# Solutions to PHY2049 Exam 2 (Nov. 3, 2017)

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1 Solutions to PHY2049 Exam 2 (Nov. 3, 207) Problem : In figure a, both batteries have emf E =.2 V and the external resistance R is a variable resistor. Figure b gives the electric potentials V between the terminals of each battery as functions of R; Curve corresponds to battery, and curve 2 corresponds to battery 2. The horizontal scale is set by R s = 0.20 Ω. What is the internal resistance of battery? Problem Solution: This problem was based on 27.4 from the text, which was assigned homework. Let us call the internal resistances r and r 2. All the batteries and resistors are in series so the current is, i = 2E R + r + r 2. This means that the voltages across the terminals of the two batteries are, V = E ir = ( R r + r 2 R + r + r 2 ) E, V 2 = E ir 2 = ( R + r r 2 R + r + r 2 ) E. Note that at R = 2 R s we have V 2 = 0. This implies, r 2 = 2 R s + r. Substituting this into the expression for V, which is 3 E at R = 2 R s, implies, R s R s + 2r = 3 = r = R s. Problem 2: In the figure, the ideal batteries have emfs E = 0 V and E 2 = 5 V, and the resistances are each 4 Ω. What is the current in R 2? Problem 2 Solution: This problem was based on from the text. Suppose we use i to label the clockwise current in the left hand loop, and i 2 to

2 label the counter-clockwise current in the right hand loop. Then Kirchhoff s Loop Law implies, E Ri R(i + i 2 ) = 0, 2 E Ri 2 R(i + i 2 ) = 0. Adding the two equations together gives, 3 2 E 3R(i + i 2 ) = 0 = R(i + i 2 ) = 2 E. Substituting this into the second equation gives, 2 E Ri 2 2 E = 0 = i 2 = 0. Problem 3: In the figure, two batteries each with emf E = 2 V and an internal resistance r = 0.30 Ω are connected in parallel across a resistance R. For what value of R is the dissipation rate in the resistor a maximum? Problem 3 Solution: This problem was based on from the text, which was assigned homework. By symmetry, the current through each battery is the same; let us call it i. This means the current through R is 2i, and Kirchhoff s loop law implies, E 2iR ir = 0 = i = The power dissipated in R is, P = i 2 R = RE 2 (2R + r) 2. E 2R + r. Differentiating with respect to R and setting the derivative to zero gives R = 2 r. Problem 4: In the figure, R = 0 kω, R 2 = 5 kω, C = 0.4 µf, and the ideal battery has emf E = 20 V. First, the switch is closed a long time so that the steady state is reached. Then the switch is opened at time t = 0. What is the magnitude of the current in R 2 at time t = 4 ms? 2

3 Problem 4 Solution: This problem was based on from the text, which was assigned homework. First, consider the long term situation with the switch closed. No more current goes into the capacitor after it has fully charged, so the circuit essentially reduces to R and R 2 in series. This means the current through R 2 and the voltage drop across it are, i = E R + R 2 = 0.8 ma = V 2 = ir 2 = 2 V. This is also the voltage drop across the capacitor, hence it carries a charge of Q max = 4.8 µc. When the switch is opened the circuit consists of just R 2 and C, and the charge on the capacitor and the current through R 2 are, Q(t) = Q max e t/r 2C = i(t) = Q max R 2 C e t/r 2C. The time constant is τ = R 2 C = 6 ms so the current at t = 4 ms is, i(4 ms) = 0.8 ma e 2/3 0.4 ma. Problem 5: A beam of protons with a velocity of v = ĵ m/s is sent into a region of uniform magnetic field of B = î T. What electric field in N/C is necessary (in magnitude and direction) such that the protons continue in a straight line without deflection by the magnetic field? Problem 5 Solution: The direction of the force caused by the B-field on the protons is ĵ î = ˆk. So the E-field has to point into the opposite direction. The magnitude has to be equal to E=vB=3000 N/C. So the answer is: E = 3 0 3ˆk N/C? Problem 6: An ion of charge +4e is sent into a region with constant magnetic field of magnitude B=0.25 T perpendicular to the plane of motion of the ion. The charge makes a U-turn. What is its mass in kg if it exits the region after a time t = s? Problem 6 Solution: The force caused by the magnetic field has to be equal to the centrifugal force: m v2 r = qvb r = mv qb 3 ()

4 We still need to replace the velocity which we can get from the pathlength and the time: πr = vt v = πr (2) t So we have r = mπr qbt m = qbt π = kg (3) Problem 7: A particle with a negative charge propagates through a uniform magnetic field. At time t=0 it has a velocity of v = 3.0î+2.0ĵ and experiences a force into the positive ˆk-direction. Which of the following directions could be correct for the magnetic field? Problem 7 Solution: The magnetic field component in the ˆk direction has to be zero otherwise the force would not point into the ˆk direction. Because the charge is negative, ˆv ˆB should point into the ˆk direction. Now draw a vector showing v and go through the three remaining possibilities for the cross product. From the given possible solutions only a B-field that points into the ĵ direction creates a cross product with v that points into the ˆk direction. Problem 8: The figure shows a wood cylinder of mass m = 0.25 kg and length L = 0.0 m, with N = 0 turns of wire wrapped around it longitudinally, so that the wire coil contains the long central axis of the cylinder. The cylinder is released on a plane inclined at an angle θ to the horizontal, with the plane of the coil parallel to the incline plane. If there is a vertical magnetic field of magnitude 0.50 T, what is the least current i through the coil that keeps the cylinder from rolling down the plane? Note that the friction between the wood and the plane keeps the cylinder from sliding. Problem 8 Solution: This problem was based on 28.5 from the text, which was assigned homework. The magnetic dipole moment of the circuit is µ = N 2RL i. The cylinder is subject to a gravitational torque, which pushes it to roll down, and a magnetic torque, which pushes it to roll up, τ grav = mgr sin(θ), τ mag = µb sin(θ) = 2NLiBR sin(θ). 4

5 For the cylinder not to roll we must have τ grav = τ mag. Cancelling the factor of R sin(θ) and solving for the current gives, i = mg 2NLB = 2.45 A. Problem 9: A current is set up in a wire loop consisting of a semicircle of radius 4.0 cm, a smaller concentric semicircle, and two radial straight lengths, all in the same plane. Figure a shows the arrangement but is not drawn to scale. The magnitude of the magnetic field produced at the center of curvature is µt. The smaller semicircle is then flipped over (rotated) until the loop is again entirely in the same plane (figure b). The magnetic field produced at the (same) center now has magnitude 5.75µT. What is the radius of the smaller semicircle? Problem 9 Solution: This problem was based on 29.8 from the text, which was assigned homework. Let us call the current through the two circuits i, and the two radii R big = 4 cm and R small. The magnetic fields of the two circuits are, B a = µ [ 0i + ] 4 R small R big, B b = µ [ 0i ]. 4 R small R big Most of the unknown factors cancel out when we take the ratio of the absolute values, Solving for R small gives, B b B a = R small R big R small + ρ. R big R small = ( ρ) Rbig = 2 cm. + ρ Problem 0: In the figure, five long parallel wires in an xy plane are separated by distance d = 50 cm. The currents into the page are i = 2.0 A, i 3 = 0.25 A, i 4 = 4.0 A, and i 5 = 2.0 A; the current out of the page is i 2 = 4.0 A. What is the magnitude of the net force per unit length acting on wire 3 due to the currents in the other wires? 5

6 Problem 0 Solution: This problem was based on from the text, which was assigned homework. Recall that parallel currents attract, while opposite ones repel. Also the force per unit length of two currents i a and i b separated by a distance r is µ 0 i a i b /2πr. Hence the force per until length on wire 3 is, F l = µ ( 0i 3 2πd 2 i + i 2 + i i ) 5 = 0.8 µn/m. Problem : Each of the four wires in the figure carries a current of magnitude i either into or out of the page. What is the value of the line integral B d s along the line and in the direction shown in the figure? Problem Solution: This problem was based on from the text. This problem tests Ampere s law. Because the red circle is directed clockwise, the positive direction is into the page. This means that the two wires out of the page contribute 2i, whereas the wire into the page contributes +i. Hence µ 0 i enc = µ 0 i. Problem 2: In the figure, current i = 56 ma is set up in a loop having two radial lengths and two semicircles of radii a = 5.7 cm and b = 9.4 cm with a common center P. What is the magnitude and direction (into or out of the page) of the loop s magnetic dipole moment in units of 0 3 Am 2? Problem 2 Solution: This problem was based on from the text, which was assigned homework. Because the current moves clockwise, the magnetic dipole is directed into the page. Its magnitude is, µ = Ai = π 2 [ a 2 + b 2] i. 0 3 Am 2. Problem 3: The figure shows two circular regions; region of radius r = 5 cm and region 2 of radius r 2 = 0 cm. separated by 60 cm. The magnetic field in region is 50mT coming out of the plane of the page, and that in region 2 is 30mT coming also out of the plane of the page. The magnitudes of both fields are decreasing at a rate of 5mT/s. Calculate the path integral E d s in mv along the path drawn in the figure; ignore the bending of the path at the crossing. Note: The direction of the path integral matters. 6

7 Problem 3 Solution: The first step to solving this is to recall that the path integral over a closed path over the electric field is equal to the change in magnetic flux through the closed path. The second step is to remember that this depends on the direction of the change in flux but also on the direction through the loop: Ed s dφ B = = A db db = πr2 (4) dt dt dt In this particular case, the left part will induce an E-field which rotates counterclockwise around the decreasing magnetic flux. This is along the direction of integration and the result will be positive. In the right half, the E-field rotates also counterclockwise to the decreasing magntic flux which is against the direction of integration. So this part will be negative. Ed s db ( ) = π r 2 dt r2 2 = 0.2 mv (5) Problem 4: Current i flows through the circular loop of wire in the center, as shown in the figure. The direction of the current is counterclockwise, when viewed from above. (The grey bars, which indicate the central axes of the loops, have been added to help you visualize the geometry of the problem.) If this current increases with time, what are the directions of the currents i 2, i 3, i 4 induced in the three adjacent circular loops of wire? Loop 3, whose current is i 3, is located above loop, whose current is i. In the answers, cw stands for clockwise, and ccw for counterclockwise, both when viewed from above or from the right, and 0 means no induced current. Problem 4 Solution: The current i creates a magnetic dipole field in which the magnetic field lines go up from the first loop, up through the third loop and then bend around to go down through loop 4 and the benc again to close in loop. The lines will be tangential to loop 2. If the current in loop one increases, the flux through loop 3 increases and the induced current has to go in the opposite direction from i, so clockwise has to be the second answer. For the same reason the induced current in loop four has to be ccw. There is no flux through loop 2, so that loop doesn t care and shows no induced current. Correct answer: 0, cw, ccw Problem 5: The rod in the figure is moving at a constant speed in a direction perpendicular to a 0.60 T magnetic field, which is directed into the 7

8 paper. The rod has a length of.6 m and has negligible electrical resistance. The rails also have negligible resistance. If the power dissipated by the light bulb is 4 W and its resistance is 00 Ω, what is the speed of the rod in m/s? Problem 5 Solution: From the power dissipation and the resistance, we can get the induced emf. From the emf, we get the change in magnetic flux which is caused by the change in area. In math: P = emf 2 emf = V = P R = 400V = 20V (6) R The change in magntic flux here is: dφ B dt = B da dt = B L v = emf v = emf BL = 20 V 0.6 T.6 m = 2 m/s Problem 6: Suppose the emf of the battery in a serial RL circuit varies with time so that the current is given by i(t) = 0.5 t where i is in A and t is in seconds. Assume R =.0 Ω and L = 6 H. What is the emf at t = 0 s? Problem 6 Solution: KVL gives with R, and L given and we have: emf = V (t) = Ri + L di dt i(0 s) = 5 ma di dt = 0.5A s (7) (8) (9) V (0 s) = Ω 5A A s = 8 V (0) Problem 7: An LC circuit has a capacitance of 0 µf and an inductance of 20 mh. At time t 0 = 0 the charge on the capacitor is 20 µc and the current is 80 ma. The maximum possible current in ma is. Problem 7 Solution: The easiest way to find the solution is energy conservation. At time t 0, the energy is split between the cap and the inductor: U C = q2 2C = ( C) 2 20µF = 20 µj U L = 0.5 L i 2 = 0 mh (80 ma) 2 = 64 µj 8 ()

9 The total energy is then 84 µj which we can use to calculate the maximum current: 2Utot 68 µj i max = L = 20 mh = A = 92 ma (2) Problem 8: The amplitude of oscillations at frequency 0 6 Hz in an RLCcircuit decreases by a factor of two in ms. If one doubles the resistance R, while keeping L and C the same, how long does it take now for the amplitude of oscillations to decrease by a factor of two? Problem 8 Solution: The damping constant is γ = R/2L. We double the damping, we damp twice as fast. So 0.5 ms is the correct answer. Problem 9: A.5 mh inductor in an oscillating LC circuit stores a maximum energy of 0 µj. If the resonance frequency of the circuit is khz, what is the maximum voltage in V across the capacitor? Problem 9 Solution: Energy is conserved. So at one point all the energy has to be in the cap. Once we know C, we can then get the voltage from the energy. To calculate C we use L and f 0 : f 0 = E Cap = V Q 2 = V 2 C 2 2π C = = 7 µf (3) LC 4π 2 f0 2 L 2ECap V = C = 2EL =. V (4) C Problem 20: Consider an serial RLC circuit with R = 0 Ω, L =.0 H, and C =.0 µf. It is connected to a voltage source which produces V 0 = 0 V sin ω 0 t where ω 0 is the angular resonance frequency of the circuit. Find the amplitude of the voltage across the inductor. Problem 20 Solution: We are on resonance which means that the voltages across L and C are equal and opposite in sign and 90 degrees out of phase with the voltage across the resistor. This also means that the impedance is just R as (ω 0 L = /ω 0 C). So the current is: i = V 0 R = I 0 sin ω 0 t I 0 = A (5) 9

10 The voltage across the inductor is then: As the angular resonance frequency is: V L = L di dt = Lω 0I 0 cos ω 0 t (6) ω 0 = L V 0 V L = LC C R cos ω 0t = 000 V cos ω 0 t (7) So 000 V is the correct answer. 0

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