Electricity-And-Magnetism · Unit 2 · Video 1 · Interactive Practice

Field Theory Versus Action at a Distance

IKey Formulas

Formula	Name	What it says
$\Delta t = \dfrac{d}{c}, \quad c \approx 3.00 \times 10^{8}\ \text{m/s}$	Finite transmission speed	Field theory: a change at one object reaches another a distance $d$ away after $\Delta t$ ; action at a distance would give $\Delta t = 0$
$\vec{B}_{\text{total}} = \vec{B}_1 + \vec{B}_2$	Superposition (total field)	Faraday's analogy is read off the field of every source added together, never one source alone
$B = \dfrac{\mu_0 I}{2\pi r}$	Field of a long straight wire	Given, not derived in this video (it comes later in the course): circles the wire, weakening as $1/r$ ; used below to add two wires' fields
$\dfrac{kQ}{r_0^{2}} = E_0 \;\Rightarrow\; r_0 = \sqrt{\dfrac{kQ}{E_0}}$	Null point of $+Q$ in a constant field $E_0$	Uses the point-charge field $kQ/r^{2}$ , given here (derived later in the course): upstream of the charge the two fields cancel, at distance $r_0$

Key Insight: In the total field, crowded lines push sideways on each other and stretched lines pull along their length, like rubber bands and strings.

IIVisualization 1 — Instantaneous or Finite Speed?

Field theory passes the force from element to element of space, so it arrives after $d/c$ , not instantly.

Model Separation Time t / (d/c) 0.40

IIIVisualization 2 — The Total Field of Two Wires

Neither wire's own circles show a force; the sum does: crowded lines repel, enclosing loops attract.

Currents Field drawn

💡 Challenge: locate the null point of the same-current total field with the probe, then explain why opposite currents have no null point anywhere.

IVVisualization 3 — A Point Charge in a Constant Field

The charge's own field is symmetric; adding $E_0$ creates a null point upstream and sweeps every line to the right.

Constant field E₀ (units of kQ/cm²) 0.25

VQuiz Questions

Problem 1 · How Long Does the Force Take?

Given: The Sun is $d = 1.496 \times 10^{11}\ \text{m}$ from Earth. Suppose the Sun's field suddenly changes. In field theory, with the change travelling at $c = 3.00 \times 10^{8}\ \text{m/s}$ , how long until the force on Earth changes?

✅ Correct!

\Delta t = d/c \approx 499\ \text{s} \approx 8.3

minutes: the field carries the change at a finite speed.

❌ That is action at a distance. Instantaneous transmission is exactly the older view field theory replaced; the stress in space travels at the speed of light.

❌ Check the unit conversion.

d/c

comes out in seconds, about

499\ \text{s}

; divide by

60

to get minutes.

❌ Not quite. Use

\Delta t = d/c

Show solution

The change is passed from element to element of space at speed $c$ , so it needs

\Delta t = \frac{d}{c} = \frac{1.496 \times 10^{11}\ \text{m}}{3.00 \times 10^{8}\ \text{m/s}} \approx 499\ \text{s}

499\ \text{s} \div 60\ \text{s/min} \approx 8.3\ \text{min}

Under action at a distance the answer would be $0\ \text{s}$ : no agent, no travel time.

Problem 2 · Nothing in the Gap?

Given: Two long parallel wires carry currents in the same direction. A student argues: "The two fields cancel at the midpoint between the wires, so nothing there can transmit a force, and the wires cannot interact." What is wrong with the argument?

✅ Correct! The total field is zero at one point, not everywhere; the loops that enclose both wires act like a rubber band around two pencils.

❌ Not quite. Same-direction currents attract: the wires bow toward each other.

❌ Check the directions. With same-direction currents the two circular fields point opposite ways in the gap, so they do cancel at the midpoint; the student's premise is right, the conclusion is not.

❌ That is action at a distance. Field theory explains the attraction through the field that fills the space around the wires.

❌ Not quite. Look at the total field away from the midpoint.

Show solution

Each wire's field circles it. In the gap, with both currents the same way, the circles point in opposite directions, so the total field vanishes at the midpoint: the premise holds.

But the force is not read from one point. Farther out, the two fields point the same way and add, and the total field lines form loops that enclose both wires (the figure-eight line through the zero point separates them from the small loops around each wire). Stretched lines pull along their length, so these loops squeeze the wires together: they attract. Neither wire's own field shows this; it appears only in $\vec{B}_1 + \vec{B}_2$ .

Problem 3 · Adding Two Wires' Fields

Given: Two long parallel wires $2.0\ \text{cm}$ apart carry $I = 10\ \text{A}$ each in opposite directions. Each wire alone produces $B = \dfrac{\mu_0 I}{2\pi r}$ , with $\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}$ . Find the total field magnitude at the midpoint, and at a point on the line through the wires $1.0\ \text{cm}$ outside the left wire.

Total field at the midpoint?

Total field 1.0 cm outside the left wire?

✅ Correct!

0.40\ \text{mT}

in the gap against

0.13\ \text{mT}

outside: the lines crowd into the gap, and their sideways pressure pushes the wires apart.

❌ Check the directions. Cancellation in the gap happens for same-direction currents; with opposite currents both fields point the same way between the wires.

❌ Close, but check the directions. Outside the pair, the two fields point opposite ways, so subtract:

0.20 - 0.067

, not

0.20 + 0.067

❌ Check the midpoint. Each wire is

r = 1.0\ \text{cm}

from the midpoint and gives

0.20\ \text{mT}

; both point the same way.

❌ Check the outside point. The distances are

1.0\ \text{cm}

and

3.0\ \text{cm}

, and the two fields point opposite ways there.

Show solution

At $r = 1.0\ \text{cm}$ one wire gives

B_0 = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi \times 10^{-7})(10)}{2\pi (0.010)} = 2.0 \times 10^{-4}\ \text{T} = 0.20\ \text{mT}

Midpoint: both wires are $1.0\ \text{cm}$ away, and with opposite currents their fields point the same way in the gap:

B = 0.20 + 0.20 = 0.40\ \text{mT}

Outside: the near wire is $1.0\ \text{cm}$ away ( $0.20\ \text{mT}$ ), the far wire $3.0\ \text{cm}$ away ( $0.20/3 \approx 0.067\ \text{mT}$ ), and the fields point opposite ways:

B = 0.20 - 0.067 \approx 0.13\ \text{mT}

The total field is three times stronger in the gap than at this outside point.

Problem 4 · Where the Total Field Vanishes

Given: A charge $Q = +2.0\ \text{nC}$ sits in a constant field $E_0 = 2.0 \times 10^{3}\ \text{N/C}$ pointing right. The charge alone produces $E = \dfrac{kQ}{r^{2}}$ pointing away from it, with $k = 8.99 \times 10^{9}\ \text{N m}^2/\text{C}^2$ . Find the distance of the null point (total $E = 0$ ) from the charge, and where it lies.

Distance from the charge?

On which side?

✅ Correct! The null point is

9.5\ \text{cm}

upstream; the lines that leave the charge leftward turn back there and stretch off to the right, pulling

+Q

along

\vec{E}_0

❌ Close, but…

kQ/E_0 = 0.00899\ \text{m}^2

r_0^{2}

; take the square root.

❌ Check the prefix.

1\ \text{nC} = 10^{-9}\ \text{C}

, not

10^{-6}\ \text{C}

❌ Check the distance. Set

kQ/r_0^{2} = E_0

and solve:

r_0 = \sqrt{kQ/E_0}

❌ Check the side. Cancellation needs the charge's field to point opposite to

\vec{E}_0

, i.e. to the left; the charge's field points left only on its left.

Show solution

The charge's field points away from it, so it points left (opposite to $\vec{E}_0$ ) only on the line to the left of the charge. There the magnitudes must match:

\frac{kQ}{r_0^{2}} = E_0 \;\Rightarrow\; r_0 = \sqrt{\frac{kQ}{E_0}} = \sqrt{\frac{(8.99 \times 10^{9})(2.0 \times 10^{-9})}{2.0 \times 10^{3}}} = \sqrt{0.00899\ \text{m}^2} \approx 0.095\ \text{m}

So the null point is about $9.5\ \text{cm}$ to the left of the charge. To the right both fields point right and reinforce, the lines crowd there, and their tension pulls $+Q$ to the right, the direction of the force $Q\vec{E}_0$ .

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