Download the App!
Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.
No Related Subtopics
Rutgers, The State University of New Jersey
(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?
(I) What force is needed to accelerate a sled (mass = 55 kg) at 1.4 m/s$^2$ on horizontal frictionless ice?
(I) A constant friction force of 25 N acts on a 65-kg skier for 15 s on level snow.What is the skier's change in velocity?
Create your own quiz or take a quiz that has been automatically generated based on what you have been learning. Expose yourself to new questions and test your abilities with different levels of difficulty.
Create your own quiz
welcome to our first section in the new unit on electromagnetic waves in this section. We're going to discuss all of the different laws we found so far that connect our electric fields E and our magnetic fields be before we get to there. There is one that we should talk about that is not necessarily always included in the list we're going to come up with, which is the Lawrence Force Law. The Lawrence Force Law is a combination a force on a charged particle due to electric fields and magnetic fields. So if you have a charged particle moving at a velocity V through both electric and magnetic fields in the same space, then you would find that this would be a total force on him. B Q times, E plus Q, Times V Cross Be noticed that this is going to give multiple components to the force, which is why we definitely need to remember that this is a force vector on the left hand side. Um, so this will be the net force on a charge Q. Due to elect external electric and magnetic fields, um, so an important one to consider, though not generally included the in the list that we're about to come up with now, Um, first, we should remember douses law, which for KAOS we found a while ago that if you have a closed surface around which you integrate electric field, then you come up with the total amount of enclosed charge divided by Epsilon. Not so. For example, if we have a box and we want to find how much electric field is coming out of it, then we could say if there's no charges in the middle queuing close with B zero, it will have zero net electric field coming out of it. Remember, that is going to be e dotted into D A where a is pointed out on each individual face. Okay, uh, on the other hand, if we had a charge in the middle and, for example, drew a sphere around it, then we would be able to say if electric field is constant at this at all distances away from this charge, we would find very quickly that is equal to the queuing. Closed just Q over epsilon, not times one over a, which is 1/4 hi r squared and that's it. We're done, and this looks a lot like the electric field due to a single electric charge. Um, we also can have a kaos is law for magnetic fields, which looks kind of similar. The way we come to this is by saying, if I take, I have a permanent magnet and I draw a Gaussian surface around it, and I integrate for all the magnetic field going through it. Well, notice here that the enclosed number of magnets is you have one north and one south east, which is like saying you have a positive and negative charge, so you end up with B dot D A. For a closed surface is equal to zero. Another way of remembering this law is simply that there are there is no such thing as a magnetic mon. A poll thes show up occasionally in popular science jokes about people searching for magnetic monta polls. And it is true people do search for them, and they theorized that they exist on that would violate this law. But as yet we have yet to discover one. Okay, so we've got two important laws here now. These don't necessarily connect e field and be field But they do connect electric field to its fundamental charge, which is the source of electric field and magnetic field to its fundamental source. Now, if we continue, we have Faraday's law as well, which is where things start to get exciting here, not a vector. Faraday's law looks like this. If we have the closed into groom of E over some path S, it's going to end up being equal to negative defy, be D t. Remember where five b is magnetic flux and we can write magnetic flux in the simplest form as equal to being b dot d a. Or we could also write it as being by B is equal to integral of b dot de a notice here that we're not necessarily considering a closed surface, simply a surface. Okay, so there's fair Days law and in meanwhile, we also had amperes law which looked like b dot Yes, so we have the same left hand side. But on the right hand side, what we have is mu not times the current enclosed by the path. So, for example, if we have a line here with current traveling through it, a wire with current traveling through it, and we were to integrate over this path that goes around it. Then we would be able to find the enclosed current. Okay, so the enclosed here in this case, would just be I. Now there is a problem with amperes law that becomes apparent as we consider the problem of the capacitor. So if we have a capacitor that's attached to some I M F stores, and at some point we were to say, I'm going to integrate over this path. That's fine, except for for two issues here. So this path could either bang be the boundary for a surface that is the smallest possible surface. Or it could be a path that is the boundary for a surface that extends into the capacitor. So think of a small basket that's open at the same point with the same radius here, but then comes down and terminates inside the capacitor. So the problem with this is that what we're really doing is we're integrating over the boundary of some surface, and the current going through that surface is the I enclosed problem is in the left hand side. Here. We definitely have a current I. That's going through the surface. But on the right hand side, we have no current passing through that surface. So how do we resolve this? This was the fundamental problem that a man named Maxwell considered was. What's going on in the right hand situation? There appears to be something wrong wrong with amperes law. So what Maxwell realized was that what if we have a time changing electric field? Okay, so we have some electric field in here, and it's a function changing as a function of time. Well, whether or not it's changing as a function of time, we know that there is this thing called electric flux, which would be exciting because we had magnetic flux. Previously. In Faraday's law, electric flux is equal to E times A and in a parallel plate capacitor, we know that E is equal to Q divided by epsilon, not times a. So if we were to plug that in here, then we would find that inside a parallel plate capacitor fi E is equal to queue over absolutely not, and then if we take the derivative of time derivative of both sides, so we have a time derivative of five e D T is equal to one over Epsilon, not times d. Q d t, which is just current. So that's current divided by Epsilon Not, but in this case that we're considering here, is there really current there? It appears there is. There's some current equal to epsilon, not times defy E DT. So what we refer to this as is the displacement current displacement current. And when we look at our total enclosed current, it has to be a some of both the i that goes through it through the surface, plus the displacement current. So when we go to write that down here, what we end up with is the integral of b dot ds is equal to mu, not times. Then we have our eye through All right, So our I enclosed as we would generally see it plus epsilon, not times defy e DT, which is our displacement current. Okay, so there we have it. This is the amount of current path, the effective current, the displacement current that's passing through this capacitor. And with this, we're able to finally complete what are known as Maxwell's equations. So if I were to write them all down very briefly for you right next to each other. They would look something like this. We have closed path closed surface integral of electric fields equal to Q. Enclosed over epsilon. Not and then we have the enclosed surface integral of be dot D A is equal to zero and then we have Faraday's law e dot ds is equal to negative defy be DT. And finally, what we just saw b dot ds This is getting close path and the girl is equal to mu, not times current through Plus you not absolutely not defy e DT. So these are Maxwell's equations and we're going to spend the next few videos looking at them and what they mean and how to use them. Mhm.