}\cdot d\vec{a} = \sigma\int\vec{E}\cdot d\vec{a}=\sigma\frac{Q}{\epsilon_0}$$ The potential difference between the spherical shells i $$V=-\int^a_b\vec{E}\cdot d\vec{r}=-\frac{Q}{4\pi\epsilon_0}\int^a_b\frac{dr}{r^2}=\frac{Q}{4\pi\epsilon_0}\left(\frac{1}{a}-\frac{1}{b}\right)$$ So, $$I=\frac{\sigma}{\epsilon_0}\frac{4\pi\epsilon_0V}{(1/a-1/b)}$$

Drift velocity

Let $n$ be the number of charge-carrier per unit volume, $A$ be the area perpendicular to the current and $q$ be the charge of the charge-carrier. Since current is defined as rate of change of charge $$\begin{align} I &= \frac{dQ}{dt} \\ &= \frac{d(nAqx)}{dt} \\ &= nAqv \end{align}$$ Define $$\bbox[5px,border:2px solid #666] { v = \frac{I}{nAq} }$$ as the drift velocity

emf

emf (electromotive force) is defined as $$\mathcal{E} = \oint \textbf{f}\cdot d\textbf{l}$$ where $\textbf{f}$ is the force pushing the charges per unit charge. Note that $\mathcal{E}$ has the dimension of energy instead of force.

Flux Rule

Consider a loop of wire moving in a magnetic field $\textbf{B}$. The area that the loop at time $t$ and $t+dt$ is $a$. The flux change is the magnetic field across this area $$d\Phi = \oint_a \textbf{B}\cdot d\textbf{a}$$ The infinitesimal area $d\textbf{a}$ can be written as $$d\textbf{a} = \textbf{v}dt \times d\textbf{l} = \textbf{v} \times d\textbf{l} dt$$ Substituting this, $$\begin{align} \frac{d\Phi}{dt} &= \oint \textbf{B} \cdot \textbf{v} \times d\textbf{l} \\ &= -\oint (\textbf{v} \times \textbf{B} ) d\textbf{l} \\ &= -\oint \textbf{f}_{mag} \cdot d\textbf{l} \\ &= -\mathcal{E} \end{align}$$ So, $$\bbox[5px,border:2px solid #666] { \mathcal{E} = -\frac{d\Phi}{dt} }$$ Note that it is the magnetic force that contributes to the emf here.

Example (Griffiths Third Edition Ex 7.4)

A metal disk of radius $a$ rotates with angular velocity $\omega$ about a vertical axis, through a uniform field $\vec{B}$, pointing up. A circuit is made by connecting one end of a resistor to the axle and the other end to a sliding contact, which touches the outer edge of the disk. Find the current in the resistor.

Solution: The speed of a point on the disk at a distance $s$ from the axis is $v=\omega s$, so the force per unit charge is $$\vec{f}_{\text{mag}}=\vec{v}\times\vec{B}=\omega s B\hat{s}$$ The emf is $$\mathcal{E}=\int^a_0f_{\text{mag}}ds=\omega B\int^a_0sds=\frac{\omega Ba^2}{2}$$ The current is $$I=\frac{\mathcal{E}}{R}=\frac{\omega Ba^2}{2R}$$

Checkpoint (Griffiths Third Edition Q7.45)

A perfectly conducting spherical shell of radius $a$ rotates about the $z$ axis with angular velocity $\omega$, in a uniform magnetic field $\vec{B}=B_0\hat{z}$. Calculate the emf developed between the "north pole" and the equator.

The speed of a point on the spherical shell at an angle $\theta$ from the $z$ axis is $\vec{v}=\omega a \sin\theta\hat{\phi}$, so the force per unit charge is $$\vec{f}=\vec{v}\times\vec{B}=\omega aB_0\sin\theta(\hat{\phi}\times\hat{z})$$ The emf is $$\mathcal{E} = \int \vec{f}\cdot d\vec{l} = \omega a^2 B_0\int^{\pi/2}_0\sin\theta(\hat{\phi}\times\hat{z})\cdot\hat{\theta}d\theta.$$ Since $$\hat{\theta}\cdot(\hat{\phi}\times\hat{z})=\hat{z}\cdot(\hat{\theta}\times\hat{\phi})=\hat{z}\cdot\hat{r}=\cos\theta,$$ we have $$\begin{align} \mathcal{E}&=\omega a^2B_0\int^{\pi/2}_0\sin\theta\cos\theta d\theta\\ &=\omega a^2B_0\left[\frac{\sin^2\theta}{2}\right]^{\pi/2}_0\\ &= \frac{1}{2}\omega a^2 B_0 \end{align}$$

Faraday's Law

Through a series of experiment, Faraday had concluded that changing of magnetic flux induces electric field $$\bbox[5px,border:2px solid #666] { \mathcal{E} = \oint \textbf{E}\cdot d\textbf{l} = -\frac{d\Phi}{dt} }$$ Applying Stoke's theorem, $$\begin{align} \oint \textbf{E}\cdot d\textbf{l} &= -\frac{d\Phi}{dt} \\ \int \nabla \times \textbf{E} \cdot d\textbf{a} &= -\int \frac{\partial\textbf{B}}{dt} \cdot d\textbf{a} \\ \nabla \times \textbf{E} &= -\frac{\partial\textbf{B}}{dt} \end{align}$$ $$\bbox[5px,border:2px solid #666] { \nabla \times \textbf{E} = -\frac{\partial\textbf{B}}{dt} }$$ Note that it is the electric force that contributes to the emf here.

Example (Griffiths Third Edition Q 7.7)

A metal bar of mass $m$ slides frictionlessly on two parallel conducting rails a distance $l$ apart. A resistor $R$ is connected across the rails and a uniform magnetic field $\vec{B}$, pointing into the page, fills the entire region. If the bar moves to the right at speed $v$, what is the current in the resistor?

Solution: The emf is $$\begin{align} \mathcal{E} &= -\frac{d\Phi}{dt}\\ &= -Bl\frac{dx}{dt}\\ &= -Blv. \\ \end{align}$$ Since $\mathcal{E}=IR$, the current is then $$I=\frac{Blv}{R}$$ As $\vec{v}\times\vec{B}$ is upward, the current direction is downward through the resistor.

Checkpoint (Griffiths Third Edition Q7.8)

A square loop of wire with side $a$ lies on a table, a distance $s$ from a very long straight wire, which carries a current $I$. The loop is pulled away from the wire at speed $v$. Find the current in the loop.

The magnetic field generated by the long straight wire is $$\vec{B}=\frac{\mu_0I}{2\pi s}\hat{\phi}.$$ So, the flux through the square loop is $$\begin{align} \Phi &= \int \vec{B}\cdot d\vec{a}\\ &=\frac{\mu_0I}{2\pi}\int^{s+a}_a\frac{1}{s}(ads)\\ &=\frac{\mu_0Ia}{2\pi}\ln\left(\frac{s+a}{s}\right)\\ \end{align}$$ The emf is $$\begin{align} \mathcal{E} &= -\frac{d\Phi}{dt}\\ &= -\frac{\mu_0Ia}{2\pi}\frac{d}{dt}\ln\left(\frac{s+a}{s}\right), \end{align}$$ where $\frac{ds}{dt}=v$, so $$\begin{align} \mathcal{E} &= -\frac{\mu_0Ia}{2\pi}\frac{d}{dt}\ln\left(\frac{s+a}{s}\right)\\ &=-\frac{\mu_0Ia}{2\pi}\left(\frac{1}{s+a}\frac{ds}{dt}-\frac{1}{s}\frac{ds}{dt}\right)\\ &=\frac{\mu_0Ia^2v}{2\pi s(s+a)} \end{align}$$ The field points out of the page, so the force on a charge in the nearby side of the square is to the right. In the far side it's also to the right, but here the field is weaker, so the current flows counterclockwise.

Inductance

Suppose there are two loops of wire. Current $I_1$ runs through loop 1 and generates $\textbf{B}_1$. Let $\Phi_2$ be the flux through loop 2. By Biot-Savart law, we know that $\textbf{B}_1$ is proportional to $I_1$. $$\Phi_2 = M_{12}I_1$$ To find $M_{12}$ $$\begin{align} \Phi_2 &= \int \textbf{B}_1\cdot d\textbf{a}_2 \\ &= \int (\nabla \times \textbf{A}_1)\cdot d\textbf{a}_2 \\ &= \oint \textbf{A}_1 \cdot d\textbf{l}_2\\ &= \oint \left( \frac{\mu_0I_1}{4\pi}\oint \frac{d\textbf{l}_1}{d} \right) \cdot d\textbf{l}_2\\ &= \frac{\mu_0I_1}{4\pi}\oint\left(\oint\frac{d\textbf{l}_1}{d}\right)\cdot d\textbf{l}_2\\ \end{align}$$ Thus, $$M_{12} = \frac{\mu_0}{4\pi}\oint\oint\frac{d\textbf{l}_1\cdot d\textbf{l}_2}{d}$$ Similarly for $\Phi_1 = M_{21}I_2$, so $$M_{12} = M_{21}$$ It means that whatever the shape of the loop 1 and 2, the flux through loop 2 due to magnetic field generated by flowing current $I$ through loop 1 is identical to the flux through loop 1 due to magnetic field generated by flowing the same current $I$ through loop 2.

Example (Griffiths Third Edition Ex 7.11)

A short solenoid (length $l$ and radius $a$, with $n_1$ turnes per unit length) lies on the axis of a very long solenoid (radius $b$, $n_2$ turns per unit length). Current $I$ flows in the short solenoid. What is the flux through the long solenoid?

Solution: Run the current $I$ through the outer solenoid. The field inside the long solenoid is $$B=\mu_0n_2I.$$ The flux through a single loop of the short solenoid is $$B\pi a^2 = \mu_0n_2I\pi a^2$$ As there are $n_1l$ turns, the total flux through the inner solenoid is $$\Phi = \mu_0\pi a^2n_1n_2lI,$$ in which the mutual inductance is $$M=\mu_0\pi a^2n_1n_2l$$ Due to equality of the mutual inductances, the flux is also the flux through the outer solenoid when the current $I$ flows through the inner solenoid.

Checkpoint (Griffiths Third Edition Q7.22)

Find the self-inductance per unit length of a long solenoid, of radius $R$, carrying $n$ turns per unit length.

The field inside the long solenoid is $$B=\mu_0nI.$$ The flux through a single turn is $$\Phi_1=\mu_0nI\pi R^2$$ In a length $l$, there are $nl$ turns. The total flux through the inner solenoid is $$\Phi = \mu_0n^2\pi R^2Il.$$ So the self-inductance per unit length is $$L=\mu_0n^2\pi R^2$$