PHY 2054 Exam 2 Spring 2011 Key Solution

PHY 2054 Exam 2 Spring 2011 Key Solution - Reitze/Kumar...

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Page 1 of 6 Reitze/Kumar PHY2054 Exam 2 March 23, 2011 ************************************************************ Useful Constants: k e = 8.99×10 9 Nm 2 /C 2 ε 0 = 8.85×10 -12 C 2 /(Nm 2 ) μ 0 = 4 π × 10 -7 Tm/A k 0 = μ 0 /(4 π ) = 10 -7 Tm/A electron charge = -e = -1.6×10 -19 C electron mass= 9.11×10 -31 kg V=Volt N=Newton J=Joule m=Meter C=Coulomb g = 9.8 m/s 2 c = 3×10 8 m/s "milli-" = 10 -3 "micro-" = 10 -6 "nano-"=10 -9 "pico="=10 -12 ************************************************************ 1. A magnetic field CANNOT: A magnetic field does not change the speed of a moving charged particle only its direction. It applies a force on a moving charged particle. It changes the velocity (direction), the momentum as well as the trajectory of motion . (1) change the kinetic energy of a charged particle (2) exert a force on a charged particle (3) change the velocity of a charged particle (4) change the momentum of a charged particle (5) change the trajectory of a charged particle 2. A proton (charge +e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge +2e) which is also traveling perpendicular to the same field. The ratio of their speeds, v proton /v alpha , is: F = q 1 v 1 B = q 2 v 2 B or v 2 /v 1 = q 1 /q 2 = 2 (1) 2 3. An infinitely long straight wire and a single circular curent loop with radius R are arranged as shown in the figure. If both the straight wire and the loop are carrying the same current I, what is the magnitude of the magnetic field at the center of the loop? There is a log wire and there is a loop. The field at point P (loop center) due to the wire is in the same direction as that due to the loop, coming out of the plane of the paper. The field due to the wire is μ I/2 π R and that due to the loop is μ I/2R. The answer follows (1) ( π +1) μ 0 I/(2 π R) R I I I
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Page 2 of 6 4. An electron (charge q = -1.6 ×10 -19 C) undergoing uniform circular motion with a radius R = 2 ×10 -8 m acts like a current loop. If the speed of the electron is 20 km/s what is the corresponding current flow around the circle (in Amps)? I = e v/2 π R = 25 nA. (1) 2.5×10 -8 5. Two isotopes of helium start from rest and pass through a potential difference V = (860, 480) V and then move out of a slit at the point S into a (0.2, 0.4 T uniform magnetic field pointing into the paper as shown in the figure. Helium-4 consists of two protons and 2 neutrons (atomic number 2, atomic mass = 4 AU) and helium-3 has two protons and one neutron
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This note was uploaded on 10/13/2011 for the course PHY 2054 taught by Professor Avery during the Spring '08 term at University of Florida.

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PHY 2054 Exam 2 Spring 2011 Key Solution - Reitze/Kumar...

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