# HW12Sol - The velocity vector can be written =(3.75x10 6...

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Jim Guinn’s PHYS2212 Assignment #12 Solutions 1. Find the magnetic force on a proton moving with velocity 4.46Mm/s in the positive x- direction in a magnetic field of 1.75T in the positive z-direction. The magnetic force is = q × , where q = 1.60x10 -19 C , = 4.46x10 6 m/s , and = 1.75T . The result is = -1.25x10 -12 N . (Remember × = - .) 2. A uniform magnetic field of magnitude 1.28T is in the positive z-direction. Find the force on a proton if its velocity is a) = 3.5Mm/s , b) = 2.5Mm/s , c) = 6.5Mm/s , and d) = 3.0Mm/s + 4.0mM/s . For each part of this problem we use = q × , where q = 1.60x10 -19 C and = 1.28T . a) = -7.17x10 -13 N , b) = 5.12x10 -13 N , c) = 0 , d) = 8.19x10 -13 N – 6.14x10 -13 N . 3. An electron moves with a velocity of 3.75Mm/s in the x-y plan at an angle of 60 o up from the x-axis. A magnetic field of 0.85T is in the positive y-direction. Find the force on the electron.
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Unformatted text preview: The velocity vector can be written = (3.75x10 6 m/s) (cos30 o + sin30 o ) . With this we find = q × = (-1.60x10-19 C) (3.75x10 6 m/s) (cos30 o + sin30 o ) (0.85T ) = -4.42x10-13 N . 4. A straight wire segment I l = (2.5A) (3cm + 4cm ) is in a uniform magnetic field = 1.5T . Find the force on the wire. With = I l × = (2.5A) (3cm + 4cm ) (1.5T ) = -0.15N . 5. A proton moves in a circular orbit of radius 65cm perpendicular to a uniform magnetic field of magnitude 0.75T . a) What is the period for this motion? Find b) the speed of the proton and c) the kinetic energy of the proton. For a proton, q = 1.60x10-19 C and m = 1.67x10-27 kg . a) The period is T = 2 π m / qB = 8.74x10-8 s . b) Since the proton completes one orbit in T seconds its speed is v = 2 π r / T = 4.67x10 7 m/s . c) The kinetic energy is K = (1/2) m v 2 = 1.82x10-12 J = 11.4 MeV ....
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## This note was uploaded on 01/23/2012 for the course PYSICS 2212 taught by Professor Jimguinn during the Fall '11 term at Georgia Perimeter.

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