 There is a circular region of radius 4 meters. It has a magnetic field pointing away from the plane, and it is increasing at a rate of 0.2 teslas per second. The question is to calculate the magnitude of induced electric field at a distance of 2 meters and 6 meters from the center. So first let's try and think about what could this circular region of radius 4 meters look like. So it could be just like this and the radius is 4 meters. And it has a magnetic field which is pointing away from the plane. So the magnetic field could look like this. It is pointing away from the plane of the screen. And it is increasing at a constant rate of 0.2 teslas per second. So this could be just a constant rate of increase. We have B versus magnetic field strength versus time graph. So it's increasing at a constant rate. So we can write this as dB by dt that is the rate at which the magnetic field is changing and this is equal to 0.2 teslas per second. Now first let's only focus on the magnitude of induced electric field at a distance of 2 meters from the center and let's keep 6 meters for later. Now we know that the magnetic field strength is changing so that means the magnetic flux must also change. And from Faraday's law we know that emf is equal to the rate of change of magnetic flux. This is equal to minus d phi by dt. But there is a problem over here. We do not see any circuit or any coil. So if we are thinking about flux, what will be the area that we will choose? And in fact the question is asking about the magnitude of induced electric field. So how do we go about this? How do we go from emf equal to minus d phi by dt to something such as magnitude of induced electric field? Now for that let's take some steps back and let's focus right over here. So let's say we have the same circular region of magnetic field and the strength of the magnetic field is increasing at a constant rate. If we had a coil that was placed inside this region and if the strength of the magnetic field passing through this coil increases that means the magnetic flux must also be increasing. And because of that there must be some induced current in the coil which will oppose that increase. Now the current should be flowing in such a direction so that it produces a magnetic field which opposes this increase. So that magnetic field should be pointing upwards opposite to this increase. And we can figure out the direction of the current using Lenz's law and right hand curl rule. So if this thumb is pointing in the direction of the magnetic field the curl of the fingers that gives the direction of the current. So in this case the direction of the current, the direction of the current comes out to be anticlockwise. So this is the direction in which the current flows in the coil. Now if there is a current that means that there must be some force on the electrons in the coil. And the coil is not moving so it cannot be a Lorentz force. A Lorentz force was qv cross b. And if there is no movement in electrons then there is no velocity. So this force cannot be pushing the electrons and it turns out that the force that is pushing the electrons is provided by the electric field. Now the direction of this induced electric field will also be in an anticlockwise direction that is because the current is flowing in an anticlockwise direction which means that the electrons are moving in a clockwise direction. And if the electrons are being pushed in a clockwise direction by the electric field. So that means that the electric field must be in a direction opposite to that. Because the force on the electron due to the electric field will be in a direction opposite to the direction of the electric field. Now we can try and relate the induced emf in this coil with the magnitude of the induced electric field. So for that we can try and recall how we defined emf. The force worked on per unit charge this was w divided by q. And if there is work done there must be a force that is doing the work. And the force is provided by the induced electric field. And in this case the force is q times the magnitude of induced electric field. Now we can write work as integration of f dot dl that is force and the displacement vector. Now in place of force we can put qe because that is the force provided by the electric field. So we can write this as w equals q which comes outside the integral and integral of e dot e dot dl. And if we divide both the sides with q to get the emf we will get emf as emf as integral of e dot dl. And because we are talking about a closed loop this is actually a closed surface integral. Now it turns out even if we remove the coil if we remove the coil there will be an induced electric field. Because the induction of electric fields it is not dependent upon electrons being there or not. Electric fields are induced in space they can be induced in a vacuum. Whenever there is a changing magnetic field there will be an induced electric field. Now we can come back to the question and try to expand this equation of Faraday's law. Do that it will become this will become integral of e dot dl and that is equal to the rate of change of magnetic flux. Now let's try and figure out the magnitude of induced electric field at a distance of 2 meters from the center. So it is at a distance of 2 meters and the induced electric field will look like this and it will be in an anticlockwise direction. This is the same case that we talked about over here. Now let's try and work out this integral. So for that let's think about what is the direction of electric field at each point. Now at each point in this loop the electric field will be a tangent to the loop. So at this point it will be a tangent at this point also it will be a tangent and the displacement vector that is dl that will be along the electric field vector. It will also be a tangent. It will always be along the electric field vector. So if you think about the dot product between electric field and dl that will just come out to be as edl because theta over here is 0 degrees that is the angle between the displacement vector dl and the electric field vector and if theta is 0 that means cos 0 is 1. So the dot product just comes out to be as edl. So let's let's write that this will be this will be closed surface integral of edl. And we are interested in the magnitude of induced electric field at a constant distance of 2 meters from the center. So at each and every point around this loop around this small red loop you will have one magnitude of induced electric field because we are at a constant distance from the center. So we can take the magnitude of electric field outside the integral. So that means it becomes e into the closed surface integral of dl. Now we can integrate dl around this path. If we do that if we add small bits of dl all around this path that will give us the circumference of this loop. So that will be 2 pi r that is the circumference of a circle. So this will be 2 pi into small r. Let's just write small r for now. So this becomes this becomes equal to e into 2 pi r. Now let's think about how the flux is changing. So flux is given by B a cos theta. And if you think about the rate of change of flux and if you think about the rate of change of magnitude flux that will be d by d 5 by d t so that becomes d by d t of B a cos theta. Now we can take a and cos theta outside because they aren't really changing. The only thing that is changing is the magnetic field. So that will give us that will give us a cos theta into Bb by d t. Now we can think about area over here. Area will be that area that is enclosed that is inside this path of integration. And flux is changing throughout this entire area that is enclosed by the small red loop. So this becomes this becomes equal to pi r square where r will be 2 meters. And db by d t is given to us that is 0.2 teslas per second. We can think about cos theta. Now to do that let's have a look at this loop from a certain angle. So if you do that it will look somewhat like this from at a certain angle. And if the loop is anticlockwise, again using the right hand curl rule we can figure out the direction of its area vector. So we can look at the hand to the right and we can curl our fingers in that direction and the direction of the thumb will give us the direction of the area vector. And that comes out to be in the upwards direction like this. And the magnetic field is pointing down. So if we think about the angle between the magnetic field and the area vector that is that is 180 degrees and cos 180 would be minus one. But we are interested in the magnitude of induced electric field. So if we take the magnitude of minus one that will just give us one. So cos theta over here is just one. So d phi by d t at the end d phi by d t becomes this becomes this becomes equal to pi r square. Pi r square that is the area multiplied by 0.2. That is the rate at which the strength of the magnetic field is changing. Now we can try and solve for e. So for that let me let me just write over here. We have electric field strength the magnitude multiplied by 2 pi r. And this is equal to this is equal to pi r square into 0.2. Now we can cancel a couple of things, we can cancel pi. We can also cancel one part of r over here. So that gives the magnitude of induced electric field to be as r divided by 2 into 0.2. Now r is two meters. So two divided by two would be one. And one into 0.2 will give us the magnitude of induced electric field to be as 0.2. Now what about the units of electric field? When we first learned about electrostatic fields, we tried and related to the change in potential per unit length. So this was given by minus delta V by delta X. And the units of potential is volts and the units of distance is meters. So units of induced electric field comes out to be as volt per meters. Now let's look at the second part where we have to calculate the magnitude of induced electric field at a distance of at a distance of six meters from the center. Why don't you pause the video and first attempt this one on your own? All right, so we start with the same circular region of four meters in radius. Even though this looks smaller, but let's say it is of the same radius, that is four meters, and it has field lines going inside. And they are increasing at the same rate, that is 0.2 teslas per second. Now we need to figure out the magnitude of induced electric field at a distance of six meters from the center. And from what we know about induced electric fields, we know that they form loops, so there will be a big red loop at a distance of six meters from the center. And even in this case, the direction of the induced electric field, it comes out to be in an anticlockwise direction. Now you can imagine that there is a coil that is kept at a distance of six meters, so the flux through the coil would be increasing because the strength of the magnetic field is increasing at a constant rate. And to oppose that, the coil will generate a current whose magnetic field will be in a direction opposite to this, that will be in the upward direction. And using the right hand curl rule, the direction of current will come out to be anticlockwise, and that will be the direction of the induced electric field. And even if we remove the coil, the induced electric field will stay there, and its direction will be anticlockwise. All right, now let's think about the magnitude of induced electric field. So from what we saw in the previous part, we saw that emf was equal to the integral, that is a closed loop integral of electric field vector dot dl. So this was dot dl. And this was equal to the rate of change of magnetic flux, minus d phi by dt. Now we can think about this closed loop integral of e dot dl. And even in this case, the electric field direction will be a tangent to this loop. So at each point, it will be a tangent. And the dl vector will also be in the same direction. It will also be a tangent. So the dot product of e and dl comes out to be just as e dl, because theta is 0 and cos 0 would be 1. So we can work out this part, this will become integral of e dl. And because we are interested in the magnitude of induced electric field at a constant distance from the center, so magnitude will be a constant. And it will also come outside the integral. That will only leave dl, the closed loop integral of dl. And if we integrate dl along this loop, that will give us the circumference of this big red loop. So let's just finally write e into e into 2 pi 2 pi r. And let's keep in mind that r over here is 6 meters. Now what about the flux? So flux is given by B a cos theta. And d phi by dt will be d phi by dt. That will be equal to d by dt of B a cos theta. Now a and cos theta can come outside d by dt. So this will become a cos theta db by dt. Now over here, cos theta would just be 1. Just like in the previous case, we will have area vector and magnetic field vector in opposite direction, but we are interested in the magnitude. So the magnitude of minus 1 would be 1. So cos theta would just be just be 1. And db by dt is given in the question that is 0.2 teslas per second. But what about area over here? What could area be? It seems that the area should be the complete area that is enclosed by this path of integration. And that would be pi into 6 square. But it turns out we should only be concerned with the area that is enclosed by the path of integration and also the area through which the magnetic flux is changing. And that area is just pi into 4 square. That is the radius of this circular region where the magnetic field lines are concentrated. And that is the area that we will take over here because it is only this area that is linked with changing magnetic flux and not the entire area that is enclosed by the path of integration. So it is worth noticing that when we integrated dL around the entire path, we took the radius as 6 meters because that is a path that is a... because we were integrating it along this entire big red loop. So that became 2 pi r where r was 6 meters. But when we are figuring out the area we are taking pi into 4 square. That is the area only through which the magnetic flux is changing. Now when we write this, this would become pi into 16. And that is multiplied by the rate of change of magnetic field strength, which is 0.2 teslas per second. Now we can try and figure out the magnitude of induced electric field. So that will be the magnitude of electric field strength e into 2 pi r. And let's plug in the values. So this will be 2 pi into 6. And that is equal to 16 pi into 0.2. Now we can cancel pi over here. And when we work out this, and I have done this, you will get the magnitude of induced electric field to be as 0.27 volts per meters.