PH202-NG Test 2 (July. 7, 2009, 3:00PM-5:05PM)

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1 name Write your name also in the back of the last page. blazer id [a] PH202-NG Test 2 (July. 7, 2009, 3:00PM-5:05PM) You may not open the textbook nor notebook. A letter size information may be used. A calculator may be used. However, mathematics or physics formula programmed in a calculator may not be used. Write down, reasoning, calculation and answer in the blank space after each problem. Use the backside of the sheet if necessary. Answers without reasonable explanation or convincing mathematical derivation will receive no point even if your answers coincide with the correct ones. Your reasoning must begin with basic physical principles. Substantial partial credit may be given to correct reasoning and mathematical procedures when your final answers are wrong. However, if your final answers are too obviously wrong, a partial credit may not be given. Important Physical Constants: electron s electric charge: e = C electron s mass: m e = kg proton s mass: m p = kg permittivity of free space: ǫ 0 = 1 4πk = C 2 /(N m 2 ) where k = N m 2 /C 2 permeability of free space: µ 0 = 4π 10 7 Tm/A acceleration due to gravity at the surface of the earth: g = 9.80 m/s 2 speed of light in a vacuum : c = m/s Total 7 8 1

2 1. [10 pts.] A proton is projected perpendicularly into a magnetic field with a certain velocity and follows a circular path. Then an electron is projected perpendicularly into another magnetic field with the same velocity. The electron follows the exact same circular path as the proton. The magnetic field for the proton has a magnitude of 0.60 T in the +z direction. Find the direction and magnitude of the magnetic field for the electron. To form a circular motion in the same direction, the direction of magnetic force exerted on the proton must be the same as that on the electron. Since electron and proton have the opposite sign of electric charge, the direction of the magnetic force on the electron must be opposite to that on proton. z direction To form a uniform circular motion, the magnitude of the magnetic force must match to the centripetal force. mv 2 = evb r = mv r eb Plug in the masses and magnetic fields for the proton and the electron, r p = m pv eb p r e = m ev eb e. Since the radius of the circle must be the same for the proton and the electron. r e = r p m ev eb e = m pv eb p m e B e = m p B p Hence, B e = m eb p m p = ( )(0.60) = T 2

3 2. [20 pts.] Two long, straight wires carry an identical identical current of I 1 =I 2 =2.5 A as shown in Figure. A 4.0-µC point charge is moving at 35m/s along the wires in direction B shown in Figure. The distance between the wires are 6.0 cm and the distance between wire 1 and the charge is 4.0 cm. Answers the following questions. The direction should be specified by a letter (U, D, F, B, L, R) shown in the figure. (a) Find the magnitude and direction of the magnetic field at the position of the charge. (b) Find the magnitude and direction of the force exerted on the charge by the magnetic field (a) Each wire generates circular magnetic fileds around it as shown in Figure. At the position of the charge, B 1 is upward and B 2 downward. Theie magnitudes are B 1 = µ 0I = 4π = T 2πr 1 2π0.040 B 2 = µ 0I = 4π = T 2πr 2 2π0.020 Since B 2 is larger than B 1, the net magnetic field is downward D. The magnitude of net magnetic field is B net = B 1 B 2 = = T (b) Since the velocity and the magnetic field are in direction B and D, respectively, RHR-1 indicates that the direction of force on the charge is L 3

4 3. [20 pts] Figure shows a rectangular N = 20-turn coil of wire (ABCD). Its dimensions is BC = 12cm by AB = 6.0 cm. It is mounted in the xy plane and hinged along one long side (AD) on the y axis. It carries a current of i = 0.10 A. The direction of the current is shown in Figure. A uniform magnetic field of magnitude B = 0.50 T, is applied at 30 from the x axis as shown in Figure. (a) Find the direction and magnitude of the force exerted on the wire AB. The direction should be expressed like +x. (b) Which edges (AB, BC, CD, DA) contribute the torque around the hinge. List all of them. (c) Find the magnitude of the torque. You must derive it from basic physics principles. (d) In which direction does the coil rotate, viewed from the top, clockwise or counterclockwise? (a) Using RHR-1, the direction is y. The magnitude is given by F = NIBLsinθ = (20)(0.10A)(0.50T)(0.060m)sin 30 = N. (b) The force on the segments AB and CD are in ±y dircetion which is parallel to the axis of rotation. Hence, the force does not contribute the torque. The force on DA has zero lever arm. Hence, it has no contribution to the torque. Only BC generates noz-ero torque. (c) Figure shows the direction of magnetic field, force and lever arm. The magnitude of the magnetic force on BC is F = NIBLsinφ = (20)(0.10A)(0.50T)(0.12m)sin 90 = 0.12 N and the lever arm Hence, the torque is l = (0.060m)cos 30 = m τ = lf = (0.052m)(0.12N) = N m (d) As shown in Figure, the loop rotates in clockwise direction. 4

5 4. [20 pts.] A metal rod of length 15cm on two parallel metal tracks is pulled with a hand to the right at a constant speed of 1.2 m/s. The tracks are connected at one end so that they and the rod form a closed circuit as shown in Figure. The rod has a resistance 10 Ω, and the tracks have negligible resistance. A uniform magnetic field of 2.0 T perpendicular to the plane of this circuit (into paper) is applied. (a) What is the magnitude and direction (up or down) of the induced current in the rod? (The direction should be one of Up, Down, Left, Right or No current.) (b) Find the magnitude of the force exerted on the rod by the hand. (a) Using RHR-1, the direction is UP. Emf generated in the loop is given by emf = vbl. Using the Ohm s law I = emf R = vbl R = (1.2m/s)(2.0T)(0.15m) = A. 10Ω (b) Since the velocity is constant, F net = F hand F mag = 0. Hence, F hand = F mag = IBL = ( A)(2.0T)(0.15m) = N. 5

5. [20 pts.] A long and narrow rectangular loop of wire is moving toward the bottom of the page with a speed of 0.025 m/s (see the drawing). The loop is leaving a region in which a 3.

6 5. [20 pts.] A long and narrow rectangular loop of wire is moving toward the bottom of the page with a speed of m/s (see the drawing). The loop is leaving a region in which a 3.4-T magnetic field (out of paper) exists. The magnetic field outside this region is zero. (a) During a time of 2.0 s, what is the magnitude of the change in the magnetic flux? (b) Find the magnitude of emf. (c) What is the direction of induced magnetic field, out of paper or into paper? Justify your answer. (d) What is the direction of induced current in the loop, Clockwise or coounterclockwise? (a) Durin the time of falling, the loop moves down by d = vt. Hence the area of the region with magnetic field decreases by A = vtl. (The negative sign is due to the decreasing are. Hence, the change in the magnetic flux is Φ = B A = Bvtl = (3.4T)(0.025m/s)(2.0s)(0.080m) = wb. (b) Using the Faraday s law emf = Φ t = Bvl = (3.4T)(0.025m/s)(0.080m) = V. (c) Since the oroginal magnetic field inside the loop is decreasaing, the Lenz s law indicates that the induced magnetic field should be in the same direction as the direction of the original magnetic field. Out of paper. (d) In order to induce the out-of-paper magnetic field, the induced current flows in the counterclockwise direction. 6

7 6 [10 pts.] (a) Which of the following electromagnetic waves have the shortest wave length in a vacuum? Visible light X-ray Gamma-ray Microwave Ultra violet (b) Which of the following electromagnetic waves have the highest frequency in a vacuum? Visible light X-ray Radiowave Infra red Gamma-ray (c) Obtain the wavelength in a vacuum for red light, whose frequency is Hz. λ = c f = m/s Hz = m. 7

7. Extra [15 pts.] (Difficult) A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 4.0 m/s perpendicular to a 0.50-T magnetic field.

8 7. Extra [15 pts.] (Difficult) A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 4.0 m/s perpendicular to a 0.50-T magnetic field. The resistance of the rod and tracks is negligible. The rod maintains electrical contact with the tracks at all times and has a length of 1.3 m. A 0.75-Ω resistor is attached between the tops of the tracks. (a) What is the mass of the rod? (b) Find the energy dissipated in the resistor while the rod falls for 0.20 s. (c) The induced current dissipates energy in the resistor. Since energy must conserve, the same amount of energy must come from another form of energy. What is it? Show that the amount of energy lost during 0.20 s matches to the dissipation obtained in part (b). (a) Since the velocity is constant, F net = F mag mg = 0. Therefore, m = F mag /g. The magnetic force is due to the current induced by the change in magnetic flux insize the loop. The current is given by I = emf R = vbl R Hence, the magnetic force given by ( ) vbl F mag = IBL = BL = vb2 L 2 R R Thus, m = vb2 L 2 Rg = (4.0m/s)(0.50T)2 (1.3m) 2 (0.75Ω)(9.8m/s 2 ) = 0.23 kg. (b) E = Pt = Iemft = (emf)2 t R = (vbl)2 t R = [(4.0m/s)(0.50T)(1.3m)]2 (0.20s) 0.75Ω = 1.8 J. (c) During the process, the height of the rod is lowered. Then, the potential energy of the rod due to gravity is reduced. The amount of the lost potential energy appears as the energy dissipated in the resistor. In fact, the loss of the potential energy mgh = mgvt = (0.23kg)(9.8m/s 2 )(4.0m/s)(0.20s) = 1.8 J which agrees with the value obtained in part (c). 8

9 8. Test 1 Bailout Problem [Maximum 20 pts.] Do not solve this problem if your test 1 score was higher than 70. The points from this problem will be added to your score of test 1. However, your test 1 score will not exceed 70 pts. For example, if your score was 60 pts, the maximum point you can earn from this problem is 10 pts. A 120-V battery and four 60-Ω light bulbs are connected as shown in Figure. (a) [5 pts.] How much current is drawn from the battery? (b) [5 pts.] How much power is consumed by the whole system? (c) [5 pts.] Find the power dissipated in the light bulb L 4. (d) [5 pts.] If the light bulb L 3 burns out, does the light bulb L 1 become brighter or darker. Justify your answer. (a) Since L 2, L 3, and L 4 are in parallel connection their equivalent registance is 1 R 234 = 1 R R R 4 = 1 60Ω Ω Ω = 1 20 R 234 = 20 Ω Now, R 1 and R 234 are in series, the net equivalent resistance is R eq = R 1 + R 234 = 80 omega. Using the Ohm s law, I = V R eq = 120V 80Ω = 1.5 A. (b) P = IV = (1.5A)(120V ) = 180 W. (c) The voltage across L 1 is V 1 = IR 1 = (1.5A)(60Ω) = 90V. Then, the voltage accross L 4 is given by V 4 = V V 1 = 120V 90V = 30V. Hence the power consumed by L 4 is P 4 = I 4 V 4 = v2 4 (30V )2 = R 4 60Ω = 15 W. (d) When L 3 burns out, the equivalent resistor of L 2 and L 4 becomes larger than R 234 obtained in part (a). Hence, the voltage accross L 1 decreases. Then, the power consumption of L 1 decreases. L 1 gets darker. 9

21 MAGNETIC FORCES AND MAGNETIC FIELDS

21 MAGNETIC FORCES AND MAGNETIC FIELDS CHAPTER 1 MAGNETIC FORCES AND MAGNETIC FIELDS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1 (d) Right-Hand Rule No 1 gives the direction of the magnetic force as x for both drawings A and B In drawing C, the