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Chaining the laws of electromagnetism: loops entering fields and leaving wires, coupled coils, LC energy, charges in combined fields, and energy balance.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to solve multi-step electromagnetic problems by choosing and chaining the right laws, and check them with energy.
This lesson closes the electricity and magnetism half of the course. You have Coulomb's and Gauss's laws for fields from charge, potential and capacitance, circuits with resistors, capacitors and inductors, the Biot–Savart and Ampère laws for fields from current, Faraday's law for induction, and Maxwell's completion. Here they work together, as they do in the free-response questions of the AP Physics C: Electricity and Magnetism exam.
| Term | What it means |
|---|---|
| Source law | A law that gives a field from its sources: Gauss, Biot–Savart, Ampère. |
| Induction law | Faraday's law, giving an emf from a changing flux. |
| Circuit law | Kirchhoff's rules with the element equations $V = IR$, $q/C$, $L\,dI/dt$. |
| Force law | $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$, and $I\vec{L} \times \vec{B}$ on currents. |
| Mutual inductance | $M = N_2\Phi_2/I_1$, the flux linkage in one coil per ampere in another. |
| Chain of reasoning | A sequence of laws, each feeding the next. |
| Resonant charging | Energy transfer between coils tuned to the same frequency. |
Most electromagnetic problems are chains. A current or charge makes a field; a changing field makes an emf; the emf drives a current through a circuit; the current feels a force, dissipates power, or stores energy. Each link has its law:
The skill is recognizing where one link ends and the next begins, and checking that energy balances across the whole chain: the mechanical or electrical power put in must equal the power dissipated or stored.
Another way: picture
Picture a relay race. The first runner, a current, hands a baton, its field, to the second. The second runner, the changing flux, hands an emf to the third, the circuit, which hands a current to the last, the force or the heat. Each runner has its own stretch of track and its own rules, and the race is lost if a baton is dropped between them.
Another way: steps
A loop moving into a region of field gains flux at the rate $BLv$, where $L$ is the side crossing the boundary. Faraday's law gives that emf, Ohm's law the current $BLv/R$, and the force on the side inside the field, $BIL = B^2L^2v/R$, opposes the motion. Once the loop is entirely inside a uniform field, the flux stops changing, and the emf, current and drag all vanish. Leaving the field, they return with the current reversed.
The energy balance is exact: the work done pushing the loop in, $F \times L$, equals the heat $I^2Rt$ produced in the time $L/v$ it takes to enter. Magnetic brakes, eddy current dampers and induction sensors all work this way.
Near a long wire the field falls off as $1/r$, so a loop moving away from the wire loses flux even though nothing about the wire changes. The flux is $(\mu_0Iw/2\pi)\ln((x + a)/x)$, from integrating the Biot–Savart field across the loop, and its rate of change, by the chain rule with $dx/dt = v$, gives the emf.
This combines a source law, an integral for flux, a derivative for Faraday's law and Lenz's law for direction. The induced current circulates so as to make its own field point the same way as the wire's inside the loop, trying to keep the flux from falling. The force on the loop pulls it back toward the wire.
Put a small coil inside a long solenoid. Ampère's law gives the solenoid's field, $\mu_0nI$; the small coil links $NA\mu_0nI$; ramping the solenoid's current induces $NA\mu_0n\,dI/dt$ in the small coil. The constant $NA\mu_0n$ is their mutual inductance $M$, and $\mathcal{E}_2 = M\,dI_1/dt$.
Mutual inductance is symmetric: the flux linkage in coil 2 per ampere in coil 1 equals that in coil 1 per ampere in coil 2, even when the coils are very different. This is the basis of transformers, wireless chargers, the current sensors that clamp around wires, and the pickups on electric guitars.
In an LC circuit, energy moves from the capacitor's electric field to the inductor's magnetic field and back: $\tfrac{1}{2}CV_0^2$ at one moment becomes $\tfrac{1}{2}LI_{\max}^2$ a quarter cycle later. Setting them equal gives $I_{\max} = V_0\sqrt{C/L}$ without solving any differential equation, and the frequency, $1/\sqrt{LC}$, follows from the loop rule.
Energy accounting is often the fastest route through a synthesis problem, and always the best check. Every joule supplied by a battery or a pushing hand must end as heat in a resistor, energy in a field, or work done on something else.
A charge in both an electric and a magnetic field feels $q(\vec{E} + \vec{v} \times \vec{B})$. With the fields crossed, a speed of $E/B$ passes straight through, the principle of the velocity selector. A charge accelerated through a potential difference $V$ gains speed $\sqrt{2qV/m}$ and then circles in a magnetic field with radius $mv/qB$: electric potential and magnetic force in sequence.
Mass spectrometers, cyclotrons, the electron beams of old television tubes and the ion engines on deep-space probes all combine electric acceleration with magnetic steering. Each stage uses one law, and the output of one, usually a speed, is the input to the next.
Checking an answer. Power in must equal power out. Every induced effect must oppose its cause. Limits must behave: far from a wire, induced emfs vanish; with zero resistance, $RL$ currents never decay. And units must reduce correctly at each step, not just at the end.
Using a static law, such as Ampère's law without the displacement term, inside a changing problem is an approximation, valid when the changes are slow enough that fields adjust everywhere almost instantly. For circuits the size of a lab bench, that holds up to millions of hertz. At higher frequencies, where the circuit is a sizable fraction of a wavelength, radiation matters and the full Maxwell equations are needed.
Superposition allows each source's field to be found separately and added, and linearity lets the emf from a sum of changing fluxes be the sum of their emfs. These two facts are what make chaining laws possible at all.
Electricity and Magnetism free-response questions typically describe one device, a capacitor with a dielectric, a loop near a wire, a rod on rails, and ask a sequence of parts that climb the chain: a field from Gauss's or Ampère's law, a flux integral, an emf, a current and its direction, a force, and often a differential equation for the motion or the current.
Graders reward stating the law and why it applies before using it, labeling directions with a reason, and checking energy. A correct number without the reasoning earns part credit; the reasoning without the number often earns more.
The course ends where its two halves meet. A rod on rails driven by a battery is a mechanics problem, with mass, acceleration and a terminal speed, and an electromagnetism problem, with a back emf, a current and a magnetic force. Its equation of motion, $m\,dv/dt = BL(\mathcal{E} - BLv)/R$, has exactly the form of the linear drag equation from the mechanics half, and its solution approaches terminal speed with the same kind of exponential.
The parallels run deeper. A capacitor's charge obeys the same equation as a falling body with drag; an LC circuit is a mass on a spring; an RLC circuit is a damped oscillator; the energy stored in fields plays the role of potential energy. Engineers exploit this by modeling mechanical systems as circuits and circuits as mechanical systems, whichever is easier to build or simulate.
When a problem mixes the two, the method is unchanged: split it into links, name the law for each, carry the quantities that pass between them, and check that energy balances. A student who can do that has the core of both halves of the AP Physics C course, and of the university electromagnetism course that follows.
A wireless charging pad, like those designed by engineers in Silicon Valley, drives a coil with current alternating at about $100$ to $200$ kHz, making a field of tens of microtesla above it. A coil in the phone links that changing flux, and Faraday's law gives a peak emf of several hundred millivolts, which circuits then boost and rectify to charge the battery.
The whole chain is visible: a current makes a field by Biot–Savart, the field's changing flux makes an emf, the emf drives current in the phone's circuit. Tuning both coils to resonance with capacitors, as in an LC circuit, raises the efficiency to about $70$ to $80$ percent, and the rest warms the phone and pad.
The rectangles cut into the pavement at many American traffic lights are coils of wire, part of an LC circuit oscillating at tens of kilohertz. When a car stops over one, its steel body links the coil's changing flux, and eddy currents in the car reduce the coil's inductance by a few percent.
That shifts the circuit's resonant frequency, $1/(2\pi\sqrt{LC})$, and the controller, detecting the shift, knows a car is waiting and changes the light. Motorcycles and bicycles, with less metal, shift the frequency less, which is why some intersections mark a spot for them to stop directly over the wires.
Because a moving rod in a field has a motional emf, it is tempting to think any loop moving through a field has one. A loop moving entirely inside a uniform field has motional emfs on its leading and trailing sides that cancel around the loop, and Faraday's law agrees: its flux does not change. Only while it is entering or leaving the field, or moving through a nonuniform one, is there a net emf.
A related error is to use one law for a whole chain. Gauss's law gives a field but no emf; Faraday's law gives an emf but no field. Each link needs its own law.
A $0.10$ m square loop with $R = 0.20$ Ω enters a $0.50$ T field at $2.0$ m/s. Find the emf.
$\mathcal{E} = BLv = 0.50 \times 0.10 \times 2.0 = 0.10\ \text{V}$
Faraday's law as the area inside grows.
Find the current.
$I = \dfrac{0.10}{0.20} = 0.50\ \text{A}$
Ohm's law.
Find the drag force.
$F = BIL = 0.50 \times 0.50 \times 0.10 = 0.025\ \text{N}$
On the leading side, opposing the motion.
Find the heat while entering.
$Q = I^2Rt = 0.25 \times 0.20 \times 0.050 = 2.5\ \text{mJ}$
It takes $0.050$ s to enter.
Check against the work done.
$W = FL = 0.025 \times 0.10 = 2.5\ \text{mJ}$
Energy balances.
A $40$ μF capacitor charged to $50$ V connects across a $1.0$ mH inductor. Find the energy.
$U = \tfrac{1}{2} \times 40 \times 10^{-6} \times 50^2 = 50\ \text{mJ}$
All electric at first.
Find the greatest current.
$I_{\max} = 50\sqrt{\dfrac{40 \times 10^{-6}}{1.0 \times 10^{-3}}} = 10\ \text{A}$
All magnetic a quarter cycle later.
Find the angular frequency.
$\omega = \dfrac{1}{\sqrt{1.0 \times 10^{-3} \times 40 \times 10^{-6}}} = 5000\ \text{rad/s}$
$1/\sqrt{LC}$.
Find when the current first peaks.
$t = \dfrac{T}{4} = \dfrac{\pi}{2\omega} = 0.31\ \text{ms}$
A quarter period.
Find the inductor's voltage then.
$V_L = 0$
The current is momentarily steady at its peak.
Add $2.0$ Ω and estimate the decay time.
$\tau = \dfrac{2L}{R} = 1.0\ \text{ms}$
The amplitude of an underdamped RLC falls as $e^{-Rt/2L}$.
A $0.25$ m rod of mass $50$ g rests on horizontal rails in a vertical $0.40$ T field. A $6.0$ V battery and $1.5$ Ω close the circuit. Find the starting current.
$I_0 = \dfrac{6.0}{1.5} = 4.0\ \text{A}$
No back emf yet.
Find the starting force.
$F = BIL = 0.40 \times 4.0 \times 0.25 = 0.40\ \text{N}$
The rod accelerates.
Write the current once it moves.
$I = \dfrac{6.0 - BLv}{1.5}$
Its motion induces a back emf.
Find the terminal speed.
$BLv = 6.0 \Rightarrow v = \dfrac{6.0}{0.40 \times 0.25} = 60\ \text{m/s}$
When the current and force fall to zero.
Write the equation of motion.
$m\dfrac{dv}{dt} = \dfrac{BL(6.0 - BLv)}{1.5}$
Newton's second law with the magnetic force.
Find the time constant.
$\tau = \dfrac{mR}{B^2L^2} = \dfrac{0.050 \times 1.5}{0.010} = 7.5\ \text{s}$
Like linear drag.
Explain what the battery's energy does.
$\text{half to kinetic energy, half to heat}$
Like charging a capacitor from a fixed voltage.
Find the rate the field rises.
$\dfrac{dB}{dt} = \mu_0n\dfrac{dI}{dt} = 4\pi \times 10^{-7} \times 1000 \times 50$
Ampère's law for the solenoid.
Multiply by the turns and area.
$\mathcal{E} = 20 \times 5.0 \times 10^{-4} \times 0.0628$
Faraday's law.
Evaluate the emf.
A square wire loop slides at $5$ m/s through a large region of uniform magnetic field, staying entirely inside the region, with its plane perpendicular to the field. What emf is induced around it?
Complete the worked solution: a $160$ μF capacitor charged to $80$ V is connected across a $1$ mH inductor with negligible resistance. Find the energy stored at the start in μJ, the greatest current in A, and the angular frequency of the oscillation in rad/s.
Find the capacitor's energy.
$U = \tfrac{1}{2}CV_0^2 =$ u
All the energy starts in the electric field.
Move all of it into the inductor.
$I_{\max} = V_0\sqrt{\dfrac{C}{L}} =$ i
A quarter cycle later the capacitor is empty.
Find the angular frequency.
$\omega = \dfrac{1}{\sqrt{LC}} =$ w
From $L\ddot{q} + q/C = 0$.
Check the pairing.
$I_{\max} = \omega Q_0 = \omega CV_0$
Like $v_{\max} = A\omega$ for a spring.
Match each task to the law best suited to it.
| Gauss's law | Ampère's law | Faraday's law | the Biot–Savart law | |
|---|---|---|---|---|
| field of a charged cylinder | ||||
| field inside a toroid | ||||
| emf from changing flux | ||||
| field at the center of a square loop |
A square loop $0.20$ m on a side, with resistance $0.50$ Ω, is pulled at $4$ m/s into a region of uniform $0.6$ T field perpendicular to it. While it is partly inside, fill in the induced emf in V, the current in A, and the force needed to keep it moving steadily in N.
| value | |
|---|---|
| emf (V) | |
| current (A) | |
| force needed (N) |
A long straight wire carries $150$ A. A rectangular loop in the same plane has sides $0.50$ m long parallel to the wire and $0.20$ m wide, and moves directly away from the wire at $5$ m/s. With $\mu_0 = 4\pi \times 10^{-7}$ T·m/A, write the emf around the loop, in μV, as a formula in the distance $x$ (m) from the wire to the near side.
Answer:
A $40$-turn coil of area $2$ cm² sits inside a long solenoid with $3000$ turns per meter, facing along its axis. The solenoid's current is ramped up at $500$ A/s. With $\mu_0 = 4\pi \times 10^{-7}$ T·m/A, what emf is induced in the coil, in mV?
Answer: mV in the inner coil
An engineer in San Jose, California, tests a wireless phone charger. The pad makes a field of amplitude $50$ μT oscillating at $140$ kHz through the phone's receiving coil, which has $25$ turns of area $6$ cm². What peak emf is induced in the coil, in mV?
Answer: mV peak in the phone coil
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A long straight wire carries $50$ A. A rectangular loop in the same plane has sides $0.50$ m long parallel to the wire and $0.20$ m wide, and moves directly away from the wire at $8$ m/s. With $\mu_0 = 4\pi \times 10^{-7}$ T·m/A, write the emf around the loop, in μV, as a formula in the distance $x$ (m) from the wire to the near side.
Answer:
You can chain the laws of electromagnetism. Explain to someone why a loop sliding through a uniform field has no emf until it reaches the edge.
20. Your turn: a $20$-turn coil of area $5.0$ cm² inside a solenoid with $1000$ turns per meter sees the solenoid's current rise at $50$ A/s. Find the emf., step 3
$\mathcal{E} = 0.63\ \text{mV}$
A fraction of a millivolt.