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1 Physics 153 Sample Examination for Fourth Unit As you should know, this unit covers magnetic fields, how those fields interact with charged particles, how they are produced, how they can produce electric fields, and related topics. (i.e., The subject matter of Chapters 28 through 30) You may use the set of notes titled "Study Guide 4", which is online, the text, and any resource papers, except those of your fellow students, or other instructors. 1. A deuteron is being accelerated in a cyclotron having a dee radius of R = 85.0 cm. The magnetic field in the region where the deuteron travels is constant. The oscillator frequency is 10.0 MHz. Determine the expression for each the first three of the following: (a) the speed of the deuteron as a function of the distance from the center of the cyclotron. (b) the magnetic force on the particle as a function of its distance from the center of the cyclotron, (c) the kinetic energy of the deuteron as it leaves the cyclotron. (d) Evaluate part (c) in J and in MeV. (e) Determine the value of the magnetic field within the cyclotron. (a) f = 1.00 x 10 7 Hz -> ω = 2πf = 2π x 10 7 rad/s, R = m, B = 2.00 T, m = 3.34 x kg. v = rω = 2πrf = (2π x 10 7 /s)r = (6.28 x 10 7 /s)r (b) F C = mv 2 /r = mrω 2 = (3.34 x kg)(2π x 10 7 /s) 2 r = 1.28 x (N/m)r (c) K = mv 2 /2 = m(rω) 2 /2 = m(rω) 2 /2 (d) R = m -> K = (3.34 x kg)(2π x 10 7 /s) 2 (0.850 m) 2 /2 = 4.76 x J = 2.97 x 10 7 ev = 29.7 MeV. (e) v is perpendicular to B in a cyclotron, so F = qvb = mv 2 /r. Thus, qb = mv/r = mω = 2πmf = (2π x 10 7 )(3.34 x ) N s/m, and B = mω/q = (2π x 10 7 N s/m)(3.34 x )/(1.60 x C) = 1.31 T. 2. An elliptical coil of wire having a semi-major axis (x-direction) of 15.0 cm and semi-minor axis of 10.0 cm is laying flat on a horizontal surface (say on a page). As you look down on the coil, + x is to the right, + y toward the top of the page and + z out of the page (toward you). The number of turns is 50 and the current is is 10.0 A in a counter-clockwise direction as viewed by you. (a) What is the magnetic moment of this coil? (Magnitude and direction) (b) Suddenly a uniform magnetic field of T is turned on. Its direction is perpendicular to the y-axis and 60.0 O from the positive x-axis. What is the magnetic potential energy associated with the dipole at that instant that the field is turned on? (c) About which axis will the dipole rotate (if permitted to do so), and in what direction (i.e. away from or toward which axis)? (a) µ = NiA, where N = 50, i = 10.0 A, a = m, b = m, and A = πab = x 10-2 m 2. µ = 50(10.0 A)(4.712 x 10-2 m 2 ) = A m 2 ~ 23.6 A m 2. Since the current is ccw (viewed from above) µ is directed upward (k). Page 1 of 2
2 (b) U B = - µ B = - (23.56 A m 2 )k (0.500 T)(i cos60 O + k sin60 O ) = - (23.56 A m 2 )k (0.500 T)(sin60 O ) = J. (c) It will rotate in the direction that will make µ align with B. This requires that it s rotation is about the y-axis. The final orientation of µ will be by 30 O from the z-axis in the x-z plane. Using the vector description for torque on a magnetic dipole, τ = µ x B. This is in the direction of k x i or + j. That means it will tilt to the right. 3. Determine the magnetic field at the center of a solenoid 5.00 m long, having a radius of 10.0 cm, a total of 20,000 windings, and a current of 4.00 A. B = µ 0 ni, where n = 20,000/5.00 m, and i = 4.00 A. B = (4π x 10-7 T m/a)(4000/m)(4.00 A) = 2.01 x 10-2 T. 4. Two parallel power lines carry currents of 200 A in opposite directions. If the two wires are 20.0 cm apart, what is the force per unit length between these wires, and is the force attractive or repulsive? Explain using the right-hand rule for the fields and forces. At the location of wire 2, B = µ 0 i 1 /2πd Using F = il x B, The magnitude of the force on wire 2 because of wire 1 is µ 0 i 1 i 2 L/2πd and the force per unit length F/L = µ 0 i 1 i 2 /2πd = (4π x 10-7 T m/a)(200 A) 2 /(2π x m) (2.00 x 10-7 T m/a)(200 A) 2 /(0.200 m) = 4.00 x 10-2 N/m. From our discussion in class we know that currents in the same direction are attractive and those in opposite directions are repulsive, so the wires repel. However, we should use our knowledge of the right-rule to confirm this. Assume the two wires are in the plane of this page, with the direction of current in the wire on the left directed toward the top of the page, and the current of the wire on the right directed toward the bottom of the page. By the RHR, the field to the left of the left wire is directed out of the page, but to the right of it into the page. With L to the bottom of the page, and B into the page, L x B is to the right. It is repelled from the left wire. Obviously the force on the left wire is to the left.
3 5. Consider a circular loop of wire having a radius of R = m. A magnetic field oriented perpendicular to the plane of the wire is described by the equation B = (5.00 x 10-3 T/m)r, where r is in the appropriate S.I. unit and measured from the center of the loop. Determine the magnetic flux through this loop. Φ = B da. Since B is perpendicular to the plane, it is parallel (or anti-parallel) to da, so Φ = BdA, where da = 2πrdr, B = Cr, and C = (5.00 x 10-3 T/m). The limits of integration are from 0 to R, with R = m. BdA = 2πCr 2 dr, and Φ = 2πCR 3 /3 = (1.047 x 10-2 T/m)(0.250 m) 3 = 1.64 x 10-4 T m 2 (or Wb). If B is in the opposite direction, the flux would be negative. 6. Consider the RL circuit shown to the right. ξ = 10.0 V, and R = 50.0 Ω. When the switch S is closed the current obeys the equation i = ξ(1 - e -t/τ )/R, where τ is the RL time constant. It is found that it takes 5.00 ms for the current to increase to 150 ma. What is the inductance of this coil? Unlike the RC time constant, the RL time constant doesn t equal RL, but L/R. ξ/r = A, or 200 ma. Thus, 150 = 200(1 - e -t/τ ) and 0.75 = (1 - e -t/τ ). Thus, e -t/τ = and t/τ = ln(4), or τ = t/ln(4) = (5.00 ms)/ln(4) = ms = L/R. Thus, L = Rτ = 180 m (Ω s) = 180 mh. 7. Find the magnitude of the magnetic field at a distance of 5.00 cm from a very long (effectively much much greater than 10 m) wire carrying a current of 8.00 A. Prove the validity of the equation you ve used by using Ampere s law. The first part of this is trivial (and a repeat of a part of 4). B = µ 0 i/2πr, where i = 8.00 A and r = m. Thus, B = (4π x 10-7 T m/a)(8.00 A)/[2π( m)] = (4 x 10-7 T m/a)(8.00 A)/[(0.100 m)] = 3.20 x 10-5 T. ξ S R L
4 The second part is also trivial, if you understand Ampere s law. Picking a circular path a constant distance from the wire makes the magnitude of B constant. Also, the direction of B is parallel to the differential arc vector. Thus, the integral defining Ampere s law yields 2πrB = µ 0 i, so B = µ 0 i/2πr. 8. Assume the wire in problem 7 is flat on this sheet of paper with the positive current directed toward the top of the page. What is the direction of the magnetic field on your side of the page, and directly above the wire. Explain clearly how you get the direction. Using your right hand, point its thumb in the direction of the current. Your figures curl in the direction of the magnetic field. That is, into the page to the right of the wire, to the left behind the wire, out of the page to the left of the wire, and to the right on your side of the wire. Thus, the field is to the right on your side of the wire. 9. Determine the magnetic field (magnitude and direction) at the geometric center of a square loop of wire of side L. Assume the current as viewed by you is in the clockwise direction and the loop is flat on this sheet of paper. Hint: Use the development methods of Equation Remember that there are four segments of wire, and the limits of integration won t be the same as for an infintely long wire. Using the right hand rule, the direction of B is into the page. The geometric center is a perpendicular distance a = L/2 from the center of each segment of the loop. The total field at the center is equal to four times that of what is produced by just one segment. Below is considered the field produced by the wire oriented in the x-dirction. Pick x = 0 at the center of the wire, and the limits of integration from - a to + a. ds x r is perpendicular to the page, ds x r = rdx(sinθ), where (sinθ) = a/r = a/(x 2 + a 2 ) 1/2 -> rdx(sinθ) = adx db 1 = (µ 0 i/4π)adx/r 3 = (µ 0 i/4π)adx/(a 2 + x 2 ) 3/2. This is an even function, so integrating from - a to + a gives twice the results from 0 to a. a r Refering to integral 19 on page A-11, B 1 = (µ 0 i/2π)ax/[a 2 (a 2 + x 2 ) 1/2 ] from x = 0 to x = a = L/2. x B = 4B 1 = (2µ 0 i/π)/[(a 2 + a 2 ) 1/2 ] = (2 3/2 µ 0 i/πl) Page 2 of 2
5 10. A circular loop of wire lays flat on a table in the presence of a magnetic field that is directed upward. The field is produced by an electromagnet that can be turned on or off. If the electromagnet is suddenly turned off, what is the direction of the current induced in the loop as viewed from above. Explain all your reasoning. The easy approach to this is to invoke Lenz s law. The original magnetic field is pointing upward, and the change will be downward. Lenz s law requires that the direction of the induced current will oppose the changing magnetic field (i.e. try to keep it the same). Thus, the induced field must point upward. To do that, its direction must be counter-clockwise when viewed from above. 11. Assume the loop in problem 10 has a radius of m, and the (uniform) magnetic field has its magnitude reduced from an initial value of B 0 = T according to the equation B = B 0 /(1 + at 2 ), where a = 1.00/s 2. Determine the amount of emf induced as a function of time, and that induced at t = 2.00 s. ξ = - dφ B /dt Since the field is uniform, Φ B = BA = B 0 /(1 + at 2 )πr 2 and ξ = - πr 2 B 0 d[(1 + at 2 ) -1 ]/dt = πr 2 B 0 [(1 + at 2 ) -2 ](2at) = 2πR 2 B 0 at/(1 + at 2 ) 2 ] 2πR 2 B 0 a = T m 2 /s 2, so ξ = ( T m 2 /s 2 )t/(1 + at 2 ) 2. At t = 2.00 s, ξ = ( T m 2 /s 2 )(2.00 s)/(5.00) 2 = (π/100)v 12. Consider a pair of coaxial cylindrical conductors. That is, each cylinder has its central axis along the same line. The radius of the inner cylinder is 2.50 cm, and the radius of the outer cylinder is 20.0 cm. The inner cylinder carries a current of 2.00 A in one direction, and the outer cylinder carries 10.0 A in the opposite direction. Use Ampere s law to find the magnitude of the magnetic field a distance of 10.0 cm from the central axis. Explain what you are doing. Integrating B ds around a circular path of radius r gives 2π rb. By Ampere s law, this must equal µ 0 times the total current enclosed. In this case, at a radius of 10.0 cm, only the inner current is enclosed by the loop. Therefore, 2πrB= µ 0 (2.00 A) -> B = µ 0 (2.00 A)/2πr = µ 0 (2.00 A)/2π(0.100m) = (2.00 x 10-7 )(2.00/0.100)T = 4.00 x 10-6 T. 13. Determine the energy density of the magnetic field on the interior of the solenoid in which the magnetic field is T. The energy density is given by u B = B 2 /2µ 0 = (6.25 x 10-4 T 2 )/(8π x 10-7 T m/a) = 249 J/m 3.
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