Physics 102 Conference 8 Magnetic Flux Conference 8 Physics 102 General Physics II Monday, March 24th, 2014 8.1 Quiz Problem 8.1 Suppose we want to set up an EMF of 12 Volts in a circular loop of wire (radius = 1 m) by changing a magnetic field that bathes the wire (the magnetic field points into the page with magnitude B(t) as in Figure 8.1). Assuming we want the twelve volt EMF over a time period of 10 s, what linear function of time should we use to modify the magnetic field if we want current to flow clockwise through the loop? B = B(t) ˆx ẑ I ˆx ŷ Figure 8.1: A magnetic field goes through a circular loop of wire. The magnetic field s magnitude changes linearly in time, so there is a time-varying flux through the loop, and hence an EMF. If we set the magnetic field to have generic linear time dependence, B(t) = B 0 α t, then the flux through the circular loop of wire is: Φ(t) =(B 0 α t) π 2 (8.1) 1 of 8
8.1. QUIZ Conference 8 and the EMF generated by the temporal dependence of the field is: E = dφ(t) If we want this to equal a particular value, then: and our magnetic field is: = π 2 α. (8.2) π 2 α = V 0 α = V 0 π 2 (8.3) B(t) = B 0 V 0 t. (8.4) π 2 To ensure that we can run the circuit for T = 10 s, we must start with B 0 = V 0 π 2 T (at least), so: B(t) = V 0 (T t). (8.5) π 2 If we insert the target quantities: B(t) = 12 V 2 (10 s t). (8.6) π (1 m) Note that we took B(t) decreasing to get current flowing clockwise (Lenz s law). 2 of 8
8.1. QUIZ Conference 8 Problem 8.2 A rectangle has side lengths that sum to 2 L the rectangle has equal side length 1 2 L at t = 0. We pull apart the rectangle with constant speed until the rectangle has two sides of length L/4 and two with 3 L/4 at time T. What current runs through the loop if it has net resistance and sits in a constant magnetic field pointing into the page with magnitude B 0? B0 into page B0 into page L/2 L/4 L/2 3 L/4 Call the horizontal length of the rectangle l(t), and the height h(t) we know that l(t) + h(t) = L so that h(t) = L l(t). We are also told that the horizontal length increases with constant speed, call it v: l(t) = L/2 + v t, and l(t ) = 3 L/4, so that ( L/2 + v T = 3 L/4 and v = L 4 T. The flux is Φ(t) = B 0 l(t) h(t) = B 0 L/2 + L 4 T t) ( L/2 L 4 T t). Then: E = dφ(t) ( ) L 2 = 2 B 0 t, (8.7) 4 T and I = E so that I = 2 B ( 0 L ) 2 4 T t the flux was calculated using an oriented boundary circulating in the clockwise direction, and the positive current here tells us that I goes in the same direction. Can also get from Lenz s law the area in the field is decreasing, so to oppose the loss of flux into the page, the wire loop wants to generate a magnetic field pointing into the page, hence current is clockwise. 3 of 8
8.2. FAADAY S LAW Conference 8 8.2 Faraday s Law Problem 8.3 A pair of conducting rails are separated by a distance d, with a uniform magnetic field B = B 0 ˆx between them. A conducting bar (of mass M) is free to slide along the rails if the bar travels to the right with initial speed v 0, find v(t), the speed of the bar as a function of time. What is the power loss of kinetic energy of the bar as a function of time? (Assume that a resistance acts in the circuit, as shown below) B = B 0 ˆx d ẑ y(t) ˆx ŷ Figure 8.2: A pair of conducting rails separated by a distance d. A bar completes the circuit, and the entire configuration is in a constant magnetic field. The movement of the bar changes the size of the circuit, and hence the magnetic flux through it the flux is given by: Φ(t) = B 0 d y(t) (8.8) where y(t) is the location of the bar at time t. The EMF generated is: so the current in the circuit is: E = dφ(t) = B 0 d v(t), (8.9) I(t) = E = B 0 d v(t), (8.10) where the sign is positive, indicating that current flows in the counterclockwise direction. Now, the bar carries this current, and is therefore a wire with current I moving in a magnetic field, so there is a force on the bar given by: F = d I B = d I(t) B 0 ŷ. (8.11) 4 of 8
8.2. FAADAY S LAW Conference 8 From Newton s second law, we have: M dv(t) B 0 d v(t) = d B 0, (8.12) or dv(t) = B2 0 d2 v(t). (8.13) M The solution to this differential equation is a decaying exponential: v(t) = A e B 2 0 d2 M t (8.14) for arbitrary A since we are given the initial speed, we have: v(t) = v 0 e B 2 0 d2 M t. (8.15) 5 of 8
8.2. FAADAY S LAW Conference 8 Problem 8.4 A time-varying magnetic field (with magnitude B(t) = α e β t ) points into the page. With what velocity v(t) must you pull a loop of wire to the right such that no current flows through the resistor? Assume at t = 0, half of the loop is in the magnetic field region. w B(t) = e t into page v(t) =? ` x(t) The total flux is: Φ(t) = B(t) l (w x(t)), and then the EMF is: E = dφ(t) = Ḃ(t) l (w x(t)) + B(t) l dx = α β e β t l (w x) + α e β t l dx. (8.16) We want E = 0 to keep current from flowing. Then: β (w x) = dx, and take z w x, then dz = β z z(t) = z 0 e β t, (8.17) and x = w z = w z 0 e β t with x(0) = w/2 giving z 0 = w/2 so x(t) = w ( 1 1 2 e β t) and v(t) = dx = w 2 β e β t. (8.18) 6 of 8
8.2. FAADAY S LAW Conference 8 The new component of Maxwell s equations is the sourcing of an electric field by a time-varying magnetic field: E dl = dφ(t). (8.19) This equation has interesting physical content (the change in flux through an arbitrary surface is related to the integral of E around the boundary of that surface), and its form is similar to another one of Maxwell s equations: B dl = µ 0 I enc. (8.20) In the next problem, you will find the electric field generated by a time-varying magnetic field, use your method from Ampere s law to help solve (8.19) by analogy with your solution to (8.20) for an infinite current carrying wire. You can check your result using Faraday s law of induction. Problem 8.5 An infinite solenoid of radius with n turns per unit length has slowly, linearly increasing, current I(t) = α t in each turn. Find the electric field outside of the solenoid. Notice that while there is no magnetic field outside the solenoid (as far as we have been concerned), you could detect the presence of a time-varying magnetic field inside the solenoid by watching a charged particle outside the solenoid. The magnetic field inside the solenoid is B = µ 0 n I(t) ẑ. Thinking about the relevant equation for E, we have: E dl = d ( ) B da. (8.21) The geometry of this setup is identical to an infinite wire sourcing a magnetic field so we use the same Amperian loop and ansatz. The electric field encircles the solenoid (outside), and has magnitude that depends only on r, the distance from the solenoid s center. Consider a circular loop of radius r centered at the center of the solenoid. Then the left-hand side of (8.21) reads (using our assumptions about the form of E): E dl = E(r) 2 π r, (8.22) and the right-hand side is: dφ(t) = d ( B(t) π 2 ) = π 2 µ 0 n di(t). (8.23) 7 of 8
8.2. FAADAY S LAW Conference 8 We can solve (8.21) for E(r): E(r) 2 π r = π 2 µ 0 n di(t) E(r) = µ 0 n α 2. (8.24) 2 r The electric field circulates in a direction opposite the current flow. The integral of this electric field, around a closed circular loop of radius r is E = µ 0 n α π 2, which is precisely what we would have gotten from Faraday s law. 8 of 8