Phys 1120, CAPA #11 solutions.

Transcription

1 Phys 1120, CAPA #11 solutions. 1) The force between wires is given by the text: F= mu0 I1 I2 l /(2 pi L) (You could work this out yourself pretty easily - we know the formula for the FIELD from a long wire (B = mu0 I / 2 pi r) and we know the formula for the FORCE given that field (ILB). You just have to think about what "r" means, and what "L" means. In this case, with the formula I've given, "l" must be the length of the wires, and "L" is the distance between them!! We don't know the length l of the wires, but it really doesn't matter. (You can pick it to be one meter, or just leave it as a symbol. It will cancel out in the end). Thus, mu_0 I1 I2 l / (2 pi L) - weight=0 (Newton's II law, net force must be zero!) The weight you have to hold up is m*g of the lower wire, and m = density * volume And Volume = area*length, so weight = (density*area*length)* g We have (mu_0) I1 I2 l / (2 pi L) = density * area * l * g. This is basically it, one equation, with only one unknown (I2), because "l" cancels. We're asked for I2, so I2 = density * area * g * 2 * pi * L / (mu_0 I1) Area is pi*radius^2 = pi*diameter^2/4, everything has been given. (Capa gave us density, diameter, L, and I1) Just plug in the numbers, being sure everything is MKS. 2) The magnetic field in a solenoid is given in your text ( have you thought about this formula? Could you derive it yourself?) In the end, it's pretty nice and simple: B = mu0*n*i/l, so that means B(A)/B(B) = mu0*n(a)*i(a)/l(a) / mu0 * N(B)*I(B)/L(B). Looks like a bit of a mess... but CAPA said the lengths are the same, so they cancel, and now B(A)/B(B) = N(A)*I(A)/ (N(B)*I(B)). In my case, NA/NB = 10 was given, and I(A)/I(B) = 5 was given. The radius is a "red herring", it does not matter. So for my numbers, B(A)/B(B) = 10*5 = 50 times greater. 3) In my case, I had "-" currents (out of the page) on the left, and "+" on the right (into the page). (If your figure is different, you'll have to adjust, but it should be very similar!) The upper left current (into the page) makes a B field at the center of the square of magnitude B0 = mu0 I*/(2 pi r), where r is the distance from the current to the center of the square. Can you work that distance out? By geometry, r = (a/2)*sqrt[2]. We have all those numbers, so we can directly compute B0. What about the direction of B0? The field lines are circles around the wire, and (using my right hand), I see that at the origin, this field will point exactly 45 degrees "up and right". I can easily get the components, then: B0_x = B0*cos(45), and B0_y=B0*sin(45). The lower right current makes the very same B field (same distance, same direction), they will just double up!

2 The remaining two currents each make a B field that points 45 degrees up and left, rather than up and right. (Use your right hand, convince yourself) So we have two "up and left" and two "up and right" B fields. Apparently, the x components will cancel, but the y components will add up. Thus, B will be in the +y direction, and the total magnitude will be B0_y +B0_y+B0_y+B0_y. The answer is 4*B0_y = 4*B0*sin(45) = 4*(mu0 I/ (2 pi r)) * sin(45), where r = (a/2)*sqrt[2]. 4) This is a hard one! Just take it one step at a time - think hard about what's given, and what the fields look like. Let me call the unknown current (in the little wire) I, and the *known* current in the fat rod "i0". I think to start with, we surely need two formulas: B(P) and B(center of rod), because the problem explicitly asks that we *set them to be equal.* Now, without knowing I, we can't figure either one out - but we can just CALL the current I, and figure them out symbolically... that seems like a good strategy to start with. Let's find B(center of rod ) first. Let's call "up" the positive direction. By superposition, B(center of rod) = B(due to I, at center of rod) + B(due to I0, at center of rod). Now, B(due to I, at center of rod) is easy, use the equation we've used many times now, the field due to a long wire (mu0 I/2 pi distance) - in this case, it's just -mu0*i / (2 pi *3 R), because it's a distance 3R from the current I to the point in question. Where did that minus sign come from? If I *assume* I is into the page, then this term is negative by the RHR (point my thumb into the page over at the wire, and the fingers curl DOWN at the rod position) (If I turns out to be OUT of the page, it's o.k, it'll just turn out to be a negative quantity!) We'll think more about directions in just a sec. Now check this out: B(due to I0, at center of rod) is exactly ZERO. You could get this from Ampere's law (Example 32.8 in Knight works this out - it gives you the B field inside the rod), or you just just think for a second: if you have a current flowing through a rod, if it was NOT zero at the center... which way could it point!? So, we add zero, and putting this together, B(center of rod)= -mu0*i / (2 pi * 3 R). Again, that minus sign means IF I is positive (i.e. into the page) THEN B(center of rod) is down the page. Now, let's get B(P). By superposition, B(P) = B(due to I) + B(due to I0). These are simply given by our favorite formula, B(long wire) = mu0 I / 2 pi *distance to center. That's true even if the wire is THICK, by Ampere's law, as long as you are OUTSIDE of it. (Check out Ex again, Knight works this out there too) B(due to I) = -mu0 * I / 2 pi R, (minus, because it's DOWN at P, assuming I is +, as I am doing.)

3 and B(due to I0) = +mu0 * I0 / 2 pi *2R. Why is it 2R? Because the "r" is supposed to be the distance of the point from the CENTER of the current. It's "+" because of the right hand rule (and we're told the current i0 is pointing into the page) Thus, B(P) = mu0 (I0/2-I)/ (2 pi R). We early had B(center) = -mu0*i/(2 pi 3R) Setting them equal, as CAPA says we should, the mu0/2 pi R cancels, and we have (I0/2-I) = -I/3, or I0/2 = I*(2/3), or I = +3/4* I0. Done. As I said, that's a fairly tricky problem. Keeping track of the signs was really the worst part, but also just figuring out what they were asking, and how to get started, is pretty challenging. PS I could actually convince myself I is into the page right from the start, and it's a nice "check" of the signs above: Imagine for a second that I were *out* of the page. That would make an "up" B field at point P. But we already know I0 makes an UP field at point P. So the two fields would add up, making an even bigger field at point P. But the field at the center of the rod is coming just from I, and it's further away than P. It's going to be smaller than the field at point P. The problem asks for them to be equal - IMPOSSIBLE!! ( if I is out of the page. ) So we conclude that I must be into the page, parallel to I0. Just as we got, mathematically, above. Cool. 5)This problem is not really all that much different from Knight's example It's a good problem - we need to figure out B due to a current, and there's LOTS of symmetry. This make me think we should use Ampere's law. Integral[B.dl] = mu0*i(enclosed). Pick, as the line to integrate around, a circular path or radius r. I'll pick r to be *anything* between a and b. Why did I pick this line? I get to make it up, it's an imaginary line, a loop - I picked it because by symmetry B is going to be the SAME everyone on this little loop! So I'll be able to pull it out of that integral. That's the trick with Ampere's law, it's really the only time this law makes it EASY to figure out B... By symmetry, B is going to make circles, and so it'll be a constant as we integrate around a circle. Thus, the left side is just B*(2 pi r). What about the right side? What's I(enclosed)? I'd expect it to be J*(area enclosed.) The "area enclosed" would have to be pi r^2 - pi a^2, because the hole in the middle has no current. The current density J = total I / total area = I / (pi b^2 - pi a^2) Putting it together, B*2 pi r = nu0*pi *(r^2-a^2) *I / pi*(b^2 - a^2), or B = mu0*i *(r^2-a^2) / (2 pi*r*(b^2 - a^2)) Does this make sense to you? The current enclosed is just a fraction of the total current. What fraction? Well, you just need the ratio of the area of the "annulus inside r" to the "annulus where the current is flowing" (OK, it's a kind of nerdy word - It's just a ring shape...) Make sure this is something you can do for yourself! 6) We know from class that a full circle of current makes B = mu0 * I / 2 R at the center. A half circle will make half that, mu0*i/ 4R From the text, an infinite line makes mu0*i/ (2 pi r), so a half infinite line makes mu0*i / (4 pi r)

4 (Well, that's technically only true at a point vertically displaced from the END of that half infinite line - but that's exactly what we have here!) We have TWO such lines, and (use your right hand, convince yourself) ALL THREE of these fields are in the same direction, they all add up. So the total results is muo*i*(1/4r + 2/ 4 pi R) That's it. 7) We kind of got a little ahead of ourselves here, I apologize - this particular problem is a Faraday's law question which we really haven't covered in class yet (though it has been on the readings). The hint pointed you here. Sorry about that, CAPA gets written weeks in advance, we thought we'd have covered this in class. Fortunately, this one is a pretty basic question, so if you've even just read section 33.5, I bet you'll be able to work this one out...)we won't put a Faraday's law question on the 3rd midterm (it'll be on the final!) Though we might have a Lenz' law question, since we did a Tutorial on that last week! Faraday's law says EMF = -N Delta Phi/Delta t. EMF is a fancy word for voltage (basically). In this formula, N is how many loops you have, Phi is the magnetic flux (just like electric flux, it's B dotted with the area vector). If you use "delta", you get average EMF. (If it's instantaneous, you could just use the derivative - see the next problem) Here, N is one (just one loop), and Delta t is given Delta Phi is the CHANGE in the magnetic flux, and flux here is simply B*Area. They gave us B, and the diameter of the circle, so use Area = Pi(diam/2)^2. But here we do need to think about signs. Flux is not a vector (B field is, but flux is a dot product.) Still, flux has a SIGN, you can have positive or negative flux, and those are different! Delta Phi = Phi_final - Phi_initial = B_final*area - B_initial*area = (area)*(b_f - B_i). Since B_f points DOWN, and B_i points UP, that "subtraction" really ends up adding magnitudes. (Think about it, convince yourself! If Bf was equal in magnitude to Bi, but opposite in direction, that would not be ZERO change in flux, that would be a *big* change!) In my case B_f =.210 T up, B_i =.690 T down, so B_f - B_i = (-.690) = +90 T. For the rest, then, EMF = Delta Phi/Delta t = Delta B * Pi*(diam/2)^2/ delta t. Don't worry about the final sign on the EMF, that's really asking which WAY the electric fields will point in the wire - that'll come soon, it's Lenz' law (see q. 8) (We just asked for magnitude) 8) Once again, a little Faraday's law problem. If you got the last one, this one is quite similar. The only difference is instead of finding Delta phi/delta t, we find d phi/dt (the derivative)... EMF = -N * d (phi) / dt = -N * d(b*a) / dt. In this case, A is constant = pi r^2, so we have EMF = -N * pi r^2 * db/dt. The problem said B = gt, so db/dt = g. (This has nothing to do with 9.8, it's given in the problem!)

5 The problem asked for the magnitude, so again we don't have to worry about the sign. But if you're curious - look at the picture, and ask yourself if current will flow from LEFT TO RIGHT or RIGHT TO LEFT through that resistor? It's a good Lenz' law exercise, think about it yourself before looking at my answer here - (I claim the answer is, the coil will try to "fight the change". The field is into the page and increasing, so the change is into the page. So we'll induce a counterclockwise current to "fight the change". That means left to right through the resistor, given the wiring shown in the picture) 9) Power = EMF^2/R, and energy = power*time, so energy = EMF^2*time/R, where EMF was the answer to #8. (Notice that because B was linearly increasing with time, db/dt was a constant, so there's no integral required, it's just steady power output!) 10) I think everyone's figure is the same here: This problem reminds me a little of the "Lenz' Tutorial"! In fig 1, if the MAGNET moves west (towards the loop), there is a flux to the left and INCREASING. The loop fights that, it will try to make a B field that points to the right, which is current labelled B. If the magnet moves east instead, the current is opposite In Figure 1, if the LOOP moves West (away from the loop), it's the opposite scenario to part A. The flux through the loop is leftward but decreasing, the loop fights the CHANGE by inducing a field to the left, i.e. current labelled A. (If the loop moves east insteast, the current is opposite this) In Fig 2, if the ring moves N (or south), nothing happens (Flux is not changing) In Fig 2, if the ring is stationary, there is zeroinduced current (Flux isn't changing) In fig 2, if the ring moves East (right), the flux is INTO the ring (x's), but DECREASING, the current will fight the change, by making a B field into the ring. I.E, Direction B. (if the ring moves west (left), it's the opposite of the previous case. The flux is now INCREASING into the ring, so to fight that you make a B field out of the ring, i.e. Direction A.