Electricity-And-Magnetism · Unit 1 · Video 4 · Interactive Practice

Closed Line Integrals: Work Around a Loop

IKey Formulas

Formula	Name	When it applies
$W = \oint_{C} \mathbf{F} \cdot d\mathbf{s}$	Work around a closed path	Zero for conservative forces (gravity, electrostatic); can be non-zero otherwise
$\mathbf{F} \cdot d\mathbf{s} = F_r\,dr + F_\theta\, r\,d\theta$	Dot product in polar coordinates	On a centred circle $dr = 0$ , so only $F_\theta\, r_0\, d\theta$ survives
$\displaystyle\oint_{C_2} \mathbf{F} \cdot d\mathbf{s} = \int_0^{2\pi} \frac{k}{r_0}\, r_0\, d\theta = 2\pi k$	Whirlpool force $\mathbf{F} = +\frac{k}{r}\hat{\mathbf{u}}_\theta$	Counterclockwise trip around any centred circle
$\oint \mathbf{F} \cdot d\mathbf{s} = F_\theta \cdot 2\pi r_0$	Azimuthal, axially symmetric field	Force times distance; for a wire, $\oint \mathbf{B} \cdot d\mathbf{s} = 2\pi r B_\theta$

Key Insight: When $F_\theta \propto 1/r$ — the whirlpool force, or the field $B_\theta$ of a long straight wire — the circumference $2\pi r$ cancels the $1/r$ , so every circle centred on the axis gives the same closed integral.

IIVisualization 1 — Adding Up $\mathbf{F} \cdot d\mathbf{s}$ Around a Loop

Carried once around a closed loop, gravity's work always returns to zero; the whirlpool's need not.

Force Loop Progress around C₁ 0%

IIIVisualization 2 — Why the Radius Drops Out

Around a centred circle $W = F_\theta \cdot 2\pi r_0$ , and only a $1/r$ force makes the radius drop out.

Force law Radius r₀ 1.50 m

💡 The magnetic field of a long straight wire also falls off as $1/r$ , so $\oint \mathbf{B} \cdot d\mathbf{s} = 2\pi r B_\theta$ is the same on every circle around it — a quantity in tesla-metres, not joules.

IVVisualization 3 — Sign Conventions

The work cannot depend on the sign convention — provided $F_\theta$ and $\theta$ use the same one.

Positive direction Trip around C₂

VQuiz Questions

Problem 1 · Work of the Whirlpool Force

Given: the azimuthal force $\mathbf{F} = +\frac{k}{r}\,\hat{\mathbf{u}}_\theta$ with $k = 3.0\ \text{N}\cdot\text{m}$ (counterclockwise positive). A body is carried once counterclockwise around a circle of radius $r_0 = 0.50\ \text{m}$ centred on the origin. Find the work $W = \oint \mathbf{F} \cdot d\mathbf{s}$ .

✅ Correct!

W = 2\pi k = 6.0\pi\ \text{J} \approx 18.8\ \text{J}

— the radius never enters.

❌ Not quite. A closed path gives zero only for a conservative force. This force points along the motion at every point, so every element

F_\theta\, r_0\, d\theta

is positive.

❌ Close, but…

2\pi k r_0

keeps an extra factor of

r_0

. The force on the circle is

k/r_0

, and its

1/r_0

cancels the

r_0

of the arc length

r_0\, d\theta

❌ Close, but…

2\pi k/r_0

uses the force

k/r_0

but forgets that each arc element has length

r_0\,d\theta

, not

d\theta

❌ Not quite. On the circle,

\mathbf{F} \cdot d\mathbf{s} = F_\theta\, r_0\, d\theta

; integrate

\theta

from

0

2\pi

Show solution

On a centred circle $dr = 0$ and $F_r = 0$ , so $\mathbf{F} \cdot d\mathbf{s} = F_\theta\, r_0\, d\theta$ with $F_\theta = +k/r_0$ :

W = \int_0^{2\pi} \frac{k}{r_0}\, r_0\, d\theta = k \int_0^{2\pi} d\theta = 2\pi k

W = 2\pi\,(3.0\ \text{N}\cdot\text{m}) = 6.0\pi\ \text{J} \approx 18.8\ \text{J}

Positive, because the body moves along the force at every point. The value $r_0 = 0.50\ \text{m}$ is not needed: every centred circle gives $2\pi k$ .

Problem 2 · Reversing the Trip

Given: the same force $\mathbf{F} = +\frac{k}{r}\,\hat{\mathbf{u}}_\theta$ (counterclockwise positive), but now the body is carried once around a centred circle of radius $r_0$ clockwise. Find the work done by the force.

✅ Correct! The body now moves against the force at every point, so every work element is negative.

❌ Close, but…

+2\pi k

survives a change of convention, which flips the signs of both

F_\theta

and the limits. Reversing the trip is a physical change: only the limits flip,

\theta

runs from

0

-2\pi

❌ Not quite. The force is non-conservative; going around the other way does not make its closed integral vanish, it reverses its sign.

❌ Close, but… The sign is right, but the

r_0

of the arc length

r_0\, d\theta

cancels the

1/r_0

in the force.

❌ Not quite. Keep the counterclockwise convention and change only the limits.

Show solution

With counterclockwise positive, $F_\theta = +k/r_0$ still, and a clockwise trip takes $\theta$ from $0$ to $-2\pi$ :

W = \int_0^{-2\pi} \frac{k}{r_0}\, r_0\, d\theta = k\,(-2\pi - 0) = -2\pi k

The force points counterclockwise while every $d\mathbf{s}$ points clockwise, so each element $\mathbf{F} \cdot d\mathbf{s}$ is negative. Choosing clockwise as positive instead would make $F_\theta = -k/r$ and the limits $0 \to +2\pi$ , giving the same $-2\pi k$ .

Problem 3 · Constant Tangential Force, Several Laps

Given: a bead on a circular track of radius $r_0 = 0.40\ \text{m}$ is pushed by a force that always points along the track in the direction of motion, with constant size $F_\theta = 2.0\ \text{N}$ . The bead goes around the track 3 times. Find the total work done by this force.

✅ Correct! Force times distance:

2.0\ \text{N} \times 3 \times 2\pi(0.40\ \text{m}) = 4.8\pi\ \text{J} \approx 15.1\ \text{J}

❌ Close, but…

5.03\ \text{J}

is one lap. Each lap adds another

F_\theta \cdot 2\pi r_0

— the force never turns against the motion.

❌ Not quite. The distance per lap is the circumference

2\pi r_0

, not the radius

r_0

❌ Not quite.

\pi r_0^2

is the area enclosed, not the distance travelled; the line integral needs the circumference

2\pi r_0

❌ Not quite. Use

W = F_\theta \cdot 2\pi r_0

per lap.

Show solution

The force is azimuthal ( $F_r = 0$ ) and has the same size at every point, so one lap gives

W_1 = \oint \mathbf{F} \cdot d\mathbf{s} = F_\theta\, r_0 \int_0^{2\pi} d\theta = F_\theta \cdot 2\pi r_0 = (2.0)(2\pi)(0.40) = 1.6\pi\ \text{J} \approx 5.03\ \text{J}

Three laps take $\theta$ from $0$ to $6\pi$ :

W = F_\theta\, r_0 \int_0^{6\pi} d\theta = 3 W_1 = 4.8\pi\ \text{J} \approx 15.1\ \text{J}

Problem 4 · Magnetic Field Around a Wire

Given: the magnetic field of a long straight wire is azimuthal with $B_\theta \propto 1/r$ . At $r = 0.10\ \text{m}$ from the wire it has size $B_\theta = 2.0 \times 10^{-5}\ \text{T}$ . Find $\oint \mathbf{B} \cdot d\mathbf{s}$ around a circle of radius $0.30\ \text{m}$ centred on the wire, taken in the direction of $\mathbf{B}$ .

✅ Correct! Because

B_\theta \propto 1/r

2\pi r B_\theta

is the same on every circle around the wire.

❌ Close, but… This uses the larger circle with the field measured at

0.10\ \text{m}

. At

0.30\ \text{m}

the field is three times weaker.

❌ Close, but check the units.

\mathbf{B}

is not a force, so

\oint \mathbf{B} \cdot d\mathbf{s}

is not work: its units are tesla-metres.

❌ Close, but… This uses the weakened field at

0.30\ \text{m}

with the circumference of the smaller circle.

❌ Not quite. Use

\oint \mathbf{B} \cdot d\mathbf{s} = 2\pi r B_\theta

with

B_\theta

evaluated on the same circle.

Show solution

On the larger circle the field is weaker by the factor $0.10/0.30$ :

B_\theta(0.30\ \text{m}) = 2.0 \times 10^{-5}\ \text{T} \times \frac{0.10}{0.30} \approx 6.67 \times 10^{-6}\ \text{T}

$\mathbf{B}$ and $d\mathbf{s}$ both point along the circle and $B_\theta$ is the same all the way round, so

\oint \mathbf{B} \cdot d\mathbf{s} = 2\pi r B_\theta = 2\pi\,(0.30\ \text{m})\,(6.67 \times 10^{-6}\ \text{T}) \approx 1.26 \times 10^{-5}\ \text{T}\cdot\text{m}

The same value comes from the $0.10\ \text{m}$ circle: $2\pi(0.10)(2.0 \times 10^{-5}) = 4.0\pi \times 10^{-6} \approx 1.26 \times 10^{-5}\ \text{T}\cdot\text{m}$ . Ampère's law will relate this number to the current in the wire.

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Closed Line Integrals: Work Around a Loop

IKey Formulas

IIVisualization 1 — Adding Up F⋅ds\mathbf{F} \cdot d\mathbf{s}F⋅ds Around a Loop

IIIVisualization 2 — Why the Radius Drops Out

IVVisualization 3 — Sign Conventions

VQuiz Questions

IIVisualization 1 — Adding Up $\mathbf{F} \cdot d\mathbf{s}$ Around a Loop