HW-10Solutions-02-08-08

HW-10Solutions-02-08-08 - Physics 21, Spring 2008 HW-10...

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Physics 21, Spring 2008 February 8, 2008 HW-10 Solutions 10-1. (HRW 28-2) An alpha particle travels at a velocity of magnitude 550 m/s through a uniform magnetic field of magnitude 0.045 T. (An alpha particle has a charge of and a mass of .) The angle between and is 52°. What is the magnitude of (a) the force acting on the particle due to the field and (b) the acceleration of the particle due to ? (c) Does the speed of the particle increase, decrease, or remain the same? Solution: 2. (a) We use Eq. 28-3: F B = |q| vB sin φ = (+ 3.2 × 10 –19 C) (550 m/s) (0.045 T) (sin 52°) = 6.2 × 10 –18 N. (b) a = F B / m = (6.2 × 10 – 18 N) / (6.6 × 10 – 27 kg) = 9.5 × 10 8 m/s 2 . (c) Since it is perpendicular to r r vF B , does not do any work on the particle. Thus from the work-energy theorem both the kinetic energy and the speed of the particle remain unchanged. 10-2. (HRW 28-4) A particle of mass 10 g and charge 80 μ C moves through a uniform magnetic field, in a region where the free-fall acceleration is . The velocity of the particle is a constant , which is perpendicular to the magnetic field. What, then, is the magnetic field?
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This note was uploaded on 02/20/2008 for the course PHYSIC 2 taught by Professor Kim during the Spring '08 term at Lehigh University .

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HW-10Solutions-02-08-08 - Physics 21, Spring 2008 HW-10...

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