WEBVTT physics/ap-physics-c-electricity-magnetism/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I'm Dan Fullerton and in this lesson we are going to talk about magnetic flux.
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Our objectives include calculating the flux of a uniform magnetic field through a loop of arbitrary orientation
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and using integration to calculate the flux of a non uniform magnetic field whose magnitude
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is a function of one coordinate through a rectangular loop perpendicular to the field.
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Sounds complicated when you see the setup in the example problem, be nice and straightforward.
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Magnetic flux sub B, you will sometimes see it written as φ sub M,
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describes the amount of magnetic field penetrating the surface just like it did for electric flux.
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The units of magnetic flux are Webber's, typically written as Wb.
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Where 1 Wb is 1 tesla m².
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Just like we did previously as we define this, we will take a small bit of area on the surface.
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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.
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The little bit of flux that we have through that little bit of DA is going to be B ⋅ DA,
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which is B DA cos θ if we are looking for magnitudes or θ is the angle between the normal NB.
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Then the total magnetic flux φ B is just the integral of all of these little bits of magnetic flux D φ B,
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which will be the integral / the open surface of B ⋅ DA.
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Same thing as we did with electric flux.
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When we talk about magnetic flux through close surfaces again, normal still, point from the inside to the outside.
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Our total magnetic flux would be the integral of D φ B which is the integral / the closed surface of B ⋅ DA,
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which for magnetic flux is going to equal 0.
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Let us talk about that.
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This brings us to Gauss’s law for magnetism.
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The total magnetic flux through any close surface is 0.
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We mentioned previously but just going a little bit more depth.
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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.
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But for now, how every draw it, whatever goes in as far as flux goes must come out so the net is 0.
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We are talking about Gauss’s law but net magnetic flux is the integral / the closed surface of B ⋅ DA = 0.
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That is the second of Maxwell’s equations.
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Let us take a look at an example where we look at flux through a circular loop.
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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.
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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.
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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°.
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Which implies then that φ B is going to be equal to, our magnetic field strength is 3.
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We have our π × 0.2 m, our radius squared × the cos 40° which implies then the flux = 0.866 Wb.
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Fairly straightforward calculation.
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We will do one more and then we will get to using magnetic flux in our next lesson.
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Example 2, we have our long straight wire here I, carrying current I as shown.
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Find the magnetic flux through the loop.
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To find that flux or total flux, φ B is going to be the integral / the open surface of B ⋅ DA.
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How we are going to find that, because we have a changing magnetic field strength here,
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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 + L
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by breaking this up into tiny little strips.
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We make those tiny little strips, we will find the flux through each one of them and
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then add the flux to the next one, the next one, the next one, to get the entire flux.
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That will be the integral from R = D to R = D + L.
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It looks like of our magnetic field strength is μ₀ I / 2 π R.
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We have done that previously.
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Μ₀ I / 2 π R.
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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.
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When we do these, how do we integrate easily?
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Φ B =... let us pull out our constants.
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We have μ₀ and I, those are not going to change.
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We have our 2 π in the denominator.
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Our H is constant for this problem, that will leave us with the integral from R = D to R = D + L of DR/ R.
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The integral of that straightforward that is to be the nat log of R.
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Then we have φ B = μ₀ I H/ 2 π log of D + L- the log of D.
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Which implies that if we want to simplify that a little bit, φ B = μ₀ IH / 2 π × the nat log,
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the difference of two logs is the log of the quotient so that will be log of D + L/ D.
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And there is our answer for the flux due to that wire.
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A lot more straightforward to do than to explain in words.
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Hopefully that gets you a good start on magnetic flux.
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We will start using magnetic flux pretty regularly in our next lesson as we talk about Faraday’s law.
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Thank you so much for your time in coming to www.educator.com.
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Make it a great day everyone.