emsol11 - Electromagnetic Theory I Solution Set 11 Due: 14...

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Unformatted text preview: Electromagnetic Theory I Solution Set 11 Due: 14 November 2011 41. In studying a quantum particle in the presence of an electromagnetic magnetic field we used the Lagrangian L = 1 2 m r 2 + q r A ( r ,t ) q ( r ,t ) (1) When there is time dependence we showed near the beginning of the course that the relation between fields and potentials is B ( r ,t ) = A ( r ,t ) , E ( r ,t ) = ( r ,t ) A t ( r ,t ) (2) a) Show that Lagranges equations of motion with this Lagrangian imply Newtons equation with the Lorentz force on the right side. F = q E + q v B (3) Solution : Lagranges equations are d dt L r = L r , L r = m r + q A , L r = q r i A i q m r + q r A + q A t = q r i A i q m r = q r A + q r i A i q q A t = q ( E + v B ) (4) b) Derive the Hamiltonian for this system. Solution : H p r L = m r 2 + q r A L = 1 2 m r 2 + q = ( p q A ) 2 2 m + q (5) c) Repeat the discussion of parts a) and b) for the relativistic Lagrangian obtained by replacing the nonrelativistic kinetic energy term in L as follows 1 2 m r 2 mc 2 parenleftBigg 1 radicalbigg 1 r 2 c 2 parenrightBigg (6) 1 Solution : The only change is L r m r radicalbig 1 r 2 /c 2 + q A , d dt m r radicalbig 1 r 2 /c 2 = q ( E + v B ) H = m r 2 radicalbig 1 r 2 /c 2 + mc 2 radicalBig 1 r 2 /c 2 mc 2 + q = mc 2 radicalbig 1 r 2 /c 2 mc 2 + q Solving for r 2 = ( p q A ) 2 c 2 mc 2 + ( p q A ) 2 , mc 2 radicalbig 1 r 2 /c 2 = radicalbig m 2 c 4 + ( p q A ) 2 c 2 H = radicalbig m 2 c 4 + ( p q A ) 2 c 2 mc 2 + q (7) 42. J, Problem 5.25. Solution : a) For the straight wire, B = I 2 / 2 with = radicalbig x 2 + y 2 . Then we can choose A = ( I 2 z ln 2 ) / 4 . Then W 12 = I 1 contintegraldisplay d l A = I 1 I 2 4 contintegraldisplay d l z ln 2 = aI 1 I 2 4 ln 2 2 2 1 (8) where 2 1 = d 2 + b 2 4 bd cos and 2 2 = d 2 + b 2 4 + bd cos . W 12 = aI 1 I 2 4 ln 4 d 2 + b 2 + 4 bd cos 4 d 2 + b 2 4 bd cos (9) b) In accord with our discussion of force from energy at constant currents, we have F = + W 12 d = 4 ab I 1 I 2 cos 2 d 2 b 2 (4 d 2 + b 2 ) 2 16 b 2 d 2 cos 2 (10) Since it is negative when cos > 0, the force is toward the wire when the current parallel to the wire is nearest the wire.parallel to the wire is nearest the wire....
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emsol11 - Electromagnetic Theory I Solution Set 11 Due: 14...

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