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A circular loop of current carries a steady current. (a) Sketch the magnetic field lines in a plane perpendicular to the plane of the loop. (b) Which side of the lo...

Question

A circular loop of current carries a steady current. (a) Sketch the magnetic field lines in a plane perpendicular to the plane of the loop. (b) Which side of the loop is the north pole of the magnetic dipole and which is the south pole?

A circular loop of current carries a steady current. (a) Sketch the magnetic field lines in a plane perpendicular to the plane of the loop. (b) Which side of the loop is the north pole of the magnetic dipole and which is the south pole?



Answers

A circular loop of wire with area $A$ lies in the $xy$-plane. As viewed along the $z$-axis looking in the -$z$-direction toward the origin, a current $I$ is circulating clockwise around the loop. The torque produced by an external magnetic field $\overrightarrow{B}$ is given by $\vec{\tau}$ = D(4$\hat{\imath}$ - 3$\hat{\jmath}$), where $D$ is a positive constant, and for this orientation of the loop the magnetic potential energy $U = -\vec{\mu}$ $\cdot$ $\overrightarrow{B}$ is negative. The magnitude of the magnetic field is $B_0 = 13D/IA$. (a) Determine the vector magnetic moment of the current loop. (b) Determine the components $B_x$, $B_y$, and $B_z$ of $\overrightarrow{B}$.

Caroling at question twenty eight seventy two, where we have a circular loop with current I too going clockwise. And then we also have ah, current a wire down here with a current, but we don't know what direction the current is going in, but we want the magnetic field at the center of the slope. So this point right here to be zero. So what that means is the magnetic field from this loop is gonna have to cancel out the magnetic field from this wire. So first, let's figure out the magnitude of the magnetic field coming from this wire. So here's the expression for the magnetic field from an infinite ah, line of current. So here we have. Ah, this we're talking about the, um this wired on here. So this is I won, and the distance that we're talking about here is going to be STI. So this is our Let's call this the one the magnetic field coming from this, A current town here again. We don't know what direction this is coming. This is going to exert on yet. So now for me, too. We have a loop of current and we're looking at the magnetic field right in the center. So the expression for ah magnetic field from a live of current is going to be, um we have mu not times. I too here says the numerator and then in the nominator is two times the radius and the radius here is R So this is going to be the magnetic field coming from this loop. And we do know the direction that this is going to be in because we can use the right hand rule. So using the right hand rule, curling our figures around where the current loop is going, that means that it's gonna point into the board or into the page here. So we want we know that this one is going to be into the page, so we know that this is going to need to be out of the page in order to cancel it out. So let's just figure out the direction of the current here right away, since we know before we get the magnitude, um, in order to, ah get a current a magnetic field out of the page from a current here, we know that this that this current is going to need to go to the right. Because if we have a current going to the right again, if we point our thumb in the direction of the magnetic field of the current here, that will give us a magnetic field coming out of the page at this point. So we now know the direction that I want has to be. So now, ah, we have to do is find the magnitude. So we know that at this point, we want I knew one to be equal to be too, because that will make the net magnetic field equal to or I'm sorry. Do you want us to be equal to negative B too? Because we know be one plus free, too past equal zero. So we wanted to set anyone equal to minus me, too. So let's sit. You not I want over here, uh, over for piety equal to minus me too. So this is just at making some equal to zero. Um, Now, what we want to do is solve for I won. So to do this, let's just cancel out the terms that we can. So we see, um, you not in both of these terms So let's get rid of that to make it easy. Um, we have, ah, he cancel out a factor of two. So let's multiply both sides by two. That's going to turn this into a two instead of before on DH. Now that's kind of all we can do. So again, we're selling fry ones. Let's just multiply both sides by two high D. So we're going to get this pride e divided by our we discovered of this part and we don't go negative sign here is a bit unnecessary because we're just looking at the magnitude we already figured out the direction, Um, so we can kind of just begin or the minus sign here, but this is going to be our final expression for the magnitude of the magnet of the current. I won in order to make the magnetic field at this point

Okay, So before we talk about how to find the direction of the current, let's talk about how do we know that if there's some induced current or not so always remember one thing that if the magnetic field is changing on Lee, then we'LL have some induced current. Okay, so if you look at the question, there are three different sections. When you have your increasing magnetic field, you're decreasing magnetic field and your constant magnetic field. So we can immediately say that your if your magnetic field is constant, there's no induced current, so they'LL be zero s o. Now let's talk about the first part where your magnet field is increasing. So a Z, I said, if you're making it feel is changing you'll have some induced current Now how do we know the direction So in that, you know, to do that, you need lenses law. So what lenses are tense? You that is that if you have some change in magnetic field, the loop will try to oppose that effect. Right? So you see your bee is increasing into the page so these magnetic fields are pointing into a page. So if your B is increasing into the page. The loop will try to oppose that because Lindsay's lawyer tells us that So if we want to oppose the effect, we need some magnetic field. Sorry, we need to have some magnetic field which is pointing towards us right, because it's opposing and now we can use our right and rule eso If you look at Chapter twenty eight Section five, it's tells you that you're managing feel will be your term And if you call your hand, that will give you the direction So you already know that the magnetic fields are pointing towards us So we point out, come towards us and then we call our fingers And if we do so, we'll have the direction this way which is anti corporates, right? Similarly, if we try to solve for part B which is when you have decreasing magnetic field so your managed fields are now decreasing. So again the Lupin traitor opposed effect so it will generate some extra magnet field towards the page or into the page. So there will be some extra magnetic fields that will be generated. So you have me overlap those right so you have some extra minute feels which are going towards the bait. So if that's the case, you again use your right and rules. So you point your thumb towards the magnetic field, which is into a page, and then you call your fingers. And if you do that, you will see the magnet field will be clockwise. Okay, so to summarize. Ah, your magnetic field. See? Yes to summarize for party here. Made it feel would be anti clockwise. You're, um, for part B, it will be clockwise. And for party, it will be zero. So if there's some changing magnetic field you'll on, only then you will have some current. And if there's no change in McGinty field, there will be no current. So, yeah, Thank you for listening.

For this problem, we have a circular loop in a magnetic field. Our goal is to find the direction of the induced current based on different situations the Byfield. So whether that's increasing, decreasing or remaining the same on this diagram, I've depicted R B field going into the page, and that's drawn with little X marks. So this problem is gonna be an exercise in Monday's law. And one version of London's law states that the induced Byfield is established and mitigate the change in the Met magnetic flux. So let's write that down here. The induced be field is established to mitigate the change in net magnetic clocks. So if we determine what that induced the field is the orientation of that, we can easily find the direction of the induced current from our right hand roll. Let's start out with part A, and I'm gonna code this green because on our diagram, I'm gonna draw green for the be induced field. So here we have our external Byfield is increasing. If that external field is increasing, we have a larger magnitude into the page. To mitigate that change we need are induced field to point out of the page. So drawing that on a diagram out of the pages depicted by this little green circle and a dot in the middle cha largest emphasize here out of age and in order to find the direction of the Indies current. With this, we call on our right hand roll, which says Wrap your four fingers around the be vector are induced. Be vector so that your thumb is pointing in the direction of the arrow. And the way that your direct the way that your forefingers are curling gives us the direction of our current. So in here we have that current being counterclockwise counter clock, eyes induced kind for the next part code. This in orange we have our external Byfield is decreasing, decreasing. So if that's decreasing, we know that the magnitude into the page is getting smaller and value. So to mitigate that change we need are induced field to be pointing into the page. So here that is into the page and same right hand rule. Here thio, figure out the direction of the current point your thumb in the direction of your be vector, your induced defector and the way that your fingers curl around that vector gives you the direction of your induced currents. So here we have a clockwise current clockwise induced Chris and Lassie. We have be remains constant, constant, be not will for external field. Our external B is remaining constant. We have no net change of flux if there's no change in the flux. There's no induced field that's established and with no induced field that we get no induced current, so you get no induced QWERTY.

Our question says it wants us to view Figure 45 as reference for this question. And it says a current I flows in the circuit. Clockwise is shown and for part eight wants his determine the magnitude in the direction of the magnetic field at the center. See? Okay. So to find the magnetic field in the direction at the center, see, we're gonna break this into two different loops are the loop which has the radius are One is the tops in my circle in the loop, which has the radius are two is the bottom large semi circle. So for part A, the magnetic field at a point C can be obtained by using the Viets of Art Law, uh, which says that which is useful for determining the magnetic field due to a known arrangement of currents. So what does this law say? We'll go ahead and write it out here. It says that the change of magnetic field D B is equal to, um, you not just the magnetic permeability of free space time to the current that's producing the field. I divided by four pi. Then this is multiplied by the cross product of D l Let's make that look like an out there We go across the unit vector are divided by the radius over considering our square So again here D B is that contribution from the magnetic field eyes the current passing through the small length Decebal So D d l is this small length that we're considering this incontestable link change and our hat is theater vector along the path the magnet magnitude in the direction of the magnetic field at point C along the path of the straight wire will be zero. So the magnetic field is only due to the semi circular portions so we can break these into the top semicircle which will call one and the bottom semester protocol tooth. Since the top one has radius are one of the bottom has radius are too so from the top would. But we have maybe should indicate this is part a. First we have B one go is equal to the integral of Devi one so we can replace above for D B. One way we can replace what we have written above. So, uh, I'm gonna go ahead and pull out some things that are constants right outside of the integral. So we have you not times I over for pie times r squared. But they are in this case is our one, cause we're considering the tops of my circle. So this is times are one squared and so in the integral we have d l cross car. Okay, But we can further simplify this even more because d l cross our hat. Remember that our has the unit vector along the path, whereas, l, um, is the small infant Esnal length. So if you do, l cross are using the right hand rule. You get, uh, what we're gonna call k hat, which is going into the page, right? So we'll say that l cross our gives u K hat or negative K hat, and we're going to find the negative K Hot direction is going into the plage. So this gives you and you're not times I divided by four pi r one squared climbers, negative K hat, and then we're still integrating over the incontestable length D l Okay, so carrying this out even further. The integral of inthe intestinal link d l gives us pi r one because its half of the circumference. So it's not two pi r one is just pie are one. Because we're considering half the circumference here. It's on a full circle. Um, So what that leaves us with is you, Not times I divided by for the pies, Cancel, right, cause there's a pie on bottom and then the deal integral gives us ah, pie. So those cancel and then one of the are ones cancel because we have in our one squared on bottom and we're gonna end up with our one from the integral of D L. So we end up with you, not I over for our wood. And then this is in the negative K hat direction. Or you could also say into the page either one. Okay, Now we need to do the same thing for the bottom semicircle. Be too. So be, too, is going to be equal to the same type of integral. It's going to be d B two. It was the same expression, except now we're replacing our one with our two. So if you go back here and look what we have, it's again the exact same expression. But we're gonna replace our one with our two. So are results of the integral are going to be the exact same. Except we're replacing our one with our two. So what's just without having to carry out all of the operations? It's just gonna be the exact same as before, except replacing our one with our two. So the result is gonna be mu. Not times I divided by four times are too acting in the negative K hat direction. Or you could also just right into the page. Either one should work. Okay, so then our total magnetic field B is going to be a linear combination of the two magnetic fields from each semicircle. So this is gonna be be one plus be too. Oh, be so one plus B. So too. Okay, so now that we know that they add together like this, Um oh, And by the way, if you wanted to figure out if you're if you're struggling to understand where the negative K hat comes from, where the into the page idea comes from, for these be values, go ahead and take your right hand and curl your fingers in the direction of the current that's running over the top semicircle and see which direction your thumb points. It's gonna be into the page, right? And then the same thing with the bottom loop curl your fingers in the direction that the current is going and see which direction of some points It's also gonna be into the page. So you could also use the right hand rule to get that justification. So that is the reason that B one plus B two is going to increase because they're both contributing to a magnetic field that's acting into the page. So we end up with here is you not times I over four cause those air constant. Also, I'm just gonna pull that out front. So you know, I over four, and then you have one over r one plus one over r two, and these are all not acting in the negative K hat direction. You could also, if you wanted to type out into the page just to be specific, you could we could go ahead and boxes in is our solution for part a Part B says he would like us to find the magnetic dipole moment of the of the circuit. So the magnetic moment is the product of area and current. The area is the sum of the half circles on by the right hand rule the direction of the current and is going to be going into the page, right. The at least the magnetic moment is gonna be going into the page because of the direction of the current from the right hand rule, as we had previously described. So for part B then will indicate that as such, we have a magnetic moment. New. It's a vector. So well indicate it is equal to the current I, as I said times the area and then the direction from the right hand rule that this is acting is into the page k at. Okay, well, the area is the area of the tops in my circle, plus the area of the bottom certainly circle. So we can call this a one plus a two. And then again, negative K hack. Okay, well, what is the area of the top of the bottom semicircle there, simply equal to pi? 1/2 pi r squared. Where are is the radius of the two semi circles, so we can replace those and we have high times are one square over too. It's 1/2 pie are once weird plus pi Times are two square over too. These air so going in the negative K hat direction. Okay. And then just to simplify this a little bit more, we can pull the pie out front as well as the two out front. So we end up with I, um actually, so we can pull the pie out, fried that too out front. And we can also pull the negative out front from the case will really make this a lot more simple to read. So we have negative 1/2 the 1/2 from the 1/2 pi r squared times. Uh, hi times I Then this is multiplied by our one squared plus are too square, And then this isn't K hat, so I called the negative from make a hat out front so we can box us in as our solution to part B


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