A uniform magnetic field is perpendicular to the plane of a single-turn circular coil. The magni

A uniform magnetic field is perpendicular to the plane of a...

Key Concepts

Faraday's Law of Electromagnetic Induction

This law states that a changing magnetic flux through a closed circuit induces an electromotive force (emf) in the circuit. It is a fundamental principle in electromagnetism and is expressed mathematically as emf = -d?/dt, where ? is the magnetic flux. The negative sign indicates Lenz's law, which ensures that the induced current opposes the change in flux that produces it.

Magnetic Flux

Magnetic flux represents the total magnetic field passing perpendicularly through a given area. It is defined as the product of the magnetic field strength and the area through which the field lines pass, taking into account the orientation of the field relative to the area. A change in the magnetic flux, due to variations in the magnetic field or the area/orientation of the loop, is what drives the induction of an emf in a coil.

Coil Geometry and Area

The shape and size of a coil determine its area, which directly influences the magnetic flux intercepted by the coil when placed in a magnetic field. When a coil is reformed into a different shape, its total length remains constant but its area changes. This change in area affects the magnitude of the induced emf as the induced voltage is proportional to the rate of change of the product of the magnetic field and the loop’s area.

Ohm's Law in Electromagnetic Induction

Ohm's Law, which relates voltage, current, and resistance, is used to determine the induced current in a circuit when an emf is generated. Once the emf is induced by the changing magnetic flux, the resulting current in the coil can be found by dividing the emf by the resistance of the coil. This relationship is essential for analyzing the performance of circuits subjected to electromagnetic induction.

Transcript

00:02 In this example, we have a circular loop here in a uniform magnetic field, and this field is changing over time, so the magnitude is changing.

00:13 And because of that, we have a change in flux.

00:17 So if we remember from faraday's law, it says that if there's a change in flux, we're going to have an induced emf, and the magnitude of that induced emf is 0 .8 volts, and consequently the current that gets induced in that loop is 3 .5 .2.

00:36 0 .2 amps.

00:39 And what we want to know is if we were to deform this loop here, the circular loop, into a square, what would our new induced emf be, and what would our new current be? okay, so first things first, let's just write off some expressions for what the induced emf should look like for each of these loops.

01:04 So for the circular one, we know that the induced emf is going to be the change in flux over the change in time, minus the change in flux over the change in time, where our change in flux is just going to be the final flux minus the initial flux, where in general our flux is just going to be b times a times the cosine of the angle between the surface normal of our loop and the magnetic field.

01:34 But in this case, since they're parallel to each other, this is always going to be one for the whole problem.

01:40 So really for flux we're just looking at b times a and for our circle the area of a circle is pi r squared so you can imagine if we define some radius r here then we could say that the flux is b times pi r squared and consequently our change in flux is going to be b final times pi r squared minus b initial times pi r squared since our area doesn't change it all right, just the magnetic field does.

02:23 Okay, so let's go down here so we have a little more room.

02:28 So the change in flux for the circular loop, so we'll say circular, we factor out a pi r squared.

02:42 It's going to be pi r squared times b final minus b initial or just pi r squared times delta b.

02:49 And if we plug that into our faraday's law over here.

02:56 We'll just have that for the circular loop, the induced emf is minus pi r squared delta b over delta t.

03:08 Now imagine we did the exact same thing with the square loop here.

03:14 Notice the only thing that would change is the area of our loop here.

03:21 So now instead of pi r squared, our area is going to be the area of square which is just the length of one side and the length of the other which is the same so just l squared so minus l squared delta b over delta t okay so we know what the induced emf is for the circular loop right we were told that it is 0 .8 volts so if we could relate the emf from the circular loop to the emf of our square loop in some simple relation, right? like if their ratio just came out to a number.

04:16 If we had one we could solve for the other.

04:19 And that's what we're actually going to show we can do.

04:22 Because remember we deformed this loop here from a circle to a square.

04:27 So what we know about this configuration is that the length of wire we are using is the same the entire time.

04:34 And the length of why are we used in the circle is just going to be the circumference of our circle 2 pi r and we know this is going to be the same that was used in the square which is just four times our side length right or if we solve this for r or l either way we'll do um we'll solve for l 5 by 4 on both sides we can see that l is just pi r over 2 so if we substitute out for l in our expression for the emf in the square here, we'll actually end up in terms of something just in r, which is going to be just like our emf for the circle.

05:26 So we could take the ratio and a bunch of things would cancel out.

05:33 So if we do that, we'll get minus ir over 2, whole thing squared times delta b over delta t, just substituting for l there, which will be minus pi r over two, squared r squared delta b divided by four delta t now notice this looks very similar to our expression for the emf in the circular loop right has a lot of the same stuff it's just different by a number constant so let's look at their ratio now so e circle to e square it's just going to be pi r squared delta b over delta t minus pi squared r squared delta b over delta t or delta t.

06:42 Okay, so a bunch of stuff cancels, right? signs a pi r squared delta b delta t, we just end up with one over pi over four.

06:55 Or if we flip that over it's just four over pi.

06:59 So these two emfs are just related to each other by this constant four over pi.

07:04 The other way we could have done this is we could have just compared these two equations, this one here and this one here, and seen that they're almost exactly the same, but there is a pi over four extra in front of the e sub s.

07:23 Okay, so now that we know this, we can get the emf in our square by having the one through the circle...

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