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Unformatted text preview: Phys. 2306 Exam 3 F09 Form A November 25, 2009 Phys. 2306 Exam 3 F09 Form A Solutions In cosmic ray experiments, particles travel downward ( j ) with speed v= 3 . 00 × 10 6 m/s. They enter a region of constant magnetic field, directed into the paper ( k ) at (x,y,z)=(0,0,0). They exit the field traveling upward at (x,y,z)=(1.00 m,0,0) with speed v’. 1) The ratio of the speeds v’/v is a) 1 b) 2 c) 1/2 d) 0 e) none of these Since F = q v × B , F · v = 0. No work is done on the particle by the magnetic force so its KE and speed don’t change, v’/v=1, (a). 2) The particle’s charge is a) positive b) negative c) 0 d) insufficient info Just after entering the field the force on the particle is in the +x ( i di rection. Thus, F = F i = qvB ( j ×  k ) = qvB i , q is positive, (a). 3) If the magnitude of the particle’s charge/mass= 0 . 240 × 10 8 C/m, what is the magnitude of the magnetic field in T? a) 0.125 b) 0.375 c) 0.5 d) 0.625 e) none of these In a constant B field with no velocity component in the field direction the mo tion is circular motion, here a semi circle of radius 0.5 m. Thus mv 2 /R = qvB or B = v/ [ R ( q/m )] so B = 3 × 10 6 / [0 . 5( . 240 × 10 8 )] T = 0 . 25 T (enone of the choices). There are three long wires that run parallel to the x axis and carry non zero constant currents. Wire 1 passing through (y,z)=(0,0) carries current I in the +x i direction. There is no force/length on wire 2 passing through (y,z)=(a,0). On wire 3 passing through (y,z)=(3a,0) there is a force/length equal and opposite that on wire 1. 1 4) The current in wire 3 has magnitude, direction a) 2I, x b) 2I, +x c) I/2, x c) I/2, +x e) none of these 0 = d F 2 = I 2 d L 2 × B 2 . d L 2 = ± dL 2 i and B 2 = B 2 , 1 + B 2 , 3 = ( B 2 , 1 ± B 2 , 3 ) out ....
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This note was uploaded on 12/24/2011 for the course PHYS 2306 taught by Professor Ykim during the Fall '06 term at Virginia Tech.
 Fall '06
 YKim

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