Physics 9 Fall 2009 Homework 8 - Solutions

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1 1. Chapter 34 - Exercise 9. Physics 9 Fall 2009 Homework 8 - s The current in the solenoid in the figure is increasing. The solenoid is surrounded by a conducting loop. Is there a current in the loop? If so, is the loop current cw or ccw? The magnetic field in the solenoid points to the right, by the right-hand rule for currents. The flux is increasing, making the magnetic field bigger. There is a current induced in the loop in such a way that it resists the change in flux. This means that the the magnetic field created by the loop has to point to the left. This requires a current moving in the counter-clockwise direction. 1

2 2. Chapter 34 - Exercise 13. A 1000-turn coil of wire 2.0 cm in diameter is in a magnetic field that drops from 0.10 T to 0 T in 10 ms. The axis of the coil is parallel to the field. What is the emf of the coil? We know that the induced emf in a coil of N turns is E = N dφ m dt, where Φ m = B A = BA cos θ is magnetic flux. Since the area is constant, then E = NA Ḃ, where a dot over a quantity denotes the time derivative. For the coil, A = πr 2, and Thus, E = πr 2 N B t = π (0.01)2 (1000) = 3.14 V. Ḃ = B t. 2

3 3. Chapter 34 - Exercise 21. How much energy is stored in a 3.0 cm-diameter, 12 cm-long solenoid that has 200 turns of wire and carries a current of 0.80 A? The energy stored in an inductor is E = 1 2 LI2, where L is the inductor. Now, for a solenoid, L = µ 0N 2 A, where A = πr 2 is the area of the solenoid, and l is the length. l So, E = 1 2 LI2 = µ 0N 2 A I 2. 2l So, this solenoid has an energy E = µ 0N 2 A I 2 = (4π 10 7 ) π (0.015) 2 (0.8) 2 = J. 2l

4 4. Chapter 34 - Problem 34. A circular loop made from a flexible, conducting wire is shrinking. Its radius as a function of time is r = r 0 e βt. The loop is perpendicular to a steady, uniform magnetic field B. Find an expression for the induced emf in the loop at time t. For a given flux, Φ m = B A = BA (t), in this case, then the emf is E = d Φ dt m = BA, where the dot denotes the time derivative, as we have seen before, A da. The area dt of the loop is A (t) = πr 2 (t) = πr0e 2 2βt, and so A = 2βπr0e 2 2βt. Thus, E = 2βπr 2 0 Be 2βt = 2βπr 2 0 Be 2βt. 4

5 5. Chapter 34 - Problem 43. A loop antenna, such as is used on a television to pick up UHF broadcasts, is 25 cm in diameter. The plane of the loop is perpendicular to the oscillating magnetic field of a 150 MHz electromagnetic wave. The magnetic field through the loop is B = (20nT) sin ωt. (a) What is the maximum emf induced in the antenna? (b) What is the maximum emf if the loop is turned 90 to be perpendicular to the oscillating electric field? (a) The area of the loop is fixed, and perpendicular to the magnetic field, so Φ m = B A = BA, and so the emf is E = dφ m dt = A Ḃ, where B = B 0 sin (ωt). Thus, E = πr 2 ωb 0 cos (ωt). The cosine varies between ±1, and has a maximum at 1. So, E max = πr 2 ωb 0 = 2π 2 r 2 fb 0, since ω = 2πf. Thus, E max = 2π 2 ( ) 2 ( ) ( ) = 0.93 V. (b) Since the electric and magnetic fields are perpendicular to each other, and mutually perpendicular to the direction of propagation of the wave, when the plane of the loop is perpendicular to the oscillating electric field, then it is parallel to the magnetic field. So, the net flux through the antenna is zero, which gives a net induced emf of zero. 5

6 6. Chapter 34 - Problem 52. You ve decided to make a magnetic projectile launcher for your science project. An aluminum bar of length l slides along metal rails through a magnetic field B. The switch closes at t = 0 s, while the bar is at rest, and a battery of emf E bat starts a current flowing around the loop. The battery has internal resistance r. The resistance of the rails and the bar are effectively zero. (a) Show that the bar reaches a terminal speed v term, and find an expression for v term. (b) Evaluate v term for E bat = 1.0 V, r = 0.10Ω, l = 6.0 cm, and B = 0.50 T. (a) When the switch closes, the battery causes a current to flow. This causes the bar to move since the current is moving in a magnetic field. The moving bar changes the flux in the loop which induces a new emf that fights the battery s emf. This builds up as the velocity gets faster until the induced emf is equal to that of the battery. Then, the net emf is zero and the bar coasts along at the terminal velocity, v term. So, E = lbv term. This gives (b) For the given numbers, This is about 70 miles per hour! v term = E lb. v term = E lb = = 33 m/s. 6

7 7. Chapter 34 - Problem 53. A slide wire of length l, mass m, and resistance R slides down a U-shaped metal track that is tilted upward at angle θ. The track has zero resistance and no friction. A vertical magnetic field B fills the loop formed by the track and the slide wire. (a) Find an expression for the induced current I when the slide wire moves at speed v. (b) Show that the slide wire reaches a terminal speed v term, and find an expression for v term. The setup is as seen in the diagram below. (a) Here, all we are doing is tilting our usual current loop. This changes the flux to Φ m = B A = BA cos θ. The induced current is I = E/R, where R is the resistance of the wire. So, I = E R = 1 R Φ m = B R cos θȧ = B R cos θlẋ = Blv R cos θ, where we set v = ẋ. Thus, I = Blv cos θ. R (b) The induced force on the bar is back to the left, pushing the bar up along the loop (along x as we ve drawn things), while gravity pulls the bar down (along the +x direction). The bar coasts along at a constant speed when the two forces are equal to each other. Looking at the forces along the x direction gives Fx = F m cos θ + F g sin θ = 0. Now, F m = IlB = B2 l 2 v term cos θ, while F R g = mg. Setting the two equal and solving for the terminal velocity gives v term = mgr l 2 B 2 tan θ cos θ. 7

8 8. Chapter 34 - Problem 65. MRI (magnetic resonance imaging) is a medical technique that produces detailed pictures of the interior of the body. The patient is placed into a solenoid that is 40 cm in diameter and 1.0 m long. A 100 A current creates a 5.0 T magnetic field inside the solenoid. To carry such a large current, the solenoid wires are cooled with liquid helium until they become superconducting (no electric resistance). (a) How much magnetic energy is stored in the solenoid? Assume that the magnetic field is uniform within the solenoid and quickly drops to zero outside the solenoid. (b) How many turns of wire does the solenoid have? (a) The energy density stored in the magnetic field is given by u B = 1 2µ 0 B 2. The total energy is the energy density times the volume of the solenoid, U B = u B (Vol) = u B Al = B2 Al 2µ 0. So, U B = Al 2µ 0 B 2 = (b) The magnetic field of a solenoid is B = µ 0NI l 100 A, we have or about 40,000 turns! π (0.2 2 ) 2 4π 10 7 (5.0)2 = J. N = Bl µ 0 I = 5 1 4π ,, and so N = Bl µ 0. For a current of I 8

9 9. Chapter 34 - Problem 66. One possible concern with MRI (see problem 65) is turning the magnetic field on or off too quickly. Bodily fluids are conductors, and a changing magnetic field could cause electric currents to flow through the patient. Suppose a typical patient has a maximum cross-section area of m 2. What is the smallest time interval in which a 5.0 T magnetic field can be turned on or off if the induced emf around the patient s body must be kept to less than 0.10 V? The emf is E = Φ. m For a constant area, oriented along the direction of the magnetic field, then Φ m = AB. So, Φm = AḂ A B A B. So, the amount of time, t. t E Plugging in the numbers give t = A B E = = 3 sec. So, we should turn the magnetic field on or off over a time interval longer than 3 seconds. 9

10 10. Chapter 34 - Problem 80. The switch in the figure has been open for a long time. it is closed at t = 0 s. (a) After the switch has been closed for a long time, what is the current in the circuit. Call this current I 0. (b) Find an expression for the current I as a function of time. Write your expression in terms of I 0, R, and L. (c) Sketch a current-versus-time graph from t = 0 s until the current is no longer changing. (a) From Kirchhoff s loop law, V = V batt IR L di = 0. When the current dt has been flowing for a long time, it no longer changes, and so I = 0. Thus, I 0 = V batt. R (b) Now, going back to the loop law, we have, upon solving for di, dt di dt = R L (I 0 I), after plugging in V batt = I 0 R. Separating this differential equation becomes = R dt. Now, integrating gives L di I I 0 I I 0 di I I 0 = R L t 0 ( ) I I0 dt ln = R I 0 L t. Solving for I (t) gives where τ L/R. I(t) = I 0 ( 1 e t/τ ), (c) The graph is seen to the right. The current starts out at zero at time t = 0, and asymptotically approaches the final current I 0 as t. 10

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