A long ideal solenoid of radius is wound with and carries a steady current

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Magnetostatics · Electromagnetic induction and alternating-current circuits

The Boundary-Value Problem, Ch The vector potential and the magnetostatic boundary-value problem

Problem

A long ideal solenoid of radius a=2.0 cma = 2.0\,\mathrm{cm} is wound with n=5.0×103 turns m−1n = 5.0\times10^3\,\mathrm{turns\,m^{-1}} and carries a steady current I=20 AI = 20\,\mathrm{A}, giving B=μ0nI≈0.1257 TB = \mu_0nI \approx 0.1257\,\mathrm{T} inside and negligible field outside. An alpha particle (charge 2e=3.204×10−19 C2e = 3.204\times10^{-19}\,\mathrm{C}, mass mα=6.644×10−27 kgm_\alpha = 6.644\times10^{-27}\,\mathrm{kg}) rests at a radial distance s=15 cms = 15\,\mathrm{cm} from the axis, outside the winding, where B=0\mathbf{B} = 0. Two claims are made about this setup. Claim 1: "A vector potential A\mathbf{A} can be written for B\mathbf{B} because B\mathbf{B} is a conservative field, in the same sense that E=−∇V\mathbf{E} = -\nabla V follows from E\mathbf{E} being conservative." Claim 2: "Outside the solenoid B\mathbf{B} vanishes identically, so by a suitable choice of gauge A\mathbf{A} can be transformed to zero everywhere in that region; consequently switching off the current can do nothing to a charge parked there." ((a)) State exactly which property of B\mathbf{B}, not curl-freeness, is what licenses writing B=∇×A\mathbf{B} = \nabla\times\mathbf{A}, and say where Claim 1 goes wrong. ((b)) Show that Claim 2 is false by computing the speed the alpha particle acquires, along the direction the winding current flowed, when the current is switched off, using the flux enclosed by the solenoid.

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