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**Example- Generator**

Over the years, we have seen that if we place a current carrying conducting loop inside of an external magnetic field region. Which can be generated by taking two bar magnets and positioning their north and south poles something like this, therefore filling the region with the magnetic field between these poles originating from north pole and entering into the south pole of the other magnet. And, if you place a current carrying loop inside of this external magnetic field region, we have seen that the magnetic forces generated by this external magnetic field along each segment of this loop ends up with a net torque which causes the loop to rotate. And as, again, we have seen earlier, that, in order to have the rotation of the loop in one direction, one had to change the direction of the flow of current in every half turn.

During this process electrical potential energy was converted into energy of motion which is kinetic energy and we call the devices which make this transformation as electric motors. Well, and again, as we recall, we ask the symmetrical question, and, we said that, okay, if we have this external magnetic field originating from the north pole of one bar magnet and entering into the south pole of the other magnet and filling the region between these poles, like this. And, if we take our loop, conducting loop, and place it into this region and instead of, in this case, letting current flow this region or through this conducting loop. If we, just, simply crank it at a certain angle of velocity too, we end up with induced current to show up along this conducting loop.

And, the answer to that question is, yes. As a matter of fact, this system is fully symmetric to the previous system. As you recall, in the previous system, in order to have the rotation of the loop in one direction only, one had to change the direction of flow of current in every half turn. And, here, if we crack the loop in one direction only, we’re going end up with induced current through out this loop, this conducting loop. And, that induced current will change direction in every half turn. In other words, if it is flowing in clockwise direction for the half of the rotation, then the next rotation, it is going to be flowing in clockwise direction.

Alright, let’s analyze this system and see why that happens. First of all, in this case, since we have the reverse process, the kinetic energy, which is the energy of motion and it is associated with the rotation of this loop. Through this process it is being converted into the energy of moving charges, electric current. In other words we end up with electrical potential energy. And, the devices which provide this conversion are called electric generators.

Since one can obtain or end up with electrical potential energy or electric current and through this process, then, the issue becomes how we rotate this coil in the external magnetic field. And, it can be done through different means, the coil can be connected to turbines. Then, the issue becomes, how to rotate those turbines. In the case of hydro-electrical power plants, the gravitational potential energy of falling water is used to rotate the turbines and eventually to rotate the coils inside of the external magnetic field. Or, one can obtain high pressurized steam to rotate the turbines. Then, the issue become how that steam is obtained. Well, the chemical potential energy stored in coal can be used to obtain the high pressurized steam to turn the turbines, which is the case of coal power plants. Or, thermal energy released as a result of fission reactions, which can be used, again, to obtain steam and then, eventually turn the turbines and, hence, the coil inside of these external magnetic fields. And, that is the case of nuclear power plants.

Alright, now, let’s see how we end up, current, induced current along this magnetic loop as we crank inside of this external magnetic field. And, as we can see, the system is going to rotate about this axis as we crank it. And, so, the position of the loop or the area surrounded by the conducting loop will change as it rotates with respect of the direction of the magnetic field. In one position, for example, it can be oriented as perpendicular to the field lines as it rotates. Something like this. And, obviously, this position corresponds to the case that we will have the maximum flux through this coil. And, as it rotates, it can be positioned as parallel to the field lines. And, so as it rotates in this clock wise direction. In that case, of course, none of the field lines will go through the area surrounded by the loop, then we’re going to end up with zero magnetic flux through the area surrounded by the loop. So, in other words, as it rotates, magnetic flux, through the area surrounded by this conducting loop will be changing. And, as we know, from Faraday’s law, if the magnetic flux is changing, then, we’re going to end up with net induced electromotive force around that conducting loop. And, that will simply go to minus number of turns times the rate of change of magnetic flux through the area surrounded by that conducting loop.

Okay, to be able to see this, let’s look at this configuration from the, again, front view. And, if you do that our loop is going to be at an arbitrary position, something like this. And, the magnetic field is pointing to the right at every point in this region. And, the system is going to be rotating about the axis passing through the center of this loop. Well, the loop is, we’re cranking the loop at a certain angle of velocity. Let’s say that this, crank this loop in counter-clockwise direction at an angle of velocity of omega. In order to represent the orientation of the loop relative to the external magnetic field let’s look at the area vector of this loop which is going to be perpendicular to the loop all the time. And, therefore, it is going to be pointed something like this. And, let’s call this angle S Theta. So, in this picture, it’ll be perpendicular to the area surrounded by this conducting loop and it will be pointing like this. And, this angle is the angle Theta, the angle between D and A.

Well, um, since this is rotating at a certain angle of velocity, it is assigned some time of DT. The loop is going to rotate by an angle of D Theta. Therefore it’s area vector will move to this new position and let’s call this angle as D Theta. Well, we know that, from the definition of angle of velocity, Omega, angle of velocity, also know as quantity as angle of frequency. That’ll be equal to change in angle of position with respect to time. And, from there D Theta is going to be equal to Omega times DT. Well, if you’ll look at this through a certain interval, starting from Theta is equal to 0 to Theta. And across the point in that B is equal to 0 to T, then we will end up, since Omega, the end of frequency is constant, the loop is cranked at a constant angle of velocity. We can take it outside of the integral, and, this expression is going to give us Theta is equal to Omega T.

Therefore, the angle that the loop makes with the external magnetic field or the loops area vector makes with the external magnetic field as it is being cranked will be a function of time. In other words, it will change with time. The magnetic flux, by definition, through the area surrounded by this loop is the magnetic field vector totaled with the area vector of this loop. At an explicit form this quantity is equal to B magnitude, A magnitude times the cosine of the angle between these two vectors and that is cosine of Theta. Here, we can express Theta in terms of the angle of velocity, or angle of frequency as B times A times cosine of Omega T.

Well, this is the flux through one turn, if we have N number of turns, N number of turns, then we’ll have total flux is equal to N times B times A cosine of Omega T at a specific instant of time. And, the reason for that, is that he flux in each turn will link to the next one, therefor generating a net flux which will be equal to this quantity. So, again, here, A is the area surrounded by the conducting loop.

Alright, therefor, this expression will give us the magnetic flux at a specific instant of time T. Since magnetic flux is a function of time, in other words, it is changing with time, then, we’re going to end up with an induced electro-motor force along this loop directed from Faraday’s law. Induced EMF is minus the rate of change of magnetic flux. D Phi D of DT. And if you take the derivative of this quantity with respect to time, then we will have NBA, we’ll have a negative sign due to the Lan’s law over here. Derivative of cosine Omega T will give us minus Omega sine Omega T. Therefor, the induced EMF of it is going to be equal to minus minus Omega plus number of turns times the magnetic field times the area surrounding by the loop times the angle of frequency or the angle of velocity of the rotating loop times sine Omega T.

And, as we see in the induced EMF will be a function of time and it is going to be changing with sine Omega T. And, if we say that that let R be the resistance of the loop. Then, from Ohm’s law, as you recall, Ohm’s law was such that the current is equal to voltage divided by the resistance. And, here, the voltage is equal to induced electro-motive force Epsilon. Therefor I becomes equal to Epsilon over R or in explicit form, NDA Omega over R times sine Omega T. Here, the area A, is the area surrounded by this loop. And, if we go back to our diagram and if we give some dimensions to the loop that we are cranking, by saying that it’s width is A and the length is D. In that case we can write down the area in explicit form as I is equal to NDA times B times Omega divided by R times sine Omega T. So, the current too, is going to be a function of sine Omega T. And, of course, it will take it’s maximum value whenever sine takes it’s maximum value. And, it will take it’s minimum value whenever sine takes it’s minimum value. The maximum value of sine is 1, and, in that case, I max will be equal to NB times AE times Omega divided by R. And, of course it occurs when sine Omega T is equal to one. It means that Omega T is equal to Pi over 2 radians. And Pi over two radians is 90 degrees. So, when this angle is 90 degrees, and that is the case that the loop is, basically, lying in the same direction with the magnetic field lines, or parallel to the magnetic field line. It means, at that instant, the flux through that loop is 0.

And, that makes sense because flux is changing with cosine whereas the current is changing with sine. Therefor, whenever one of them takes it’s maximum value the other one will take it’s minimum value. So, sine and cosine are out of place of Pi over 2 radians. So, we can write down this current in a compact form as I is equal to I max times sine Omega T. And, here, Omega is the angle of frequency again. And, if you recall, the relationship between the angle of frequency and the linear frequency is such that Omega is equal to 2 Pi times F. Therefor we can express this as also in terms of linear frequency as I max times sine of 2 pi FT.

And, when we look at the mathematical form of this current, we see that it is changing as a function of sine. And, if you plot this current as a function of, let’s say, Omega T it will look like something like this sine function. And, in other words, it will oscillate. If we look at one full oscillation, which we can pick up the same two successive points that the, functions taking the same value. And, we can see that this is the half-cycle and half-cycle as the current it varies as a negative pass, for the next half-cycle it varies as a positive pass. In other words, it changes direction in every half-cycle. And, that’s why we call these type of currents as alternating currents. This is the type of current that we obtain from our power outlets. And, in the US, the frequency of this oscillating current is 60 hertz.