154CHAPTER 19 19.28Since the path is circular, the particle moves perpendicular to the magnetic field, and the magnetic force supplies the centripetal acceleration. Hence, 2vmqvr=B, or mvqr=B. But the momentum is given by ()2pmvm KE==, and the kinetic energy of this proton is 19611.6010J10.010 eV1.6010J1 eVKE−−×=×2. We then have ()27121219102 1.6710kg1.6010J27.8810T1.6010C5.8010mmKEBqr−−−−××=×19.29For the particle to pass through with no deflection, the net force acting on it must be zero. Thus, the magnetic force and the electric force must be in opposite directions and have equal magnitudes. This gives , or which reduces to mFF=eqvBqE=vEB=19.30The speed of the particles emerging from the velocity selector is B=(see Problem 29). In the deflection chamber, the magnetic force supplies the centripetal acceleration,
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