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.
- Fall '11