Main content

### Course: AP®︎/College Physics 2 > Unit 5

Lesson 4: Magnetic flux and Faraday's law# What is magnetic flux?

Learn what magnetic flux means and how to calculate it.

# What is magnetic flux?

Magnetic flux is a measurement of the total magnetic field which passes through a given area. It is a useful tool for helping describe the effects of the magnetic force on something occupying a given area. The measurement of magnetic flux is tied to the particular area chosen. We can choose to make the area any size we want and orient it in any way relative to the magnetic field.

If we use the field-line picture of a magnetic field then every field line passing through the given area contributes some magnetic flux. The angle at which the field line intersects the area is also important. A field line passing through at a glancing angle will only contribute a small component of the field to the magnetic flux. When calculating the magnetic flux we include only the

**component**of the magnetic field vector which is**normal**to our test area.If we choose a simple flat surface with area $A$ as our test area and there is an angle $\theta $ between the normal to the surface and a magnetic field vector (magnitude $B$ ) then the magnetic flux is,

In the case that the surface is perpendicular to the field then the angle is zero and the magnetic flux is simply $BA$ . Figure 1 shows an example of a flat test area at two different angles to a magnetic field and the resulting magnetic flux.

**Exercise 1:**

If the blue surfaces shown in Figure 1 both have equal area and the angleis $\theta $ , how much smaller is the flux through the area in Figure 1-left vs Figure 1-right? ${25}^{\circ}$

# How do we measure magnetic flux?

The SI unit of magnetic flux is the Weber (named after German physicist and co-inventor of the telegraph Wilhelm Weber) and the unit has the symbol $\mathrm{Wb}$ .

Because the magnetic flux is just a way of expressing the magnetic field in a given area, it can be measured with a $0.5\text{}{\mathrm{m}}^{2}$ area near a large sheet of magnetic material and indicates a constant reading of $5\text{}\mathrm{mT}$ . The magnetic flux through the area is then $(5\cdot {10}^{-3}\text{}\mathrm{T})\cdot (0.5\text{}{\mathrm{m}}^{2})=0.0025\text{}\mathrm{Wb}$ . In the event that the magnetic field reading changes with position, it would be necessary to find the average reading.

*magnetometer*in the same way as the magnetic field. For example, suppose a small magnetometer probe is moved around (without rotating) inside aA related term that you may come across is the $\mathrm{Wb}/{\mathrm{m}}^{2}$ . Because we are dividing flux by area we could also directly state the units of flux density in Tesla. In fact, the term magnetic flux density is often used synonymously with the magnitude of the magnetic field.

*magnetic flux density*. This is measured in**Exercise 2:**

Figure 2 shows a map of a non-uniform magnetic field measured near a sheet of magnetic material. If the green line represents a loop of wire, what is the magnetic flux through the loop?

# Why is this useful?

There are a couple of reasons why the description of magnetic flux can be more useful than that of a magnetic field directly.

- When a coil of wire is moved through a magnetic field a voltage is generated which depends on the magnetic flux through the area of the coil. This is described by
*Faraday's law*and is explored in our article on Faraday's law. Electric motors and generators apply Faraday's law to coils which rotate in a magnetic field as depicted in Figure 3. In this example the flux changes as the coil rotates. The description of magnetic flux allows engineers to easily calculate the voltage generated by an electric generator even when the magnetic field is complicated. - Although we have thus-far only concerned ourselves with magnetic flux measured for a simple flat test-area, we can make our test-area a surface of any shape we like. In-fact, we can use a
*closed surface*such as a sphere which encloses a region of interest. Closed surfaces are particularly interesting to physicists because of*Gauss's law for magnetism*. Because magnets always have two poles there is no possibility (as far as we know) that there is a magnetic monopole inside a closed surface. This means that the**net**magnetic flux through such a closed surface is always zero and therefore all the magnetic field lines going into the closed surface are exactly balanced by field lines coming**out**. This fact is useful for simplifying magnetic field problems.

# Magnetic flux around a current-carrying wire

**Exercise 1:**

Figure 4 shows a square loop of wire placed near a current carrying wire. Using the dimensions shown in the figure, find the magnetic flux through a coil. If you don't know how to calculate the magnetic field around a wire, review our article on the magnetic field. Hint: it may be useful to plot the magnetic field vs vertical distance from the wire.

## Want to join the conversation?

- Why we use cos( theta ) instead of sin (theta) in magnetic flux density's equation? I mean what is the reason behind using cos ( theta )(11 votes)
- Hello Faria,

Consider COS(θ) as shown in Figure 1. If θ = 0° as shown in part B then the maximum number of flux lines pass through the blue coil. When the coil has physically rotated 90 degrees we find θ = 90° and there are no flux lines passing through the coil.

We could have used SIN(θ) but we would need to rotate our plane of reference (green) to be vertical instead of horizontal.

Regards,

APD(35 votes)

- Solution to How do we measure magnetic flux? Exercise 2

I agree with the calculations up until they say

Magnetic Flux Φ = (0.05 m ⋅ 0.06 m) ⋅ (4.1 mT) = 0.451 mWb

As far as I can tell Magnetic Flux Φ should = 0.0123 mWb? I could be wrong, if so please explain what I'm doing wrong. Thanks so much for taking the time to address this!(12 votes)- Yes, it looks like they added .05 and .06 rather than multiplying.

I have notified the physics specialist at KA.

Good catch. Thanks.(5 votes)

- I'm confused as to how the solution for the last exercise was derived. Firstly, how that graph was plotted seems out of context to me, but since it's one of the ways this question can be solved I would think of a different approach. Secondly, the SI unit for magnetic fields are supposed to be in Tesla, but the solution above says Tm (Tesla meters). Why is that? And what other approach can be used to solve this problem?(10 votes)
- The graph was plotted by taking the equation for the magnitude of the magnetic field B = (μ0I)/(2πr) and treating it as a function of distance r from the wire:

B = f(r) = (μ0I)/(2πr) [assume I is constant].

To get the flux we need to multiply B times A, but the value of B is changing over the length of the side of area A which is perpendicular to the wire.

This is why we use the area under the curve of the function (the integral) to get a precise sum of the varying magnitudes of B over the length of the side of A moving away from the wire, rather than just multiplying by the value of B at some arbitrary point.(4 votes)

- I don't get why closed surfaces have zero magnetic flux. Also, how is it different from magnetic flux in a non closed area?(4 votes)
- A closed surface is something like a ball or balloon (even if its a weird shaped one). If a magnetic flux line cuts the surface going in, then it must cut the surface on its way out. So the net 'in' = zero

non-closed is like one side of a window or a sheet of paper.

flux lines usually all go into your side of the paper or come out of it. (Sometimes, if its a loop of flux it goes in then comes back out. )

ok??(11 votes)

- Third Paragraph.

Where do we get the Magnetic Field Vector? Is it the Magnetic Flux Density?

I feel like Magnetic Flux is defined as Magnetic Flux Density times the area, and that Magnetic Flux Density is defined as Magnetic Flux divided by the area.(5 votes)- The magnetic field vector B is defined by the force it would apply to a charge moving in it.

B = F/qv, so the units are (N *s) / (C *m), which we call a Tesla, or T. This is also called flux density. Flux is then given by B*A, so it is in T*m^2, which we also call a Weber.(2 votes)

- How would you use integration to solve the last problem?(2 votes)
- You would take the definite integral of the equation:

F(r) = ∫ (μ * I)/(2 * π * r) dr evaluated between 25 and 75.

This can be simplified because μ, I, 2, π are all not dependent on r giving you:

F(r) = (μ * I)/(2 * π) * ∫ 1/r dr

F(r) = (μ * I)/(2 * π) * ln(r) + c

This is the indefinite integral.

Now to evaluate the definite interfral you evaluate it from 25 to 75 which just means that you caulate:

F(75) - F(25)(7 votes)

- Please explain more in detail of how exercise 1 in magnetic flux around a current-carrying wire is calculated? How is the total area calculated?(5 votes)
- I also in other post try to explain the solution(1 vote)

- How has the y axis been ploted in the last graph? The axis representing magnetic field that is...(4 votes)
- In exercise two, What is mT? Why is it not just Tesla..? What is m?(2 votes)
- when do we use sin and when do we use cos?(0 votes)

- In the solution for the final exercise, more specifically its final paragraph, it is stated: "The field is not varying along the horizontal direction so we can simply multiply by the width of the coil." What does this mean?

Thanks!(3 votes)