Electricity-And-Magnetism · Unit 2 · Video 4 · Interactive Practice

Curl: Measuring the Circulation of a Field

IKey Formulas

Formula	Name	What it measures
$(\nabla\times\vec{v})_z = \lim_{A\to 0}\frac{1}{A}\oint_C \vec{v}\cdot d\vec{s}$	Curl (loop definition)	Counterclockwise circulation per unit area around a point
$(\nabla\times\vec{v})_z = \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y}$	Curl in the $xy$ plane	The same quantity, from partial derivatives
$\nabla\cdot\vec{v} = \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y}$	Divergence	Spreading out from (or into) a point
$\nabla\times\vec{E} = 0$ (static), $\ \nabla\cdot\vec{B} = 0$	Field structure	Static $\vec{E}$ spreads from charges; $\vec{B}$ loops

Key Insight: Any field is a curl-free part plus a divergence-free part: $\vec{F} = \vec{F}_1 + \vec{F}_2$ with $\nabla\times\vec{F}_1 = 0$ and $\nabla\cdot\vec{F}_2 = 0$ .

IICirculation per Unit Area on a Small Square

For the source, rotation and spiral, $\oint \vec{v}\cdot d\vec{s}$ divided by the enclosed area gives the curl, wherever the loop sits.

Field Side s 1

IIICurl Is Local: The 1/r Whirlpool

Both flows move in circles, yet only one has curl away from the centre: the speed profile decides.

Flow Loop r₁ 1 r₂ 2 Δθ 45° R 1.5

💡 The magnetic field of a long straight wire has the same $1/r$ shape: its curl is zero everywhere outside the wire and sits only in the wire, where the current flows.

IVSource Plus Rotation: Splitting a Field

$\vec{F} = a(x\hat{i} + y\hat{j}) + b(-y\hat{i} + x\hat{j})$ : the first part carries all the divergence, the second all the curl.

a (source) 1 b (rotation) 1 Show

💡 In electrostatics $\nabla\times\vec{E} = 0$ , so a static $\vec{E}$ is all curl-free part; $\nabla\cdot\vec{B} = 0$ , so $\vec{B}$ is all divergence-free part and always loops.

VQuiz Questions

Problem 1 · Curl from Partial Derivatives

Given: $\vec{v} = (2x + y)\,\hat{i} + (3x - y)\,\hat{j}$ — find $(\nabla\times\vec{v})_z$ .

✅ Correct!

\partial v_y/\partial x - \partial v_x/\partial y = 3 - 1 = 2

: a counterclockwise circulation.

❌ Not quite. The curl is a difference,

\partial v_y/\partial x - \partial v_x/\partial y

, not a sum.

❌ That is the divergence.

\partial v_x/\partial x + \partial v_y/\partial y = 2 - 1 = 1

; the curl uses the cross partials

\partial v_y/\partial x

and

\partial v_x/\partial y

❌ Check the order.

(\nabla\times\vec{v})_z = \partial v_y/\partial x - \partial v_x/\partial y

; reversing it flips the sign.

❌ Not quite. Differentiate

v_y

with respect to

x

and

v_x

with respect to

y

Show solution

Here $v_x = 2x + y$ and $v_y = 3x - y$ . The cross partials are

\frac{\partial v_y}{\partial x} = 3, \qquad \frac{\partial v_x}{\partial y} = 1.

(\nabla\times\vec{v})_z = \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} = 3 - 1 = 2

The curl is positive, so the circulation is counterclockwise. (For comparison, the divergence is $2 + (-1) = 1$ .)

Problem 2 · Straight Streamlines, Nonzero Curl

Given: the shear flow $\vec{v} = 2y\,\hat{i}$ . Every particle moves in a straight horizontal line, faster the higher it is. Find $(\nabla\times\vec{v})_z$ .

✅ Correct!

0 - 2 = -2

. The fluid does not travel in circles, but a small loop still collects clockwise circulation.

❌ Straight paths do not mean zero curl. Curl is measured point by point with a small loop: here the top side of a loop sees faster flow than the bottom side, so the two do not cancel.

❌ Check the sign.

\partial v_x/\partial y = 2

is subtracted:

(\nabla\times\vec{v})_z = 0 - 2

❌ Check the derivative.

\partial(2y)/\partial y = 2

, not

1

, so

(\nabla\times\vec{v})_z = 0 - 2 = -2

❌ Not quite. Use

\partial v_y/\partial x - \partial v_x/\partial y

with

v_x = 2y

and

v_y = 0

Show solution

With $v_x = 2y$ and $v_y = 0$ :

(\nabla\times\vec{v})_z = \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} = 0 - 2 = -2

Loop check on the unit square $0 \le x \le 1$ , $0 \le y \le 1$ , counterclockwise: the bottom side ( $y = 0$ ) has $v_x = 0$ and contributes $0$ ; the vertical sides contribute $0$ because $v_y = 0$ ; the top side ( $y = 1$ ) is travelled in the $-x$ direction against $v_x = 2$ and contributes $-2$ . So $\oint \vec{v}\cdot d\vec{s} = -2$ over area $1$ , matching the curl of $-2$ (clockwise).

Problem 3 · Circulation of a Whirlpool

Given: a whirlpool with counterclockwise speed $v = k/r$ , $k = 3$ . Loop 1 is an annular wedge off the centre with $r_1 = 1$ , $r_2 = 2$ , $\Delta\theta = \pi/3$ . Loop 2 is a circle of radius $R = 5$ around the centre. Find the counterclockwise circulation $\oint \vec{v}\cdot d\vec{s}$ around each.

Circulation around the wedge?

Circulation around the circle $R = 5$ ?

✅ Correct! Off the centre the arcs cancel; around the centre the circulation is

2\pi k = 6\pi

for any radius.

❌ That is only the outer arc. The inner arc is travelled against the flow and contributes

-k\Delta\theta = -\pi

❌ That is only the inner arc. The outer arc runs with the flow and contributes

+k\Delta\theta = +\pi

, cancelling it.

❌ The speeds differ on the two arcs. The inner fluid is faster:

(k/r_1)(r_1\Delta\theta) = k\Delta\theta

, exactly the same size as the outer arc's.

❌ That divides by the area.

6\pi/(25\pi) = 6/25

is circulation per unit area; the question asks for the circulation itself.

❌ The speed falls as $1/R$ . On the circle

v = k/R

, so

\oint = (k/R)(2\pi R) = 2\pi k

;

R

cancels.

❌ Zero curl off the centre does not make this loop zero. The circle encloses the centre, where all the circulation sits.

❌ Check the wedge. Add the outer arc, the inner arc (travelled against the flow) and the two radial sides.

❌ Check the circle. Speed

k/R

times circumference

2\pi R

Show solution

Wedge. The radial sides are perpendicular to the flow and contribute $0$ . The arcs:

\text{outer: } \frac{k}{r_2}\,(r_2\Delta\theta) = k\Delta\theta = 3\cdot\frac{\pi}{3} = \pi, \qquad \text{inner: } -\frac{k}{r_1}\,(r_1\Delta\theta) = -k\Delta\theta = -\pi

\oint_{\text{wedge}} \vec{v}\cdot d\vec{s} = \pi - \pi = 0

Circle. The flow is tangent to the circle with constant speed $k/R$ :

\oint_{\text{circle}} \vec{v}\cdot d\vec{s} = \frac{k}{R}\cdot 2\pi R = 2\pi k = 6\pi

The radius cancels: every loop around the centre collects $6\pi$ , while every loop that misses the centre collects $0$ . The curl is zero everywhere except at the centre.

Problem 4 · Which Fields Can Be E or B?

Given: a static electric field must satisfy $\nabla\times\vec{E} = 0$ , and a magnetic field must satisfy $\nabla\cdot\vec{B} = 0$ . Identify the field in each list that cannot play the role.

Which cannot be a static electric field?

Which cannot be a magnetic field?

✅ Correct! The rotation has curl

2

, so it cannot be a static

\vec{E}

; the source has divergence

2

, so it cannot be a

\vec{B}

❌ A source is exactly what a static E looks like.

x\hat{i} + y\hat{j}

has curl

0 - 0 = 0

; its nonzero divergence just means charge is present.

❌ Check its curl.

y\,\hat{i} + x\,\hat{j}

has

\partial(x)/\partial x - \partial(y)/\partial y = 1 - 1 = 0

, so it is allowed as a static

\vec{E}

❌ Check its curl.

2x\,\hat{i}

has

v_y = 0

and

\partial(2x)/\partial y = 0

, so its curl is

0 - 0 = 0

; it is allowed as a static

\vec{E}

❌ The rotation is the magnetic pattern.

-y\,\hat{i} + x\,\hat{j}

has divergence

0 + 0 = 0

, so it is allowed as a

\vec{B}

; its nonzero curl is what loops look like.

❌ Check its divergence.

y\,\hat{i} + x\,\hat{j}

has

\partial(y)/\partial x + \partial(x)/\partial y = 0 + 0 = 0

, so it is allowed as a

\vec{B}

❌ Check its divergence.

y\,\hat{i}

has

\nabla\cdot\vec{B} = \partial(y)/\partial x = 0

, so it is allowed; its curl of

-1

only means a current flows there.

❌ Check the curl. Compute

\partial F_y/\partial x - \partial F_x/\partial y

for each field; only a nonzero result rules out a static

\vec{E}

❌ Check the divergence. Compute

\partial F_x/\partial x + \partial F_y/\partial y

for each field; only a nonzero result rules out a

\vec{B}

Show solution

Field	$\nabla\cdot\vec{F}$	$(\nabla\times\vec{F})_z$
$x\,\hat{i} + y\,\hat{j}$	$1 + 1 = 2$	$0 - 0 = 0$
$y\,\hat{i} + x\,\hat{j}$	$0 + 0 = 0$	$1 - 1 = 0$
$-y\,\hat{i} + x\,\hat{j}$	$0 + 0 = 0$	$1 - (-1) = 2$
$2x\,\hat{i}$	$2$	$0 - 0 = 0$
$y\,\hat{i}$	$0$	$0 - 1 = -1$

Static E: needs curl $0$ . Only $-y\,\hat{i} + x\,\hat{j}$ fails (curl $2$ ).

B: needs divergence $0$ . Only $x\,\hat{i} + y\,\hat{j}$ fails (divergence $2$ ).

Note that $y\,\hat{i} + x\,\hat{j}$ has neither curl nor divergence, so it passes both tests.

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