AP Physics C: Electricity & Magnetism Magnetic Flux

Hello, everyone, and welcome back to www.educator.com.0000

I'm Dan Fullerton and in this lesson we are going to talk about magnetic flux.0003

Our objectives include calculating the flux of a uniform magnetic field through a loop of arbitrary orientation0009

and using integration to calculate the flux of a non uniform magnetic field whose magnitude0015

is a function of one coordinate through a rectangular loop perpendicular to the field.0020

Sounds complicated when you see the setup in the example problem, be nice and straightforward.0025

Magnetic flux sub B, you will sometimes see it written as φ sub M,0033

describes the amount of magnetic field penetrating the surface just like it did for electric flux.0037

The units of magnetic flux are Webber's, typically written as Wb.0042

Where 1 Wb is 1 tesla m².0048

Just like we did previously as we define this, we will take a small bit of area on the surface.0051

We will define the normal to it DA, going from inside to outside if it is a close surface and some magnetic field that is penetrating through that.0057

The little bit of flux that we have through that little bit of DA is going to be B ⋅ DA,0068

which is B DA cos θ if we are looking for magnitudes or θ is the angle between the normal NB.0077

Then the total magnetic flux φ B is just the integral of all of these little bits of magnetic flux D φ B,0086

which will be the integral / the open surface of B ⋅ DA.0095

Same thing as we did with electric flux.0107

When we talk about magnetic flux through close surfaces again, normal still, point from the inside to the outside.0111

Our total magnetic flux would be the integral of D φ B which is the integral / the closed surface of B ⋅ DA,0119

which for magnetic flux is going to equal 0.0132

Let us talk about that.0138

This brings us to Gauss’s law for magnetism.0140

The total magnetic flux through any close surface is 0.0143

We mentioned previously but just going a little bit more depth.0147

that could not be true if magnetic monopoles were ever found, if you can find a north without a south, a south without a north.0150

But for now, how every draw it, whatever goes in as far as flux goes must come out so the net is 0.0156

We are talking about Gauss’s law but net magnetic flux is the integral / the closed surface of B ⋅ DA = 0.0163

That is the second of Maxwell’s equations.0175

Let us take a look at an example where we look at flux through a circular loop.0180

Find the flux of the 3 tesla uniform magnetic field through the circular loop of radius 0.2 m with 3 turns of wire as shown in the diagram.0184

Our total flux φ B is going to be the integral / the open surface of B ⋅ DA, which will be constant integral of DA is A, dot product.0194

The cos of the angle between them θ is going to be B × our area is just going to be π R² cos θ, cos 40°.0211

Which implies then that φ B is going to be equal to, our magnetic field strength is 3.0228

We have our π × 0.2 m, our radius squared × the cos 40° which implies then the flux = 0.866 Wb.0235

Fairly straightforward calculation.0258

We will do one more and then we will get to using magnetic flux in our next lesson.0260

Example 2, we have our long straight wire here I, carrying current I as shown.0266

Find the magnetic flux through the loop.0271

To find that flux or total flux, φ B is going to be the integral / the open surface of B ⋅ DA.0275

How we are going to find that, because we have a changing magnetic field strength here,0289

is what we are going to do is we are going to break this up and we are going to integrate from D to D + L0294

by breaking this up into tiny little strips.0299

We make those tiny little strips, we will find the flux through each one of them and0302

then add the flux to the next one, the next one, the next one, to get the entire flux.0307

That will be the integral from R = D to R = D + L.0314

It looks like of our magnetic field strength is μ₀ I / 2 π R.0321

We have done that previously.0328

Μ₀ I / 2 π R.0330

Now for our DA, that little bit of area, it looks like we have got the height of these little rectangles H × their width DR.0335

When we do these, how do we integrate easily?0349

Φ B =... let us pull out our constants.0353

We have μ₀ and I, those are not going to change.0356

We have our 2 π in the denominator.0360

Our H is constant for this problem, that will leave us with the integral from R = D to R = D + L of DR/ R.0362

The integral of that straightforward that is to be the nat log of R.0375

Then we have φ B = μ₀ I H/ 2 π log of D + L- the log of D.0380

Which implies that if we want to simplify that a little bit, φ B = μ₀ IH / 2 π × the nat log,0399

the difference of two logs is the log of the quotient so that will be log of D + L/ D.0411

And there is our answer for the flux due to that wire.0425

A lot more straightforward to do than to explain in words.0427

Hopefully that gets you a good start on magnetic flux.0430

We will start using magnetic flux pretty regularly in our next lesson as we talk about Faraday’s law.0433

Thank you so much for your time in coming to www.educator.com.0440

Make it a great day everyone.0443