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Unformatted text preview: HW#4Solutions Phys410Fall 2011 Prof. Ted Jacobson Room 4115, (301)405-6020 www.physics.umd.edu/grt/taj/410a/ firstname.lastname@example.org 7.49 ( charged particle in uniform magnetic field polar gauge ) (a) Just evaluate. (b) A = 1 2 B so v A = 1 2 B 2 , hence L = 1 2 m ( 2 + 2 2 + z 2 ) + 1 2 qB 2 . (1) The EL eqns are m = m 2 + qB (2) p = m 2 + 1 2 qB 2 = const. (3) m z = 0 (4) (c) The z component of velocity is constant in all solutions. If is constant then m + qB = 0, so =- qB/m. (5) All orbits, regardless of radius, have this same angular frequency, the cyclotron fre- quency . The general orbit is a spiral that is counterclockwise looking in the positive z-direction, with constant velocity in the z direction, and angular velocity qB/m . (d) A calculation shows the Hamiltonian is equal to the kinetic energy: H = 1 2 m ( 2 + 2 2 + z 2 ) . (6) (In general, H = q i L/ q i- L , so terms in L independent of the velocity appear in H with the opposite sign, terms linear in the velocity do not appear, and terms quadratic in the velocity appear unchanged between L and H .) (e) p in (3) is the angular momentum about the z axis. (f) For orbits of constant in the xy- plane, the energy is E = 1 2 m 2 2 = 1 2 ( qB ) 2 /m . For these orbits, using (5) the angular momentum is p =- 1 2 qB 2 , so E =- ( qB/m ) p . (g) Dimensional analysis: [ m ] = M , [ qB ] = [ MT- 1 ], [ p ] = ML 2 T- 1 , and energy = ML 2 T- 2 , so the only way to make energy is by the combination ( qB/m ) p . (h) If p =- n h is quantized (the sign of the allowed integers n is the sign of qB ), then E n = n hqB/m . [Actually there is a zero-point energy in quantum mechanics, and...
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