Physics 9 Fall 2009 Homework 7  Solutions


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1 Physics 9 Fall 009 Homework 7  s 1. Chapter 33  Exercise 10. At what distance on the axis of a current loop is the magnetic field half the strength of the field at the center of the loop? Give your answer as a multiple of R. The magnetic field on the axis of a current loop is given by B (z) = µ 0 IR (z + R ) 3/. We want a distance z, such that B (z) = 1 B (z = 0). The field at the center of the loop is B (z = 0) = µ 0I. So, we want R µ 0 IR (z + R ) 3/ = 1 µ 0 I R. Solving, we find R 3 = (z + R ) 3/, or z = 4 1/3 1R. So, z 0.77R. 1
2 . Chapter 33  Exercise 5. Magnetic resonance imaging needs a magnetic field strength of 1.5 T. The solenoid is 1.8 m long and 75 cm in diameter. It is tightly wound with a single layer of.0 mmdiameter superconducting wire. What size current is needed? The magnetic field of a solenoid is B = µ 0NI BL, and so the current is I =. What s L µ 0 N N? It s just the length of the wire, divided by the diameter of the wire  this is how many times you can wind the coils together for a given length. So, there are N = 1.8/0.00 = 900 turns of the wire. Thus, I = BL µ 0 N = π = 400 A.
3 3. Chapter 33  Exercise 9. Radio astronomers detect electromagnetic radiation at 45 MHz from an interstellar gas cloud. They suspect this radiation is emitted by electrons spiraling in a magnetic field. What is the magnetic field strength inside the gas cloud? The frequency of the emitted radiation is equal to the frequency of the acceleration of the electron (i.e., the frequency of the revolutions). The revolution of the frequency is just the cyclotron frequency, which is f = qb. Solving for the magnetic field gives πm B = πfm. Plugging in the values gives q B = π = T. 3
4 4. Chapter 33  Problem 45. The element niobium, which is a metal, is a superconductor (i.e., no electrical resistance) at temperatures below 9 K. However, the superconductivity is destroyed if the magnetic field at the surface of the metal reaches or exceeds 0.10 T. What is the maximum current in a straight, 3.0 mmdiameter superconducting niobium wire? The magnetic field of a long straight wire is B = µ 0I, where r is the distance away πr from the center. So, at the surface of the wire, when r = d, the current in the wire is I = πbd µ 0. Plugging in the values gives the maximum current as I = πbd µ 0 = π π 10 7 = 750 A. 4
5 5. Chapter 33  Problem 46. (a) Find an expression for the magnetic field at the center (point P) of the circular arc in the figure. (b) Does your result agree with the magnetic field of a current loop when θ = π? (a) Recall the BiotSavart law, B = µ 0 I s ˆr, where s points along the current, and 4π r ˆr points long the direction from the current to the point P. Now, because the current and ˆr point along the same line in both straight segments of the wire, their magnetic field contributions to the point P are zero. So, we only need to find the magnetic field from the curved part of the wire. Along the arc, the current points along the wire, which is always at right angles to ˆr. So, along the arc, s ˆr = s ˆr = s, since ˆr is a unit vector. So, along the arc the little bit of the magnetic field from the little section s is B = µ 0I 4π s r. Now, the segment is always a distance r = R from the point P. So, we just have to add up the distance, s, along the arc, which is s = Rθ. Thus, B = µ 0Iθ 4πR. (b) Recall that the magnetic field of a loop of wire is B = µ 0 IR (z + R ) 3/, which gives the field at the center, when z = 0, as B center = µ 0I, which is precisely R the correct result in part a when θ = π. So, our results agree. 5
6 6. Chapter 33  Problem 56. An electron orbits in a 5.0 mt field with angular momentum kg m /s. What is the diameter of the orbit? Recall that the electron makes an orbit in a magnetic field of radius r = mv eb. The angular momentum of a particle in an orbit is L = mvr. So, multiplying the expression for the radius of the orbit by r gives and so the radius can be expressed as r = mvr eb = L eb, r = L eb. With numbers, we find that r = 6 = 0.01 m So, the electron orbits at a distance of about 1 cm. 6
7 7. Chapter 33  Problem 61. An antiproton (same properties as a proton except that q = e) is moving in the combined electric and magnetic fields of the figure. (a) What are the magnitude and direction of the antiproton s acceleration at this instant? (b) What would be the magnitude and direction of the acceleration if v were reversed? (a) The antiproton is negatively charged, so the electric force is F E = ee, and points up, while the magnetic force is F M = e v B, and points down. So the two forces work ( against) each other. The net force is just the sum of the forces, = e E v B. The magnitude of the force is F net = e (E vb) = F net ( ) N. Because the magnetic force is stronger, the force points down. Finally, since a = F/m, where m is the mass of the proton, the acceleration is a = m/s, and still points down. (b) If v was reversed, then both forces would point up. The forces would add, such that F ( ) net = e E + v B. In this case, the magnitudes would add, instead of subtract, giving F net = N, and points up. Again, the acceleration is just a = F/m, giving a = m/s upwards. 7
8 8. Chapter 33  Problem 73. In the semiclassical Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius m with speed m/s. According to this model, what is the magnetic field at the center of a hydrogen atom? Hint: Determine the average current of the orbiting electron. The moving charge orbiting the nucleus constitutes a current, I. The magnitude of the current is just the charge on the electron, divided by the time it takes to orbit the nucleus, i.e., it s orbital period, T. The period is the total distance around πr, divided by the speed of the electron, v. So, T = πr/v, which gives I = e T = ev πr. The magnetic field at the center of the loop is, as we ve seen, B = µ 0I, which gives R upon plugging in for the current, B = µ 0ev 4πR. With numbers, we find B = µ 0ev 4πR = ( ) = 1.5 T, which is a huge field! 8
9 9. Chapter 33  Problem 79. A flat, circular disk of radius R is uniformly charged with total charge Q. The disk spins at angular velocity ω about an axis through its center. What is the magnetic field strength at the center of the disk? The plate is uniformly charged, so it s surface charge density, η = Q = Q, is constant. A πr Consider a small bit of charge, dq. From the BiotSavart law, the tiny bit of magnetic field from the this charge is db = µ 0 ˆr dq v = µ 0 4π r 4π dq v r ˆk, since v points along the direction of spin, which is tangent to the disk, while ˆr points towards the z axis. This means that v and ˆr are perpendicular. Now, for an object spinning in a circle, v = rω, where ω is the angular velocity. Therefore, db = µ 0 4π dq ω r. Next, we integrate, writing dq = ηda = πrηdr, since we are on a disk, as we ve seen before. Then, db = µ 0ωη dr. Then, integrating both sides gives B = µ 0ωη R 0 dr = µ 0ωηR. We can express this in terms of the total charge, Q, on the disk by recalling that η = Q, which gives πr B = µ 0ωQ πr, and points up. 9
10 10. Chapter 33  Problem 81. A long, straight conducting wire of radius R has a nonuniform current density J = J 0 r/r, where J 0 is a constant. The wire carries total current I. (a) Find an expression for J 0 in terms of I and R. (b) Find an expression for the magnetic field strength inside the wire at radius r. (c) At the boundary, r = R, does your solution match the known field outside a long, straight currentcarrying wire? (a) For a varying current density, the total current in the wire is I = JdA. For a disk, da = πrdr, as we ve seen before. Integrating over the total surface area of the disk gives the total current, I, on the wire. So, I = and so J 0 = 3 πr I. JdA = πj 0 R R 0 r dr = πj 0 R R 3 3 = πr 3 J 0, (b) Here we will use Ampere s law, which says that B d s = µ0 I encl. Because of the symmetry of the wire, the magnetic field circles around inside the wire. So, we choose a circle of radius r R for our Amperian loop. Since the magnetic field is constant along that loop, and points along d s, the lefthand side of Ampere s law says B d s = B (πr). Now we just need the enclosed current. We can use the same method as in part (a) to determine the enclosed current  only now we integrate out to radius r, since we only want the enclosed current. Thus, I encl = JdA = πj 0 R r 0 ( ) r dr = πr r 3 J 0 = I, R where in the last step we inserted our value for J 0. Thus, Ampere s law gives B (πr) = µ 0 Ir /R, or B = µ 0I πr r. (c) The field outside a long straight wire with current I is given by B = µ 0I. At the πr boundary, we have r = R, and so B = µ 0I, which is precisely the correct answer πr that we get from part (b) at the boundary. 10
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