# 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 ) − qφ ( 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 ) = −∇ φ ( r ,t ) − ∂ A ∂t ( r ,t ) (2) a) Show that Lagrange’s equations of motion with this Lagrangian imply Newton’s equation with the Lorentz force on the right side. F = q E + q v × B (3) Solution : Lagrange’s equations are d dt ∂L ∂ ˙ r = ∂L ∂ r , ∂L ∂ ˙ r = m ˙ r + q A , ∂L ∂ r = q ˙ r i ∇ A i − q ∇ φ m ¨ r + q ˙ r ·∇ A + q ∂ A ∂t = q ˙ r i ∇ A i − q ∇ φ m ¨ r = − q ˙ r ·∇ A + q ˙ r i ∇ A i − q ∇ φ − 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 + qφ = ( p − q A ) 2 2 m + qφ (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 + qφ = mc 2 radicalbig 1 − ˙ r 2 /c 2 − mc 2 + qφ 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 + qφ (7) 42. J, Problem 5.25. Solution : a) For the straight wire, B = I 2 μ ˆ φ/ 2 πρ with ρ = radicalbig x 2 + y 2 . Then we can choose A = − ( I 2 ˆ z ln ρ 2 ) / 4 π . Then W 12 = I 1 contintegraldisplay d l · A = − I 1 I 2 μ 4 π contintegraldisplay d l · ˆ z ln ρ 2 = aI 1 I 2 μ 4 π ln ρ 2 2 ρ 2 1 (8) where ρ 2 1 = d 2 + b 2 4 − bd cos α and ρ 2 2 = d 2 + b 2 4 + bd cos α . W 12 = aI 1 I 2 μ 4 π ln 4 d 2 + b 2 + 4 bd cos α 4 d 2 + b 2 − 4 bd cos α (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 μ π cos α 2 d 2 − b 2 (4 d 2 + b 2 ) 2 − 16 b 2 d 2 cos 2 α (10) Since it is negative when cos α > 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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