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$\vec{J} = \sigma\vec{E}$, resistance and Joule heating, slow drift and fast signals, electromotive force, and the motional EMF $BLv$ of moving conductors.
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By the end of this lesson you will be able to relate conductivity to resistance and drift speed, explain what drives a steady current, and compute motional EMF and its energy balance.
You know DC circuits, resistance and power from Physics C, the Lorentz force, and the current density $\vec{J}$. In electrostatics the field inside a conductor was zero. Now charges keep flowing, the field inside conductors is not zero, and something must keep the charges moving around a circuit.
| Term | What it means |
|---|---|
| Conductivity | $\sigma$, with $\vec{J} = \sigma\vec{E}$; copper's is $5.96 \times 10^{7}$ S/m. |
| Resistivity | $\rho = 1/\sigma$; copper's is $1.68 \times 10^{-8}$ Ω m. |
| Drift velocity | The average velocity of charge carriers along a conductor, $v_d = J/(ne)$. |
| Joule heating | Power dissipated as heat, $\vec{J} \cdot \vec{E}$ per volume or $I^2R$ in total. |
| Electromotive force | $\mathcal{E} = \oint\vec{f} \cdot d\vec{l}$, the circulation of force per unit charge around a circuit. |
| Motional EMF | $\mathcal{E} = BLv$, from a conductor moving through a magnetic field. |
| Flux rule | $\mathcal{E} = -d\Phi/dt$ for a circuit whose flux changes. |
In most conductors the current density is proportional to the force per charge; for an ordinary circuit that is the electric field:
$$\vec{J} = \sigma\vec{E}.$$
This is Ohm's law at a point. For a uniform wire of length $L$ and area $A$, $E = V/L$ and $I = JA$, giving $V = IR$ with $R = L/(\sigma A)$. The field does work on the moving charges, which lose it in collisions: the power per volume is $\vec{J} \cdot \vec{E}$, totaling $I^2R$.
An electrostatic field alone cannot drive a steady current around a closed loop, since $\oint\vec{E} \cdot d\vec{l} = 0$. Something else must push: chemistry in a battery, light in a solar cell, motion through a magnetic field in a generator. Its strength is the electromotive force,
$$\mathcal{E} = \oint\vec{f} \cdot d\vec{l},$$
the circulation of force per unit charge. For a rod of length $L$ moving at $v$ perpendicular to a field $B$, the force per charge is $vB$ along the rod, and
$$\mathcal{E} = BLv.$$
More generally, whenever the magnetic flux through a moving circuit changes, $\mathcal{E} = -d\Phi/dt$: the flux rule, which the next lesson extends to circuits at rest.
Another way: picture
Picture a circuit as a loop of pipe full of water, the free electrons. A pump — the battery — lifts the water, and it flows back down through narrow sections, the resistors, losing its energy to friction. The pump's lift per unit of water is the EMF. The water itself moves slowly, but pressure changes travel through the full pipe almost instantly, which is why a light switches on at once.
Another way: steps
Checks. Units: $\Omega = $ V/A, and $BLv$ gives T m²/s $=$ V. The force on a moving rod must oppose its motion; otherwise energy would be created. Mechanical power $Fv$ must equal $I^2R$. And drift speeds must come out tiny, fractions of a millimeter per second in household wires.
Free electrons in a metal race around at about $10^6$ m/s, colliding with lattice imperfections and vibrations every $10^{-14}$ s or so. An applied field accelerates each electron slightly between collisions, adding a small average drift along the field. Because each collision resets the direction, the drift is proportional to the field, and so is the current density: $\sigma = ne^2\tau/m$ with $\tau$ the time between collisions.
Ohm's law is therefore an empirical rule about materials, not a fundamental law. It fails for semiconductor diodes, whose current rises exponentially with voltage; for gases, which conduct only above a breakdown field; and for superconductors, which carry current with no field at all. But for metals over an enormous range it holds to high accuracy, which is why resistors are such dependable components.
In a copper wire $1$ mm in diameter carrying $1$ A, electrons drift at about $0.09$ mm/s, slower than a crawling ant. Yet when you flip a switch, the light turns on at once. The field that drives the electrons spreads along the wire at a large fraction of the speed of light, set up by surface charges that rearrange almost instantly. Every electron in the circuit starts drifting nearly simultaneously.
The same distinction holds in signal cables: data travel along a coaxial or twisted-pair cable at about two thirds of the speed of light, while the electrons themselves barely move. What travels is energy and information, carried by the electromagnetic field around the conductors — a point the lesson on the Poynting vector makes precise.
The word force in electromotive force is historical; the EMF is a voltage, work per unit charge. What distinguishes it from an ordinary potential difference is that it is not the line integral of an electrostatic field, which vanishes around any loop. A battery uses chemical energy to push charges uphill inside itself against the electrostatic field; the EMF is the work per charge of that push.
In a circuit with a battery of EMF $\mathcal{E}$ and internal resistance $r$ driving an external resistance $R$, the current is $\mathcal{E}/(R + r)$ and the terminal voltage is $\mathcal{E} - Ir$. Lithium-ion cells have EMFs near $3.7$ V from the chemistry of their electrodes; solar cells produce about $0.6$ V each from the built-in field of a p–n junction; thermocouples a few microvolts per kelvin from temperature differences.
A conductor moving through a magnetic field carries its charges along, and the magnetic force $q\vec{v} \times \vec{B}$ pushes them along the conductor. For a rod sliding on rails, the EMF is $BLv$, and the circuit's flux $\Phi = BLx$ changes at the rate $BLv$: the EMF equals $-d\Phi/dt$. The same flux rule holds for any circuit whose shape or position changes in a static field.
Every rotating generator uses this. A coil turning in a magnetic field sees its flux oscillate, $\Phi = BA\cos\omega t$, and produces an alternating EMF $BA\omega\sin\omega t$. The generators in American power plants spin at $3600$ rpm, sixty times a second, which is why the grid runs at $60$ Hz. Wind turbines, hydroelectric dams and car alternators all work the same way.
Push a rod along rails in a field, and the current it drives feels a force $BIL$ opposing the motion. To keep the rod moving at speed $v$ you must supply power $Fv = BILv = I\mathcal{E}$, exactly the electrical power delivered, $I^2R$. The magnetic field does no work itself; it converts your mechanical work into electrical energy.
If the retarding force pointed the other way, the rod would accelerate on its own, generating current that accelerated it further: a perpetual motion machine. That this never happens is Lenz's law, and it is why eddy-current brakes work: a metal disk spinning through a magnetic field has currents induced in it that slow it down, used in roller coasters, trains and exercise bikes.
A metal's resistance rises with temperature, because hotter lattice atoms vibrate more and scatter electrons more often. For copper, resistance rises about $0.4$ percent per kelvin near room temperature. Platinum resistance thermometers, calibrated standards for industrial temperature measurement, rely on this steady rise.
Cooling does the opposite. At liquid-helium temperatures pure copper's resistivity falls by a factor of hundreds, limited only by impurities. And some materials lose resistance entirely below a critical temperature: superconductors. Niobium-titanium, used in MRI magnets, becomes superconducting below $9$ K; the copper-oxide superconductors discovered in 1986 do so above $77$ K, the boiling point of cheap liquid nitrogen.
The power lost in a transmission line is $I^2R$. To deliver power $P$ at voltage $V$ requires current $I = P/V$, so the loss is $P^2R/V^2$: raising the voltage tenfold cuts losses a hundredfold. That is why long-distance lines run at hundreds of kilovolts and why transformers, which change voltage easily, made alternating current the choice for the grid.
Even so, the U.S. Energy Information Administration estimates that about five percent of the electricity generated in the country is lost in transmission and distribution. Very long lines increasingly use high-voltage direct current, which avoids the reactive losses of AC lines; new HVDC links are being built to carry wind power from the Great Plains to cities in the east.
A long conducting wire trailing from a spacecraft in low Earth orbit sweeps through Earth's magnetic field at about $7.5$ km/s. The motional EMF, $BLv$, reaches kilovolts for a tether kilometers long. In NASA and Italy's Tethered Satellite System mission of 1996, a $20$ km tether deployed from the Space Shuttle Columbia generated about $3.5$ kV — less than the ideal $BLv$ because the tether was not perpendicular to the field — and drove currents of about an ampere through the ionosphere, before the tether broke.
Such tethers can do more than generate power. The current flowing in the tether feels a force $BIL$ that opposes the motion, slowly lowering the orbit without fuel, and engineers have proposed tethers to deorbit dead satellites and reduce space debris. Driving current the other way with solar power could raise an orbit instead, a propellant-free thruster built from the Lorentz force.
A power plant must deliver hundreds of megawatts through lines with resistance of a few ohms per hundred kilometers. At $10$ kV, delivering $100$ MW would take $10{,}000$ A, and the $I^2R$ losses in a $5$ Ω line would be $500$ MW — more than the power sent. At $500$ kV the current is $200$ A and the loss $0.2$ MW, a fifth of a percent.
That arithmetic is why transmission lines run at $115$ to $765$ kV in the United States, stepped up at power plants and down at substations. The same $I^2R$ reasoning sets the thickness of household wire, the size of electric-vehicle charging cables and the design of the copper busbars in data centers. Joule heating is the unavoidable cost of pushing charge through real conductors.
Because a light turns on the instant the switch closes, it is natural to think electrons rush from the switch to the bulb. They drift at a fraction of a millimeter per second; an electron would take hours to travel the length of a household circuit. What moves fast is the electric field, which is established along the whole wire almost instantly by surface charges, setting every electron drifting at once.
A second misconception is that a battery supplies electrons or charge. It supplies energy: the same charge circulates around the circuit, raised in energy inside the battery and losing it in the resistors. A dead battery still contains all its electrons; it has run out of chemical energy to push them.
Find the cross section of a wire $1.0$ mm in diameter.
$A = \dfrac{\pi(1.0 \times 10^{-3})^2}{4} = 7.85 \times 10^{-7}\ \text{m}^2$
Circular.
Write the resistance of $10$ m of copper.
$R = \dfrac{\rho L}{A} = \dfrac{1.68 \times 10^{-8} \times 10}{7.85 \times 10^{-7}}$
$\rho = 1/\sigma$.
Evaluate the resistance.
$R = 0.21\ \Omega$
Small, as copper wiring should be.
Find the voltage drop at $10$ A.
$V = IR = 2.1\ \text{V}$
About two percent of $120$ V.
Find the power lost as heat.
$P = I^2R = 100 \times 0.21 = 21\ \text{W}$
Why electrical codes limit current in thin wires.
A rod $0.50$ m long slides at $4.0$ m/s through a $0.20$ T field on rails joined by a $2.0$ Ω resistor. Find the EMF.
$\mathcal{E} = BLv = 0.20 \times 0.50 \times 4.0 = 0.40\ \text{V}$
Motional EMF.
Check with the flux rule.
$\dfrac{d\Phi}{dt} = BL\dfrac{dx}{dt} = BLv = 0.40\ \text{V}$
The circuit's area grows at $Lv$.
Find the current.
$I = \dfrac{0.40}{2.0} = 0.20\ \text{A}$
Ohm's law.
Find the force on the rod.
$F = BIL = 0.20 \times 0.20 \times 0.50 = 0.020\ \text{N}$
Opposing the motion.
Find the mechanical power needed.
$P_{\text{mech}} = Fv = 0.020 \times 4.0 = 0.080\ \text{W}$
Work done by whoever pushes the rod.
Find the electrical power dissipated.
$P_{\text{elec}} = I^2R = 0.04 \times 2.0 = 0.080\ \text{W}$
Energy is conserved.
Poorly conducting material fills the space between cylinders of radii $a$ and $b$, length $L$. Let a current $I$ flow radially.
$J = \dfrac{I}{2\pi sL}$
The current spreads over cylindrical surfaces.
Find the field.
$E = \dfrac{J}{\sigma} = \dfrac{I}{2\pi\sigma sL}$
Ohm's law at a point.
Integrate for the voltage.
$V = \displaystyle\int_a^bE\,ds = \dfrac{I}{2\pi\sigma L}\ln\dfrac{b}{a}$
From the inner to the outer cylinder.
Find the resistance.
$R = \dfrac{V}{I} = \dfrac{\ln(b/a)}{2\pi\sigma L}$
Longer cylinders leak more, so $R$ falls with $L$.
Relate to the capacitance.
$C = \dfrac{2\pi\varepsilon L}{\ln(b/a)} \quad\Rightarrow\quad RC = \dfrac{\varepsilon}{\sigma}$
The same geometry for both.
Evaluate the leakage time for polyethylene.
$\dfrac{\varepsilon}{\sigma} = \dfrac{2.3 \times 8.85 \times 10^{-12}}{10^{-15}} \approx 2 \times 10^{4}\ \text{s}$
A charged coax holds its charge for hours.
Compare with copper.
$\dfrac{\varepsilon_0}{\sigma_{\text{Cu}}} = \dfrac{8.85 \times 10^{-12}}{5.96 \times 10^{7}} = 1.5 \times 10^{-19}\ \text{s}$
Charge inside a metal vanishes almost instantly.
Interpret the two times.
$\tau = \varepsilon/\sigma$
The relaxation time separates conductors from insulators.
Write the motional EMF.
$\mathcal{E} = BLv$
All perpendicular.
Substitute the values.
$\mathcal{E} = 0.5 \times 0.8 \times 5$
SI units.
Evaluate the EMF.
A copper wire has resistance $43$ mΩ. What is the resistance of a wire of the same copper and thickness but twice as long?
Complete the worked solution: a rod $0.9$ m long slides at $6$ m/s on rails through a field of $0.4$ T, closing a circuit of total resistance $2$ Ω. Find the EMF in V, the current in A, and the retarding force in N.
Find the EMF.
$\mathcal{E} = BLv =$ e
Motional EMF.
Divide by the resistance.
$I = \dfrac{\mathcal{E}}{2} =$ i
Ohm's law for the circuit.
Find the force on the current.
$F = BIL =$ f
It opposes the motion: Lenz's law.
Match each quantity to its expression.
| $\vec{J} = \sigma\vec{E}$ | $L/\sigma A$ | $BLv$ | $I^2R$ | |
|---|---|---|---|---|
| Ohm's law in a material | ||||
| a wire's resistance | ||||
| motional EMF | ||||
| dissipated power |
A wire has resistance $365$ mΩ. Fill in the resistance, in mΩ, of wires of the same material that are twice as long, have twice the cross-sectional area, or have twice the diameter.
| resistance (mΩ) | |
|---|---|
| twice as long | |
| twice the area | |
| twice the diameter |
A metal rod $0.4$ m long moves at $1$ m/s perpendicular to itself and to a field of $868$ mT. What EMF appears between its ends, in mV?
Answer: mV
A copper wire of diameter $1$ mm carries $1$ A. Copper has $n = 8.49 \times 10^{28}$ free electrons per cubic meter. What is the electrons' drift speed, in mm/s?
Answer: mm/s
A spacecraft in low orbit, moving at $7.5$ km/s, trails a conducting tether $18$ km long perpendicular to its motion and to Earth's field of $29$ μT. What EMF develops along the tether, in volts?
Answer: V
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A wire has resistance $52$ mΩ. Fill in the resistance, in mΩ, of wires of the same material that are twice as long, have twice the cross-sectional area, or have twice the diameter.
| resistance (mΩ) | |
|---|---|
| twice as long | |
| twice the area | |
| twice the diameter |
You can analyze currents and EMF. Explain to someone why a light turns on instantly even though electrons drift so slowly.
19. Your turn: what EMF appears across a $0.8$ m rod moving at $5$ m/s through a $0.5$ T field?, step 3
$\mathcal{E} = 2.0\ \text{V}$
Enough to light a small bulb.