Magnetic Flux vs Magnetic Field: The Difference

A magnetic field is a vector defined at a point in space, measured in tesla. Magnetic flux is a scalar attached to a surface: how much of that field passes through it, measured in webers. So flux depends on the area and the orientation of the surface as well as on the field.

AP Physics: Unit 12 (topics 12.1 Magnetic Fields, 12.4 Electromagnetic Induction and Faraday's Law). In AP Physics 2 the field is Topic 12.1 (Magnetic Fields) and flux appears inside Topic 12.4 (Electromagnetic Induction and Faraday's Law), both in Unit 12, Magnetism and Electromagnetism, weighted at 12 to 15 percent of the multiple-choice section over a suggested 10 to 14 class periods. EK 12.1.A.1 defines a magnetic field as a vector field that determines the magnetic force on moving charges, currents or magnetic materials, and EK 12.1.A.2 records that it is a vector quantity representable by vector field maps. The AP Physics 2 sheet prints the flux twice, as the dot product and as its expansion with the cosine. AP Physics C: Electricity and Magnetism promotes flux to its own topic, 13.1 Magnetic Flux, opening Unit 13 (Electromagnetic Induction) at 10 to 20 percent. Learning objective 13.1.A is to describe the magnetic flux through an arbitrary area or geometric shape. EK 13.1.A.1 defines the flux of a field constant across an area as the dot product of the field and the area vector, EK 13.1.A.1.i defines the area vector as perpendicular to the plane of the surface and outward from a closed surface, EK 13.1.A.1.ii states that the sign of the flux is given by that dot product, and EK 13.1.A.2 defines the total flux as the surface integral of the field over the surface area. The identical structure appears on the electric side at EK 8.5.A.2 and EK 8.5.A.3.

The distinction, stated once

A magnetic field is defined at a point. Ask where, and you get a vector: a magnitude in tesla and a direction. Ask again a metre away and you get a different vector. A field is a whole map of these, one arrow per point.

Magnetic flux is defined on a surface. You cannot have a flux at a point, because there is nothing there for the field to pass through. Give the surface a shape, a size and an orientation, and the flux is one signed number in webers for the whole thing.

The AP Physics 2 sheet prints the connection for a uniform field:

\Phi_B = \vec{B} \cdot \vec{A} = \lvert \vec{B} \rvert \cos\theta \lvert \vec{A} \rvert

and AP Physics C: Electricity and Magnetism generalises it to a field that varies across the surface, with EK 13.1.A.2 defining the total flux as the surface integral of the field over the surface area:

\Phi_B = \int \vec{B} \cdot d\vec{A}

Three ingredients go into that number and only one of them is the field. Change the size of the surface and the flux changes. Turn the surface and the flux changes. Leave the surface alone and change the field and the flux changes. The field itself knows about none of this, and that is the distinction the rest of this page works out.

Side by side

	Magnetic field $\vec{B}$	Magnetic flux $\Phi_B$
Belongs to	A point in space	A surface
Scalar or vector	Vector	Scalar, signed
Unit	Tesla (T)	Weber (Wb), which is $\mathrm{T \cdot m^2}$
Depends on	The sources and where you stand	The field, the area, and the orientation
Defined without a surface?	Yes	No
Uniform field, flat surface	One arrow per point	$\Phi_B = BA\cos\theta$
Non-uniform field	Still one arrow per point	$\Phi_B = \int \vec{B} \cdot d\vec{A}$
Sign comes from	Nothing; it is a direction, not a sign	The dot product with the area vector
Turning the surface 90 degrees	No effect	Takes the flux to zero
What drives induction	Nothing on its own	Its rate of change, $\Delta \Phi_B / \Delta t$

The belongs to row is the whole thing. Every other row is a consequence.

The turning the surface row is the one that produces exam questions, because it is a change in flux with no change in field anywhere, which sounds impossible until you have seen it once.

The what drives induction row is why flux exists as a named quantity at all. Nobody would bother with it if it were only a bookkeeping device. It earns its place because the induced emf depends on the rate of change of flux and on nothing else, so the whole of Faraday's law is written in terms of a quantity that has no value at any point.

The angle is measured from the area vector, and this is where the marks go

In $\Phi_B = \lvert \vec{B} \rvert \cos\theta \lvert \vec{A} \rvert$ , the angle $\theta$ sits between the field and the area vector, not between the field and the surface.

EK 13.1.A.1.i defines the area vector as perpendicular to the plane of the surface, and outward from a closed surface. So the area vector sticks out of a loop like a nail through a picture frame, not around its rim.

Work through what that means for a flat loop in a uniform field.

Loop orientation	Angle from the area vector	$\cos\theta$	Flux
Face-on to the field	$0$	$1$	Maximum, $BA$
Tilted at 60 degrees	$60^\circ$	$0.5$	Half the maximum
Edge-on to the field	$90^\circ$	$0$	Zero
Face-on, flipped over	$180^\circ$	$-1$	$-BA$

Students who measure from the plane of the loop get every one of those rows backwards: they put maximum flux at edge-on, where the true answer is zero. It is the single most productive error in this whole topic, and it is invisible in the final number, which comes out looking perfectly reasonable.

The fix is mechanical. Before touching a calculator, draw the arrow that comes out of the loop face, and measure to that.

The sign comes from the same dot product. EK 13.1.A.1.ii says the sign of the flux is given by the dot product of the magnetic field vector and the area vector, so flux running along the area vector is positive and flux running against it is negative. That sign is what makes "the flux is decreasing" a statement with content rather than a comparison of magnitudes, and it is what Lenz's law needs in order to give a direction.

The case that separates them: turn the loop and change nothing else

Take a rectangular loop, $0.20 \ \mathrm{m}$ by $0.15 \ \mathrm{m}$ , so its area is $0.030 \ \mathrm{m^2}$ . Put it in a uniform field of $0.40 \ \mathrm{T}$ . Now rotate the loop, and nothing else.

Orientation	Field at every point of the loop	Flux through the loop
Face-on	$0.40 \ \mathrm{T}$	$0.012 \ \mathrm{Wb}$
Tilted 60 degrees	$0.40 \ \mathrm{T}$	$0.0060 \ \mathrm{Wb}$
Edge-on	$0.40 \ \mathrm{T}$	$0$

One column never moves. The other runs from its maximum to zero. Not a single field vector anywhere in the room changed while that happened, and if a magnetometer were sitting at the centre of the loop it would read $0.40 \ \mathrm{T}$ throughout.

Swap the loop for a smaller one, $0.20 \ \mathrm{m}$ by $0.060 \ \mathrm{m}$ , and hold it face-on. Same field, area now $0.012 \ \mathrm{m^2}$ , flux now $0.0048 \ \mathrm{Wb}$ . Again the field is untouched.

The reverse case exists too and is worth holding alongside it. A small loop in a strong field and a large loop in a weak field can carry exactly the same flux: $0.80 \ \mathrm{T}$ through $0.0075 \ \mathrm{m^2}$ and $0.10 \ \mathrm{T}$ through $0.060 \ \mathrm{m^2}$ both give $0.0060 \ \mathrm{Wb}$ . A flux value is not a measure of how strong the field is, and two situations with equal flux can be physically nothing alike.

Only one of the two can induce anything

This is what the distinction buys you, and it is where it stops being a definitional nicety.

A loop sitting in the strongest steady magnetic field you can build has an emf of exactly zero across it. A loop in a weak field that is changing has an emf. The AP Physics 2 sheet prints the rule as

\lvert \mathcal{E} \rvert = \left\lvert \frac{\Delta \Phi_B}{\Delta t} \right\rvert

and there is no $B$ in it anywhere. It is the rate of change of flux, and only that.

So the useful question is never "how big is the field?" It is "is the flux changing, and how fast?" Three things can make it change, which come straight from the three factors in $\Phi_B = BA\cos\theta$ :

Change $B$ . Move a magnet toward the loop, or change the current in a nearby coil.
Change $A$ . Stretch, shrink or slide the loop so that more or less of it is in the field. This is where the motional emf $\mathcal{E} = B\ell v$ on the sheet comes from.
Change $\theta$ . Rotate the loop. This is a generator, and the field never changes at all.

Route 3 is the one that cannot be described without the flux concept. A generator produces electricity in a constant, unchanging field, and any explanation phrased in terms of the field alone has nothing to say about it. Faraday's law and Lenz's law both take flux as their input for exactly this reason.

One consequence worth stating because it is counterintuitive: a large flux does not mean a large emf, and a zero flux does not mean a zero emf. A loop passing through the edge-on position while rotating has zero flux at that instant and is changing flux at its fastest rate, so the emf is at a maximum precisely where the flux is zero.

Units, and the arithmetic they enforce

The tesla and the weber are different units and a numerical answer in the wrong one loses the mark on its own.

A tesla is the field unit. Rearranging the printed force law $F_B = I\ell B\sin\theta$ gives $B = F/(I\ell)$ , so one tesla is one newton per ampere metre.

A weber is the flux unit, and it is a tesla times a square metre. That factor of area is the whole difference between the two quantities written as units, and it is a useful check: if your flux answer has no square metre buried in it, you have written down a field.

A third unit appears the moment flux starts changing. Dividing a weber by a second gives a volt, because $\lvert \mathcal{E} \rvert = \lvert \Delta \Phi_B / \Delta t \rvert$ has to come out in volts. So a weber is also a volt second, and an emf in volts times a time in seconds is a change in flux in webers. That is a fast way to sanity-check an induction answer without redoing it.

AP Physics 2 does not print the weber in its unit symbols table; the table lists the tesla, and the flux equations then carry $\mathrm{T \cdot m^2}$ implicitly. Writing your answer as $0.012 \ \mathrm{T \cdot m^2}$ rather than $0.012 \ \mathrm{Wb}$ is correct and unambiguous, and it makes the area dependence visible on the page.

When it costs a mark

Measuring the angle from the plane of the loop. Covered above, and worth repeating because it is so common. The angle is from the area vector, which is perpendicular to the loop.

Dropping the cosine. Writing $\Phi_B = BA$ when the loop is tilted. Right answer only when the loop is face-on, and there is no warning in the arithmetic that it has gone wrong.

Answering a flux question in tesla. Or a field question in webers. Both are visible to a grader without reading the working.

Assuming a strong field means a big flux. It does not, if the surface is small or edge-on. The worked example above has a $0.40 \ \mathrm{T}$ field producing zero flux.

Assuming a big flux means an induced emf. It does not, unless the flux is changing. A steady flux of any size induces nothing at all.

Treating flux as a vector. It is a signed scalar. Resolving it into components, or drawing an arrow for it, is a defect even when the number is right. Its sign already carries all the directional information it has.

Forgetting that a curved or tilted surface still has one flux. Flux belongs to the whole surface, so the answer is one number, not a distribution. If your answer varies from place to place on the surface, you have computed a field.

When they look interchangeable

In the most common textbook picture they are proportional, which is exactly why the distinction slips.

Hold a fixed flat loop face-on to a uniform field and $\Phi_B = BA$ with a constant $A$ . Double the field and the flux doubles; halve it and the flux halves. In that setup flux is nothing more than the field wearing an area as a coefficient, and every sentence about one can be turned into a sentence about the other. Most introductory diagrams are drawn exactly this way, with the field arrows piercing a loop head-on.

The agreement breaks in three places, and those three are the exam:

The surface is tilted, so the cosine appears and the proportionality picks up a factor that has nothing to do with the field.
The surface changes size or moves partly out of the field, so the flux changes with the field held constant.
The field varies across the surface, so there is no single $B$ to multiply by, and only an integral will do. AP Physics 2 keeps to uniform fields; AP Physics C: Electricity and Magnetism does not.

A fourth, subtler case: the field can be nonzero everywhere on a surface and the flux still zero, if the field enters part of the surface and leaves the rest. That is not a special trick, it is what always happens on a closed surface, and it is the subject of electric flux vs magnetic flux rather than this page.

How the exam frames it, and what to read next

In AP Physics 2 there is no separate flux topic. Flux appears inside Topic 12.4, Electromagnetic Induction and Faraday's Law, while the field itself is Topic 12.1. Both are in Unit 12, weighted at 12 to 15 percent of the multiple-choice section over a suggested 10 to 14 class periods. The placement tells you what flux is for in this course: it is the input to induction and it is not examined for its own sake.

AP Physics C: Electricity and Magnetism gives it a topic of its own, Topic 13.1 Magnetic Flux, opening Unit 13 (Electromagnetic Induction) at 10 to 20 percent. Learning objective 13.1.A asks students to describe the magnetic flux through an arbitrary area or geometric shape, EK 13.1.A.1 gives the dot product form for a constant field, and EK 13.1.A.2 gives the surface integral for the general case. The word "arbitrary" in that objective is the difference between the two courses: Physics 2 stays with flat loops in uniform fields.

For the definitions on their own, see magnetic field and magnetic flux. For what a field does to a moving charge, and how it differs from an electric field, see electric vs magnetic field. For the closed-surface behaviour that separates the two kinds of flux, see electric flux vs magnetic flux.

One field, three orientations, two loops

A rectangular loop measuring $0.20 \ \mathrm{m}$ by $0.15 \ \mathrm{m}$ sits in a uniform magnetic field of $0.40 \ \mathrm{T}$ . Find the magnetic flux through the loop when it is (a) face-on to the field, (b) tilted so that its area vector makes $60^\circ$ with the field, (c) edge-on. (d) Then replace it with a loop measuring $0.20 \ \mathrm{m}$ by $0.060 \ \mathrm{m}$ , held face-on, and find the flux. (e) State the field at the centre of the loop in each case.

Area of the first loop: $A = (0.20)(0.15) = 0.030 \ \mathrm{m^2}$ .
(a) Face-on means the area vector is along the field, so $\theta = 0$ and $\cos 0 = 1$ . $\Phi_B = BA\cos\theta = (0.40)(0.030)(1) = 0.012 \ \mathrm{Wb}$ .
(b) At $\theta = 60^\circ$ , the AP trigonometric values table gives $\cos 60^\circ = 1/2$ . $\Phi_B = (0.40)(0.030)(0.5) = 0.0060 \ \mathrm{Wb}$ , exactly half.
(c) Edge-on means the area vector is perpendicular to the field, so $\theta = 90^\circ$ and $\cos 90^\circ = 0$ . $\Phi_B = 0$ .
(d) The smaller loop has $A = (0.20)(0.060) = 0.012 \ \mathrm{m^2}$ , and face-on gives $\Phi_B = (0.40)(0.012) = 0.0048 \ \mathrm{Wb}$ .
(e) The magnetic field is $0.40 \ \mathrm{T}$ at the centre of the loop in all four cases, and at every other point too, because the field is uniform and nothing has been done to it. Rotating a loop does not disturb the field it sits in, and neither does swapping the loop for a smaller one.

$0.012 \ \mathrm{Wb}$ , $0.0060 \ \mathrm{Wb}$ , $0$ , and $0.0048 \ \mathrm{Wb}$ . The field is $0.40 \ \mathrm{T}$ throughout. Four different fluxes, one field.

Rotating the loop: an emf out of a field that never changes

The same $0.030 \ \mathrm{m^2}$ loop starts face-on in the same uniform $0.40 \ \mathrm{T}$ field. (a) It is rotated to edge-on in $0.050 \ \mathrm{s}$ . Find the average induced emf. (b) It is instead held still, face-on, for the same $0.050 \ \mathrm{s}$ . Find the induced emf. (c) Explain the difference in one sentence.

(a) Initial flux, face-on: $\Phi_i = BA = (0.40)(0.030) = 0.012 \ \mathrm{Wb}$ . Final flux, edge-on: $\Phi_f = 0$ .
Change in flux: $\Delta \Phi_B = 0 - 0.012 = -0.012 \ \mathrm{Wb}$ .
Average emf from the sheet: $\lvert \mathcal{E} \rvert = \lvert \Delta \Phi_B / \Delta t \rvert = 0.012/0.050 = 0.24 \ \mathrm{V}$ .
Units check: webers per second is volts, since a weber is a volt second. $0.012 \ \mathrm{V \cdot s}$ divided by $0.050 \ \mathrm{s}$ gives $0.24 \ \mathrm{V}$ .
(b) Held still, the flux stays at $0.012 \ \mathrm{Wb}$ , so $\Delta \Phi_B = 0$ and the emf is zero. The field is still $0.40 \ \mathrm{T}$ , and the flux is still at its maximum value, and neither of those facts induces anything.
(c) The field was identical in both parts, so the field cannot be what produced the emf. What differed was the rate of change of flux, which was $0.24 \ \mathrm{Wb/s}$ in the first case and zero in the second.

$0.24 \ \mathrm{V}$ while rotating and $0$ while held still, in the same unchanging $0.40 \ \mathrm{T}$ field. This is a generator in miniature: an emf produced without changing the magnetic field anywhere.

Same flux, different physics; and finding B from a flux

(a) Show that a field of $0.80 \ \mathrm{T}$ through a face-on area of $0.0075 \ \mathrm{m^2}$ and a field of $0.10 \ \mathrm{T}$ through a face-on area of $0.060 \ \mathrm{m^2}$ give the same magnetic flux. (b) A circular loop of radius $0.10 \ \mathrm{m}$ is held face-on in a uniform field and the flux through it is $6.0 \times 10^{-3} \ \mathrm{Wb}$ . Find the field.

(a) First case: $\Phi_B = BA = (0.80)(0.0075) = 0.0060 \ \mathrm{Wb}$ .
Second case: $\Phi_B = (0.10)(0.060) = 0.0060 \ \mathrm{Wb}$ . Identical.
The two situations are not remotely alike. The first has a field eight times stronger; the second has an area eight times larger. Flux cannot distinguish them, so a flux value on its own tells you nothing about the strength of the field.
(b) The area of a circle is $A = \pi r^2$ , which the AP geometry table prints. $A = \pi (0.10)^2 = \pi (0.010) = 3.14 \times 10^{-2} \ \mathrm{m^2}$ .
Face-on means $\cos\theta = 1$ , so $B = \Phi_B / A = (6.0 \times 10^{-3})/(3.14 \times 10^{-2})$ .
$B = 0.191 \ \mathrm{T}$ , which rounds to $0.19 \ \mathrm{T}$ at two significant figures.
Check the units: webers divided by square metres is $\mathrm{T \cdot m^2 / m^2}$ , which is tesla. Dividing a flux by an area gives a field, which is the arithmetic statement of everything on this page.

(a) Both give $0.0060 \ \mathrm{Wb}$ , from completely different fields and areas. (b) $B = 0.19 \ \mathrm{T}$ .

Frequently asked questions

What is the difference between magnetic flux and magnetic field?

A magnetic field is a vector defined at a single point in space, measured in tesla, and it exists whether or not any surface is nearby. Magnetic flux is a signed scalar defined on a surface, measured in webers, and it counts how much of the field passes through that surface. Because flux is field times area times the cosine of the angle to the area vector, it changes when the surface is resized or rotated even though the field is untouched. You can have a flux only if you have named a surface.

Can magnetic flux be zero when the magnetic field is not?

Yes, and it is easy to arrange. Turn a loop edge-on to a uniform field and the angle between the field and the area vector is 90 degrees, so the cosine is zero and the flux is zero, while the field at every point of the loop is exactly what it was. A 0.40 T field through a loop of area 0.030 square metres gives 0.012 Wb face-on and zero edge-on, with no change to the field anywhere.

Is the angle in the magnetic flux formula measured from the loop or from the area vector?

From the area vector, which points perpendicular to the plane of the surface. AP Physics C: E&M EK 13.1.A.1.i states this directly, and the same convention applies in AP Physics 2. A loop lying face-on to the field has an angle of zero and maximum flux; a loop edge-on has an angle of 90 degrees and zero flux. Measuring from the plane of the loop instead reverses every answer you will get, and the resulting numbers look perfectly plausible.

What are the units of magnetic flux?

The weber, which is one tesla times one square metre. It is also one volt second, because the induced emf is the rate of change of flux, so webers divided by seconds must come out in volts. That gives you two independent checks: a flux answer should carry an area factor, and a flux divided by a time should come out in volts.

Why does induction depend on flux rather than on the magnetic field?

Because a loop can produce an emf without the field changing at all. Rotate a loop in a steady uniform field and the flux through it varies from maximum to zero and back, so an emf appears, and that is how a generator works. The printed relation is that the magnitude of the emf equals the magnitude of the rate of change of flux, with no B in it. Three things can change the flux and only one of them is the field: the field, the area, or the orientation.

Does a strong magnetic field mean a large magnetic flux?

No. Flux is the product of field, area and the cosine of the angle to the area vector, so a strong field through a small or tilted surface can give less flux than a weak field through a large face-on one. A field of 0.80 T through 0.0075 square metres and a field of 0.10 T through 0.060 square metres both give 0.0060 Wb. Flux is not a measure of field strength, and a given flux does not identify the situation that produced it.

Is magnetic flux a vector?

No, it is a signed scalar. It comes from a dot product, and a dot product of two vectors is a scalar. The sign is supplied by that dot product: flux running along the area vector counts positive and flux running against it counts negative. That sign is doing real work, since it is what lets Lenz's law say whether the flux is increasing or decreasing rather than merely comparing magnitudes.