<p> Practice Exam 3</p><p>PHY 2140 Sp/Su 2007</p><p>Solutions</p><p>1. A 10.0 F capacitor and a 2.00 H inductor are connected in series with a 60.0 Hz source whose rms output is 150 V. What is the value of the rms current in the circuit?</p><p>Z = XL-XC XL = 2f L = 754 , XC = 1/2fC = 265  so Z = 489  therefore Irms = Vrms/Z = 0.307 A</p><p>2. A step-up transformer is designed to have an output voltage of 5500 V (rms) when the primary is connected across a 110 V (rms) source. If there are 100 turns on the primary winding, how many turns are required on the secondary winding?</p><p>Ratio = 5500V/110V = 50 so we require 50 x 100 turns = 5000 turns on the secondary.</p><p>3. An electromagnetic wave in vacuum has a magnetic field amplitude of 7.00 x 10-7 T. What is the amplitude of the corresponding electric field?</p><p>E/B = c so E = cB thus E = 210 V/m</p><p>4. An electron has a speed of v = 0.90c. At what speed will a proton have a momentum equal to that of the electron?</p><p>-22 p = mv and electron = 2.29 so for the electron: p = 5.64 x 10 Kg m/s -22 we desire that pproton = pelectron = p = pmpvp = 5.64 x 10 Kg m/s 2 2 p v pe c e p v 5 so pvp = pe/mp then = solve for vp: p= = 3.37x10 m/s 2 2 2 p2+ m 2 c 2 mp1- v p / c e p 5. In a certain x-ray machine electrons are accelerated through a potential difference of 2 MeV before striking a metal target. With what speed do the electrons strike the x-ray target?</p><p>KE KE = 2 MeV, mc2 = 0.511 MeV, KE = (-1) mc2 solve for g = +1. mc2 Then expand : 1 2 2 KE v= c1 -2 = 0.979 c g =1/ 1 -v / c =2 + 1 solve for 骣KE mc 琪 +1 桫mc2</p><p>6. A muon formed high in Earth’s atmosphere travels at a speed v = 0.99c for a distance of 4.6 km before it decays. How long does the muon live, as measured in its reference frame?</p><p>3 g =1 = 7.09 In earth frame:  = 4.6x10 m/0.99c = 15.5 sec, 2 1- 0.99</p><p>The proper time is then: p = / = 15.5 sec/7.09 = 2.2 sec</p><p>7. What is the minimum accelerating voltage required to produce an x-ray with a wavelength of 0.0300 nm?</p><p>-25 min = hc/eV using hc=1.989x10 Jm 4 Then V = hc/emin = 4.14x10 V = 41.4 kV</p><p>8. Assuming that the tungsten filament of a lightbulb is a blackbody, what is its peak wavelength if its temperature is 2900 K?</p><p>-2 -2 max T = 0.2989 x 10 mK so max = 0.2989 x 10 mK/2900 K = 999 nm</p><p>9. A radio wave has a wavelength of 6 m. What is the frequency of this radio wave? </p><p> c = f so f = c/ = (3.0x108 m/s) / 6 m = 50 x 106 /s = 50 MHz 10. What is the energy in electron volts of a photon having a wavelength of 5.00 cm?</p><p>E = hc/ = 3.978x10-24 J, convert to eV by dividing by e gives: E = hc/e = 2.49x10-5 eV</p><p>11. What is the value the inductor which must be inserted in a 60 Hz circuit in series with a generator of 170 V maximum output voltage to produce an rms current output of 0.75 A? </p><p>Z = Vrms/Irms = 160.3 </p><p>Vmax = 170 V, so Vrms = 170V2 = 120 V I = V/XL = 0.75 A as stated in the problem, thus </p><p>XL = 2f L = V/I, giving L = (1/2f) Vrms/Irms = 0.425 H</p><p>12. The proper length of one spaceship is three times that of another. The two spaceships are traveling in the same direction and, while both pass directly overhead, an observer on Earth measures the two spaceships to be the same length. If the slower ship is moving with a speed of 0.350c, determine the speed of the faster spaceship.</p><p>Let’s use f as the subscript for the fast ship and s for the slow ship. Then the lengths of the ships when measured in their rest frames are related by: Lp,f = 3 Lp,s. We choose the proper length of the fast ship to be the larger since it will be length contracted more than the slow ship so that from earth both can appear the same length, Lf = Ls = L.</p><p>Then by the length contraction definition, Lp,s = s L and Lp,f = f L </p><p>Eliminating L gives: L = Lp,s/s = Lp,f/f</p><p>Then solving for f = (Lp,f / Lp,s) s = 3 s Expanding the ’s we find: 1 3 = 2 2 and finally, solving for vf we find: vf v s 1- 1- 2 c2 c</p><p> v2 1- s 2 = 0.95c v= c 1 - c f 9</p>