Let's go over the whole of AQA magnetic fields and electromagnetism for A-level physics. We're going to start off with a force on a current carrying wire in a magnetic field. Now, something which is really interesting, imagine that I have a wire which is in a magnetic field and in this case they are perpendicular. So, we can have the field into the board or out of the board. Now when I switch this circuit on, maybe I have a switch or something like this, we're going to see a force which will be acting on the wire. The magnitude of that force will be proportional to the current. It will also be proportional to the length of the conductor within the magnetic field and will also be proportional to a new quantity which is equivalent to the magnetic field strength but we call it magnetic flux density. This here is just B. We measure that with Tesla. Here's a little tricky point. So let's have a magnetic field and let's say that we place that at an angle. Let's also have our conductor be at an angle to the magnetic field. So let's say that this here is our conductor, which could be part of a some sort of a circuit. The actual circuit is pretty much irrelevant. So when that's the case, we only really need to consider the angle which is perpendicular to the conductor. So in other words, if this here was theta, the angle between the magnetic field and the actual conductor, then the equation is going to change to F is equal to bill sin theta. Why is that? If we imagine each of those field lines, they can be represented by a vertical component and we only really care about the component that is perpendicular to the field. This actually has some really interesting consequences. So if we were to have a conductor which is fully perpendicular to the field the force in this case will simply just be equal to bill in a way sine of 90 which is just one. So this will be the maximum magnetic force that's acting on the conductor. And let's call this one sort of case one. Case two, the one that we've already just discussed, the case in which there is a perpendicular component. We only take that into consideration. And the final possibility is if they're actually parallel. So if that's the case in case three, we are actually not going to have any magnetic force because there is no component that is perpendicular to the conductor. The angle between them is either zero or 180°. So in a way we have something like F is equal to bill sine of 0 which is just 0 newtons. So so far we have the magnitude of the force and this is equal to bill for perpendicular conductor. What about the direction of the force? Well, this is found out using Fleming's left hand rule. So I've drawn the hand over here with the thumb pointing in the direction of the force. The first finger is the field and then the second finger is the current. A really good way of remembering it is simply FBI FBI. Let's do an example problem which is the electric motor. We have a square coil that's been placed within a uniform magnetic field. Please note that these wires here are just drawn like this. But in reality, we assume that this distance here is a lot smaller than the actual length. Each square coil has a side L and the magnetic flux density is uniform and it's 2 mill teslas. Something which is very useful is to just transfer some of that information into the diagram. So I'm just going to call this side L L. And this one here is also L. And flux density is just 2 mill Tesla. What we need to do is to express the force on the side in terms of L and then the direction of the force. So in this part of the circuit we are actually given the terminals and remember the bigger one is the positive one and the smaller one is the negative one. You can just about see them over here. So this means that the current is going to go through here through here through here and then back through there. If we know the directions for the current and the magnetic flux density, then we're in business. We can just use Fleming's lefthand rule to determine the direction of the force. So for side A, if you were to apply Fleming's left hand rule, remember FBI, what we're going to get is that the force, I'm going to write it like this, is into the page or the board or the screen. Similarly for position C or side C, the current is going down. The field is going to the right, which means that the direction of the force for C will be out of the page. A good rule of thumb, no pun intended, is to just reverse the direction if one of the quantities reverses. Just one of them though. So for instance in this case the current has reversed in its direction which means that the force will also have to reverse and it's just going to be out of the page. What about the magnitude of those forces? Well all three are perpendicular to one another. So the force is just going to be equal to B I L. Let's just write it over here. That will be 2 mill Tesla. That would be 2 * 10^ - 3 * 2 * L. So what is that going to give me? 4 * 10 ^ -3 L. And this question wanted us to write it in terms of L. So right over here for the force I'm just going to write 4 * 10 ^ of -3 L. Now look at this. For B, the magnetic flux density is going this way, but so is the current. So there is no component of the magnetic flux density that is perpendicular to the conductor. Therefore the force is going to be zero and it's not going to have any direction. For C the magnitude of the force will be exactly the same. F is equal to bill which is going to be 4 * 10 ^ -3 L. And for D, we're once again going to have zero force, no direction in a way because the magnitude of the component of the magnetic flux density perpendicular to the length L is zero. Let's do part two of this question. Calculate the net moment with respect to the axis of rotation XY. You can see that I've added an axis of rotation around here. Well, the net moment is going to be the sum of the clockwise moments and the anticlockwise moments. The clockwise moment is just going to be equal to the magnitude of this force multiplied by its distance to the pivot, which in this case is just half of the length, which is going to be L / 2. The magnitude of this force was 4 * 10 ^ of -3 L multiplied by its distance to the axis which is just L / 2 plus the magnitude of the other force which is exactly the same 4 * 10 ^ -3 multiply by its distance to the axis of rotation which is also um L / 2. So it's going to be 4 * 10^ - 3 L * L / 2. Adding those up, we are just going to get 4 * 10 ^ -3 L². In a way, this here is general result that the net moment with respect to this axis of rotation is proportional to the square of the side of the coil for a square coil. One more example problem which is sort of based on one of the required practicals but we have a conductor of mass 200 gram and length 75 cm is placed at right angles in a uniform magnetic field with a given flux density. Calculate the current required for the conductor to be suspended and in equilibrium in the magnetic field. So in order for the conductor to be suspended, the force due to gravity known as the weight will need to be balanced out by the magnetic force. So in other words, F is equal to bill. I'm going to set F to be equal to mg. So this here will be equal to B I L. We're just looking for the current I, which is just going to be MG divided by B L. Okay. So 200 g is equal to 0.2 kg multiplied by 9.81 divided by the flux density which is 0.95 and then we are given the length in cm. So let's put that as 0.75 in m. And this here is around 2.75 amps. We're given mostly two significant figures. So should we just say 2.8 amps. And let's talk about magnetic flux density. So an equation for it can be found from F is equal to bill. And then all we need to do is simply rearrange for B. And we're going to get that this here is equal to F divided by IIL. As we mentioned the unit for that is the Tesla. Sometimes we write units in brackets such as these. Now how do we define this? Well, magnetic flux density is defined as the force per unit current per unit length that acts on a current carrying conductor which is perpendicular to the magnetic field. Let's just add the word magnetic over here. Sometimes we're also asked to define units. And if we have to define the Tesla, then all we need to do is take this equation and plug in ones for all the units. So one Tesla will be equivalent to 1 Newton per 1 amp per 1 meter. You can also be asked to express the SI base units of Tesla. Well, how would we do that? My first step would be just to simplify the expression a little bit with the simplest possible equation. So, for instance, B will be equal to F over IIL. Now, force is just mass time acceleration. And then we have IIL. Now, let's plug in some units. Mass is in kilog. Acceleration is in m/s squared. We're going to divide this by amps and then by meters. So what is this going to give me? Well, just kilogram amps to the^ of -1 s to a power of -2. Now let's talk on the force on charged particles moving in a magnetic field. There is an equation given in the formula booklet that is that the force is equal to magnetic flux density multiply by the charge of the particle multiplying by its speed. It's important to say that we can only use this equation when the velocity is perpendicular to the field. If it's not, we actually have to find the perpendicular component, which depending on where the angle is, might turn the equation into F is equal to BQ V sin theta. Maybe a little bit unlikely to appear, but let's just put it out there. Where does this equation actually come from? If this was to appear for instance in a show question to show that this equation is true. Well, it typically comes from F is equal to bill. Now remember that the current I is equal to the amount of charge over the amount of time multiplied by L. And then if we have a look this quantity L / T well this here is actually the speed. So this here it will be then equal to B QV for a charge that is moving perpendicularly to a magnetic field. Another important fact to mention is the charge can be either positive or negative. So positive and negative charges will have forces from the magnetic field act in opposite directions. And now let's have a look at the circular path of charged particles within a magnetic field. We're going to imagine first of all that we have a positively charged particle which is moving to the right. If that's the case, the direction of conventional current is exactly the same as the direction of the motion of the particles. Both of them are to the right. We're going to have a field which is into the screen in this region across here. By Fleming's left hand rule, remember FBI, I've kind of tried to drawn it in 3D. The force will be upwards. So, the particle will start will start curving upwards. As it starts curving upwards though, the force will remain perpendicular to the direction of motion and this is the recipe for motion in a circle. So they're going to follow a circular path. Now very interestingly, what if the particle was negative? So we have this tricky point where the direction of conventional current in this case is opposite to the direction of motion. So let's see maybe I'm going to draw the motion just like this in a in a line. Well, this is not the direction of conventional current. And in Fleming's left hand rule, we're going to take the direction of conventional current. We're still going to have the field, the magnetic field, which is into the board. But now our current will be going the opposite way. Even if the particle is actually going to the right, our current is to the left. Oh, magnetic field is still kind of into the board. I'm going to try and draw that in 3D. Well, by Fleming's left hand rule, now the force is going to be downwards. So, the particle will start moving in circular motion going downwards rather than upwards because the two forces will act in different directions by Fleming's left hand rule. Mathematically we can actually find the radius of this motion. And I really recommend as soon as you see a particle moving in a circle, I would literally write that the magnetic force is the only force which is acting. So that's going to be the net centripedal force mv^ 2 / r. So QVB is going to be equal to MV^ 2 / R. We can cancel out those and then what we're going to get is that the radius is equal to MV. Divide that by QB. Sometimes you may get asked to explain why the particle travels in circular motion in the region of the magnetic field. This is due to the force of the uniform magnetic field being at 90 degrees to the direction of travel. Since this is the only force acting on the particle, it will act as the net centrial force. And this here is a relatively standard answer. And now let's practice a show question. Show that the kinetic energy for a particle of charge Q, mass M, moving in a magnetic field with flux density B, and radius R is given by the following expression. Because this is a show question, we're going to try and showcase as much of the work as possible. What I would start off with is simply the fact that kinetic energy is equal to a half mv squ. Then somewhere on the side, I would think about the velocity B. Well, because we have a charged particle moving in a magnetic field, we know that QVB is equal to MV² / R, meaning that V will be given by QB and then multiplied by R / M. Then what I'm going to do is I'm going to plug this expression into here. Let's see what we're going to get. We're going to get a half random v^ 2. I'm going to say Q 2 B² R² / M 2. So this mass and this one here like that will end up cancelling. So I think we've got it. This is Q ^ 2 B ^ 2 R 2 over 2 M. And this is virtually what's required. Oops. I've just noticed that my Q's should have been lowercase Q's. So, let me just change that. That's a bit better. And here's an important distinction to make. Sometimes I've seen this in multiple choice questions, but the path of a charged particle moving in a uniform electric field is actually a parabola. It is not a circle. On the other hand, the path of a charged particle moving in a uniform magnetic field is exactly a circle. So if we are asked to describe it, we cannot be not very specific and say that it's just a curve. We have to actually quote that it is moving in circular motion and hence the path is a circle. This is the only thing that allows us to essentially use the centropedal acceleration equation mv^² / r. On the other hand, we cannot use this for the uniform electric field path because it is a parabola similar to the one for a particle moving in a gravitational field. So, electric field parabola circular motion circle. Moving on to magnetic flux and magnetic flux linkage. So, magnetic flux is defined as the product of the magnetic flux density perpendicular to an area and the area itself. Here is our area represented by this rectangular coil for instance or rectangular square coil and here is our magnetic flux density which let's say it goes this way and it cuts through this area. So the magnetic flux is typically given the symbol phi and it's defined as the magnetic flux density perpendicular to the area multiplied by a. It's really important this is not be in general but is the magnetic flux density which is perpendicular or normal to the area. Well, where is the actual normal? So the normal is just a line which is perpendicular to the surface. I'm just going to draw it let's say along here. Typically in terms of the magnetic flux density B which are these blue lines we're given the angle to the normal itself which is theta. So that makes the vertical component of the magnetic field the perpendicular component to be given by B cos theta if B is the overall magnetic flux density. Let's just be even more specific. So I'm going to call this B sort of perpendicular. This here is a little perpendicular sign. Because of that our flux density will be given by B cos theta multiplied by the area. In practice this is very often written as BA cos of theta. Now something which is really important which doesn't happen very often but if we're given this angle here let's call it alpha then phi will just be given by b a sine of alpha or you could find alpha by taking theta away from 90° and this will also work. And what about flux linkage? Well, if we had multiple coils, in fact, let's say we had n number of turns to the actual coil, then our flux linkage is defined as n multiplied by the magnetic flux. So, magnetic flux linkage is simply defined as MBA cos of theta where theta is the angle to the normal. Let's talk about units briefly as well. The unit for magnetic flux is actually known as the Weber. This is exactly the same unit for flux linkage as well. In some textbooks, particularly all the textbooks, you might come across Weber turns, but the turns is just simply this number over here. In practice, this leads us to a couple of extreme cases. We're going to have maximum magnetic flux whenever the magnetic flux density is parallel to the normal. In other words, perpendicular to the area. If that's the case, what we're going to get is that our magnetic flux is just equal to ba cos of theta. Now, theta in this case is just 0° because they're just parallel. and cos of 0 is just equal to 1. So this here will be ba cos of 0. Meaning that phi will just be given by ba. On the other hand, we will get zero magnetic flux whenever our magnetic flux density is parallel to the area perpendicular to the normal. Now phi will be just b a cos of theta. But remember theta is the angle between the field and the normal which is just this angle right over here which is just 90°. So phi will be ba cos of 90°. cos of 90 is just zero. So this here is just zero. And let's do an example problem. We want to calculate the magnetic flux through this square coil of size 3 cm and we're given the magnetic flux density 5 m Tesla. We're also given the angle between the coil and the flux density which is 55°. Okay. Now the first thing that I want you to do in those problems is to simply draw the normal. So the angle that we want for our BA cos theta equation is actually this one here which is going to be 90 - 55 which is 35°. And now we're ready to apply our equation. So phi will be given by b a cos of theta. B is 5 * 10 ^ -3. Now our area is going to be 3x3 cm. Don't forget the units. 3 * 10 ^ -2 all of it squared. Multiply this by cosine of 35. And this here is going to give me around 3.68. So let's call it 3.7 multiply by 10 ^ of -6. And now let's talk about electromagnetic induction. Starting with the famous Faraday law. It states that the induced emf is directly proportional to the rate of change of magnetic flux linkage. Now what does that mean? Imagine that we have a magnet. This magnet will have a magnetic field around it. So it's going to look something like this going from north to south. But now we've got a magnetic field that is present next to a conductor. According to Faraday's law, if we were to move this magnet, we are going to get an emf induced within this conducting ring. We should probably say that it's actually a conductor. So that means that if I was to move this left or right, there will be emf and potentially current running through it. Let's turn Faraday's law into actually an equation. Well, it says that the magnitude of the induced emf will be directly proportional to the rate of change. Now, rate of change is change of something divided by delta t. That something is magnetic flux linkage, which is ni. In this case, n is just equal to one because we only have a single coil. If we had, I don't know, 10 coils for instance, then each of them will have its own magnetic flux. So n here will be equal to 10. Please note that this here is simply the magnitude of the induced emf. And once we have a look at lenses law in a second, we're also going to have a look at the direction of this. So we know that current is going to flow within this ring. The question is which way is it actually going to go? Well, this is answered by lenses law. It states that the direction of the induced EMF is such as to oppose the change that is producing it. Now in practice, what does that mean? Let's have a look at a couple of possible scenarios. So one very simple one. So what if the magnet is actually stationary with respect to the ring? Well, there's not going to be a rate of change of magnetic flux flux linkage. it will remain exactly the same. Therefore, the induced emf will be equal to zero because there is no rate of change of magnetic flux linkage. Another possibility would be for the magnet to be going towards it. Well, if that's the case, the current will run in such a direction. Actually, if we apply a different rule, we're going to see that this direction for the current is simply downwards. such as to produce a north pole onto here because now this here will be a wire and each current carrying wire produces its own magnetic field. So this is going to be a coil which is going to have a north here and a south onto the other side and those two will repel one another. Next possible case. What if I was moving my magnet this way away from the conductor? Well, once again, we want to oppose the change that is actually producing it. So that would mean that if we're pulling it to the right, Lens's law wants to keep the magnet to left. So this here will be a south pole, meaning that the current will be going in the opposite direction, kind of this way. In all of those cases, lenses law states that the direction of the induced EMF and the word EMF is typically underlined in mark schemes is such that to oppose the change that produced it. Because of that, essentially the induced EMF is always in the opposite direction. Then other exam boards, not AQA, but often include a little minus sign here, which I'm just going to add in essentially in brackets. The questions are likely only going to ask about the magnitude, though. Let's do a really interesting problem. We have a Boeing 747 that has a wingspan of 68.4 m flies at a constant speed of 250 m/s for the Earth's magnetic field. Okay, we can assume that the field is vertically downwards at that section of the flight and the Earth's magnetic flux density is 2.2 * 10 ^ - 5 Tesla. We need to figure out the induced emf between the wing tips. Okay, now let's have a look at a picture from above. So, we're going to imagine that the field is going down into the page. Let's see if I can draw an airplane. I don't know, maybe this here is the tail. And then something like this. And then I don't know, something like that. And then let's draw some wings. I am pretty sure that this is not how an airplane looks, but you know what? It will do. So we have this distance here is how many? 68.4 m. In order to figure out the induced emf, we have to use a Faraday's law. And I'm going to show you an equation which is incredibly useful. So E is equal to the rate of change of magnetic flux linkage. In this case, N is just equal to one. So we're just going to write delta 5 over delta T. This here will be equal to the change of ba cos of theta and then divide that by delta t. Now in this particular case cos of theta will just be equal to one since the line of the normal is actually going to be into the board or out of the board as well. So the angle between the normal and the flux density is 0°. So that just turns our emf equation into just simply delta ba delta t. Okay. Well, let's think what's actually changing. The field is uniform in this case. So we can take it sort of outside of the change. How is the area that cuts through the field actually changing? The wingspan of 68.4 m remains absolutely the same. Because we're going to derive a general equation, I am just going to call this as a variable L. So this here will be equal to the area which is going to be L and then multiplied by another distance which I'm just going to call should we just call that X which is the distance traveled in 1 second. So let's say that this part of the airplane sort of the wing moves from here to here in 1 second. And that's the overall change. We're going to be dividing that by delta t. And look at this. We can spot that delta x over delta t is actually the velocity. So this means that our induced emf is simply going to be equal to bl multipli by the speed v which is going to be equal to well our magnetic flux density is 2.2 2 * 10 ^ -5. Our length of the conductor is 68.4 and then that's multiplied by the speed or in this case the distance traveled per second which is going to be 250. Okay, let's put this into a calculator. If I've done this correctly, we're going to get up to two significant figures 0.38 volts. The answer kind of makes sense. I mean, it's probably not going to be a th00and volts generated, but we're going to have some small voltage between the tips. This formula can be applied any time we have a vertical conductor that is moving at right hand angle. So even if we take a wire and we start to move that through a magnetic field, we are going to generate some emf across its ends. This is because each of the individual charges going to be is going to be experiencing a force due to Fleming's left hand rule. This equation can be applied anytime we have a straight conductor moving in a magnetic field. Here's another example. So we have this 2 m long conductor that is moving in a uniform magnetic field with a flux density of 4 m Tesla constant speed of 1 m/s. Want to find the induced emf. Now we can remember this formula. It is not given in the formula booklet but very very useful and we can just say that the induced emf is equal to B LV. Okay. So this here will be 4 * 10 ^ -3 * this by 2 * this by 1 which is what 8 * 10 ^ of -3 vol. The next case that we're going to look at is emf induced in a coil that's rotating uniformly in a magnetic field. So imagine that we're looking at the coil from above and in this case the magnetic flux density is pointing to the left. What will the magnitude of the magnetic flux linkage actually be? Well, we can write down that ni will just be equal to nba cos of theta. But the angle theta remember will be the angle between the normal and the flux density lines. In other words, it's just this angle across here. This is changing because the field will be because the conductor will be rotating uniformly. And if it's rotating uniformly, we'll have some angular velocity omega which will be equal to the change of angle the rate of change of angular displacement. In other words, if we were to rearrange for theta and we are going to get that the cosine will just be equal to omega t. If the angle is the only aspect which is actually changing the emf will be equal to the rate of change of magnetic flux linkage and this leads us to a really interesting equation. So the induced emf sh just write in a new line over here will actually be equal to b a n then we get a factor of omega then we get sine of omega t. Where does this equation actually come from? If you're not doing a level maths you don't need to worry about this. Well even if you do a level maths you don't you still don't need to worry about this but it might be interesting to see where it comes from in reality. So this is outside of the syllabus. Considering the direction with lenses law, we can say that the induced emf is equal to minus because it's in the opposite direction and the little deltas turn into derivatives. So this is the negative derivative of the magnetic flux linkage. Okay. So this here will then be equal to minus NBA. All of these are actually constants. But what is not constant is the angle. So we need to differentiate cosine of omega t by dt. The derivative of cosine is of course negative sign. So this will cancel out all of the functions. So this here will be equal to nba sin of omega t. But then the inside function omega t also depends on t. So we need to differentiate with with so we need to differentiate it as well. So the derivative of omega t is just omega and we get one more factor over here. We can just rewrite this as NBA omega sin omega t. This is where it comes from. But once again outside of the specification. What you do have to know though is that their magnitudes will be out of phase by 90°. So if you were to plot them, those two will be out of sync by 90° or pi / 2 radians. Let's apply this equation to a question. We have a rectangular square coil of size 0.75 m 100 turns rotating at the frequency of 20 Hz at right angles to a uniform magnetic flux density of 200 mil Tesla. What is the peak induced emf of the coil? So the word peak here is really important. We'll first write our emf equation that it's simply equal to b a n and then we have omega and then we have sine of omega t and this is really important. So e is at its peak or its max when sine of omega t is just equal to one because s is between 1 and minus one. So it can never exceed one. It will be at its maximum when this thing here is equal to one. Which means that e shall we just call it maximum the peak emf is just equal to b a n omega. Now we can figure out omega relatively easily because we're just given the frequency of rotation. So this here will be equal to b a n. Remember though, omega is just equal to 2 pi f. So this here will just be b a n and then 2 pi multiplied by the frequency. Okay, so let's plug in some numbers. B is just 200 m tesla. The area is 0.75 squared multiplied by the number of turns which is 100 multiply this by 2 pi multiply this by 20 and this here is going to give me around 1,413 around 1,400 volt. So that's a lot of voltage in this case. One of the tricks here is to determine our angular frequency omega as fast as possible and then apply it to this equation. If we were given a specific time t, for instance, I don't know, 0.2 0.3 seconds after the start, we just plug this into this equation. We may also be given things like revolutions per minute. And we may have to convert those to radians/s for omega. Let's talk about alternating currents next. So, an alternating current is current which changes its direction with time, and we tend to measure them with an oscilloscope. I've got lots of separate videos on oscilloscopes that I will link into the description. What we need to be aware of is the graph for voltage against time. So, whereas with direct current, this would have been a nice straight line, in this case, this will be sinosoidal. There are several important features. For instance, this distance almost like the amplitude is known as V KN which is the peak voltage. From the peak voltage, mathematically you can calculate the root mean square voltage or V RMS. You can kind of think of it as the most likely value for the voltage to be over time. And this is simply equal to V KN / <unk>2. If you have the RMS value, you have the peak voltage. It's important not to confuse this with the peakto peak voltage, which is the length of this entire height over here from one peak to the other. So we can say that V peakto peak not just the peak voltage is equal to 2 * V KN. Exactly the same equations will also apply for the current. So we can say that I RMS is equal to I over<unk>2 and I peak to peak is equal to 2 * IN. Sometimes we get questions on the average power that's been dissipated. So we can write down an equation that P average in this case is simply going to be I RMS multiplied by V RMS. Okay, let's do a little example question. So we have the mean power dissipated in electrical component is 25 mwatt and the RMS current is 5 milliamps. We're looking for the peak voltage and then the peakto peak voltage. First things first, so the power that's been dissipated, the average power will just be equal to Irms multiplied by V RMS. So we're looking for the peak voltage which we're going to find from Vrms. So V RMS will just be equal to the power divided by the current. So this here will be 25 * 10 ^ of -3 / I RMS which is 5 milliamps. And this here will just give us 5 vol. Okay, we've got the RMS value. Now we can figure out the peakto peak voltage and the peak voltage. Let's say the peak voltage is part one. Let's say this one here is part two. So V RMS is equal to the peak voltage V V over <unk>2 meaning that V is V RMS multiplied by <unk>2 so this here will be 5 * <unk>2 and that's around 7.07 let's call it 7.1 volts. Now the peakto peak voltage will just be twice that value. So this here will be 2 * 7.1 which is what 14.2 vol. And now let's talk about transformers. They can be split into several components. What we have over here on the left is the primary coil. Then we also have the core of the transformer. And then right over here we have the secondary coil. For the primary coil we have AC current which runs through it. Since it's alternating current, that means that the magnetic field around this coil over here is also changing. The function of the core is to provide greater magnetic flux linkage from the primary to the secondary coil. This part can be a little bit tricky to remember, but without the core, the field might look something like this all the way around like that. A lot of it will be lost into this region over here. This is just a rough sketch of course, but if you were to build a transformer in the lab without the core, you're going to notice that the efficiency is awful. With the core, those magnetic field lines are going to remain like this all the way within the core, improving the efficiency greatly. Onto the secondary coil now that has a changing magnetic flux linkage passing through it and this induces an emf which is determined by the number of turns in both the primary and the secondary coils. Let's word this. This wording here of the explanation of how a transformer works will typically give you full marks. Of course, depending on the mask scheme, you don't need to worry about this diagram here. It was only for explanation purposes. The output voltage will be determined by the transformer equation. So you would have encountered this in your GCSEs but NS divide that by NP will be equal to VS divide that by VP. In this case NS is the number of turns in the secondary coil. P stands for the primary coil and those are the respective voltages. Please note that for an ideal transformer the power is conserved. So that means that P primary will be equal to the power across the secondary coil. So this here means that V primary multiplied by I primary will be equal to V S I. Okay. So from this equation we can just rearrange that to get that V S over VP will be equal to IP over S. And now we can bring this back into here to expand our transformer equation for the current as well. So then we're going to get IP divide that by S. And notice how the current is flipped in our ideal transformer equation. Very often we're also asked to calculate the efficiency of a transformer. In this case we have a non ideal transformer. So let's do this in a different color on the side. This will be equal to the amount of power in the secondary coil divided by the amount of power in the primary coil. In a way this is useful divided by total power. Because of that it will be equal to I multiply this by VS and then I multiply this by VP. Finally, we're also going to be talking about inefficiencies in a transformer. So, the magnetic flux actually cuts through the metallic core and this induces an emf and hence current. These are known as eddy currents. They dissipate heat and reduce the efficiency in the transformer. So, how do we deal with this? We can improve the efficiency by laminating the core. We have layers of insulator within the core which limit the effects of eddy currents. Well done guys. You have revised the whole of magnetic fields for AQA. Have you also revised capacitors? If not, you should definitely click at this video right over here. Good luck with your revision.