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2 1: A zero resistance rod slides to the right on zero resistance rails separated by.3 m. The rails are connected by a 5 Ω resistor. A constant magnetic field of.5 T permeates the region, coming straight out of the page. Find the speed at which the bar must be moved to produce a current of.1a in the resistor. a) 34 m/s b) *17 m/s c) 89 m/s d) 5 m/s e) 8 m/s R = 5 Ω v.3 m You know that the EMF will be EMF = dφ B /dt = BL I = EMF / R = BLv / R v = RI / BL = 16.7 m/s : Four different loops of wire will move to the right with the same velocity from a region of zero field, into a region with constant field B. Rank them from least to greatest according to the maximum magnitude of the induced EMF they will feel. Relative heights and widths are listed below each loop. a) 1 is least, then and 4 tie, and 3 is greatest b) and 3 are tied for least, 1 and 4 are tied for greatest c) 3 is least, 4 is larger, and 1 and are tied for greatest d) *3 is least, and 4 are tied and larger, 1 is greatest e) and 4 are tied for least, 3 is larger, 1 is greatest. 1 W x 3H 3W x H 3 3W x 1H 4 W x H This problem is very related to the above. The EMF in each loop will be dφ B /dt = BHv. So if they move at the same speed the determining factor will be the height. This means the EMF in 3 is least, and 4 are tied, and 1 is the most.

3 3: A small bar magnet is dropped from the same position below the centers of several different loops of wire, each of which is held fixed in place. Loop A has resistance R, loop B has resistance R, and loop C has a small break in it. Which of the following statements is correct? This is just like the practice test, except now the magnet is falling away instead of towards the loops. Lenz s law says the induced magnetic force will still resist the change. a) The magnet accelerates downward at the same rate in each case b) The magnet accelerates down away from loop C, but accelerates up towards loops A and B c) The magnet accelerates down away all loops, most rapidly from B, most slowly away from C d) The magnet accelerates down away slowly from loop A, down more rapidly from loop C, and up towards loop B e) *The magnet accelerates down away from all loops, most rapidly from C, most slowly from B A N S B S N C N S 4: Consider the circuit shown, which contains a 6 V battery, a 5 Ω resistor, a light bulb with resistance 5 Ω, and a 1 H inductor. What is the current through the bulb long after the switch is closed? Assume the inductor has zero resistance. a) 1. A b).4 A c).6 A d) *Zero e) 1.6 A Long after the switch is closed, current runs freely through the inductor. Since it has zero impedance, and the light bulb has 5 Ω of impedance, all the current runs through the inductor and none through the bulb. 6 V 5 Ω 5 Ω 1 H

4 5: Most contemporary high voltage power lines operate at 38, V RMS. There are efforts to build new power lines which operate at 76, V RMS. If we switch to these new lines, keeping the resistance of the cables themselves the same and still transmitting the same total power, how will this affect power transmission? a) Current in the lines will increase by a factor b) Ohmic energy losses in the wires will decrease by a factor of c) Ohmic energy losses in the wires will increase by a factor of d) *Ohmic energy losses in the wires will decrease by a factor of 4 e) Ohmic energy losses in the wires will remain the same If we increase the voltage by a factor of two, we cut the current by a factor of two. This reduces losses in transmission by a factor of 4 (I R) 6: The circuit shown is in a uniform magnetic field which points into the page. The magnitude of this field is decreasing at a rate of 15 T/s. What is the current in this loop while this is happening? a) *.18 A b). A c).4 A d).6 A e).34 A 1 cm 1 cm 1 Ω There is an EMF in this loop caused by the change in magnetic flux: EMF = dφ B / dt =.1 *15 T/s =.16 V The direction is clockwise, to resist the change in flux. Putting this together with the 4V which are there, we have a total EMF of 1.84 V, and a current of.18 A. 4 V

5 7: A resistor, an inductor and an initially charged capacitor are wired together in series. Charge on the capacitor then oscillates and gradually drains away. We could slow down the exponential decay of the maximum charge on the capacitor plates by: a) Decreasing the inductance b) Increasing the resistance c) *Decreasing the resistance d) Increasing the capacitance e) Decreasing the capacitance The decay here is governed by the L/R decay time, so we could slow this by either increasing the inductance or decreasing the resistance. 8: An AC generator producing a maximum voltage of 1 V at an angular frequency of rad/s is connected in series with a 5 Ω resistor, a 4 mh inductor, and a μf capacitor. What is the peak current in this circuit? a).15 A b) *.135 A c).18 A d). A e).4 A Impedance is Z = (R + (ωl 1/ωC) ) 1/ = 74.3 Ω Maximum current is 1 V / 74.3 Ω =.135 A

6 9: In an AC circuit, a.5 H inductor is connected to a generator that has an RMS voltage of 5 V and operates at 5 Hz. What is the RMS current through the inductor? a).6 A b). A c) *3. A d) 7.1 A e) 14 A First find the angular frequency: ω = πf = 314 rad/s Then find impedance: Z = ωl = 7.85 Ω Then find current: I RMS = V RMS / Z = 3. A 1: The electric field within the circular region of radius r shown in the picture points out of the page, and is increasing in magnitude according to the equation: E = C + Dt where C and D are constants. What is the magnitude of the magnetic field at the point P, a distance R (which is greater than r) from the center of the circle? a) C B = μ ε r b) D B = μ R c) R B = μ ε D r r d) * B = μ ε D R e) D B = μ ε R Here we have a displacement current producing the magnetic field, so we use the relation: r r dφ E d B dl = με = με [( C + Dt ) πr ] dt dt B πr = μ ε Dπr ( ) r B = με D R R P r

7 11: A current sheet is constructed by flowing current uniformly through a thin sheet of metal aligned with the x-axis. The sheet is m wide and 1 m long and carries a total current of 3. A. Estimate the magnitude of the magnetic field a distance of 3 cm above the center of the plate. a) 3.3x1-8 T b) 5.1x1-4 T c) *9.4x1-7 T d) 6.9x1-11 T e) 8.9x1-5 T I y x z 1 m m What you need to see here is that when you are close to the plate, it looks infinite. When it is infinite, you know the magnetic field will be constant and in the +z direction above the plate and the z direction below the plate. Now draw a rectangular loop above and below the plate, and use Ampere s law: r r B dl = μ i enclosed = μ I B = μ = 9.4x1 W 7 A I W L BL = μ I W L 1: A metal rod with length L and electrical resistance R moves through a constant uniform magnetic field B coming out of the page. What force must be applied to the rod to keep it moving at a constant velocity v? a) * b) BLv c) BLv/R d) B L v/r e) B L v /R Once this rod is in motion it requires no force to keep it in motion. F? v

8 13: A single circuit element is connected to an AC generator. It is driven at an angular frequency of 4 rad/s. The voltage across and current through this circuit are shown in the graph. Which of the following statements about the circuit element is true? a) It must be a capacitor with C =.5x1-5 F b) *It must be an inductor with L =.5x1-3 H c) It must be a resistor with R = 1 Ω d) It must be a capacitor with C = 4x1-7 F e) It must be an inductor with L = 4x1-5 H In this circuit you have voltage leading current, so you re dealing with an inductive circuit. To find the inductance, note that: I max = V max / Z or Z = V max / I max = 1 Ω ωl = 1 Ω L =.5x1-3 H 14: An 8. mh inductor and a. Ω resistor are wired in series to an ideal battery. A switch in the circuit is closed at time t =, at which time the current is zero. The current reaches half its final value at time: a) *.8 ms b) 4. ms c).5 ms d) 17 μs e) 3 s In this case you have i(t) = i final (1 e -Rt/L ) and you want to know when i(t) =.5i final. So you have:.5 = 1 e -Rt/L.5 = e -Rt/L ln(.5) = -Rt/L t = -(L/R)*ln(.5) = -(8.x1-3 /.)*ln(.5) =.8 ms

9 15: In an LC circuit in which oscillations of charge and current are occurring, at the moment when the current is zero: a) the charge on the capacitor is zero. b) *the energy in the electric field is a maximum. c) the energy in the magnetic field is a maximum. d) the charge is moving through the inductor. e) the energy is shared equally between electric and magnetic field. When the current is zero the capacitor is has its maximum charge, and the energy in the electric field is maximum. 16: A circular coil has 75 turns and a radius of.45 m. The coil is used as an AC generator by rotating it in a.5 T magnetic field. At what angular speed should the coil be rotated to achieve a maximum EMF of 175 V? a) 8 rad/s b) 5 rad/s c) 13 rad/s d) * rad/s e) 49 rad/s EMF = -dφ B / dt = -d/dt(n turns *A*B*cos(ωt)) = N turns ABωsin(ωt) The maximum occurs when sin(ωt) = 1, so we have 175V = 75 turns * π * (.45 m) *.5 T * ω or ω = 175V / (75 turns * π*(.45 m) *.5T) = rad/s

10 17: The figure below shows the total impedance of a series RLC circuit as a function of the driving angular frequency ω. What can you say about the reactance of the resistor R in this circuit? a) Its value depends on frequency ω b) It dominates the impedance at low frequency c) It dominates the impedance at high frequency d) Its value is 5 Ω e) *Its value is 1 Ω When you run this circuit at resonance, the impedance reaches its minimum value, which is the resistance R. So for this, the resistance R = 1 Ω Impedance Z (in Ω) Angular frequency ω 18: In the same circuit described in the previous problem, at which of the following driving frequencies does the current lead the voltage by the largest amount? a) *1 rad/s b) 356 rad/s c) 5 rad/s d) 78 rad/s e) 18 rad/s Current will lead voltage when the capacitive reactance is bigger than the inductive reactance. This happens at frequencies below the resonant frequency. For this case, the resonant frequency is 5 Hz (where you see the minimum in the impedance). So the current will lead by the most when you have the lowest frequency.

11 19: In class we viewed a demonstration in which a small magnet was slid down two different slabs, each inclined to the same slope. We saw that the magnet accelerated rather freely down the insulating slab, but slid at a slow and constant rate down the nonmagnetic aluminum slab. Which of the following statements is correct? a) The magnet slid freely down the insulating slab because no EMF was induced in the insulating slab b) The magnet more slowly down the metal slab because of the larger surface friction between the magnet and the metal slab c) *The magnet slid at a constant rate down the metal slab because it was subject to a velocity dependent force caused by eddy currents in the metal slab d) The magnet slid at a constant rate down the metal slab because it was subject to a velocity independent force caused by eddy currents as it slid down the slab e) The magnet slid rapidly down the insulating slab it was subject to a downward force caused by eddy currents The same EMFs are produced no matter what the magnet slides over, but that EMF produces larger currents when it slides over a conductor, like the metal. The size of the force created by the eddy currents depends on how rapidly the magnetic flux through parts of the metal is changing. This means it is a velocity dependent force. You can be sure it is velocity dependent, because this is what causes it to reach a terminal velocity. It accelerates downward until the eddy current generated force balances the downward force of gravity, after which it travels at a constant speed. : Two closed loops lie near one another on a table. Each loop has a total resistance of 15 Ω. When the current in loop A is increased smoothly from zero to. A over an interval of 3 ms, a current of ma is induced in loop B. What is the mutual inductance of these two loops? a) 6.x1-3 H b) *5.x1 - H c).9x1-4 H d) 7.1x1-5 H e) 3.1x1-7 H EMF B = M (di A / dt) = I B R B Here we have di B / dt =. A /.3s = 66.7 A/s M * 66.7 A/s =. A * 15 Ω M =. A * 15 Ω / 66.7 A/s = 5.x1 - H

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