 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2016-resit/5/ii/solution

Outside the wire, the Amperian loop encloses the entire electric current, so

B (r) = \frac{μ _{0} I}{2 π r}, r > a .

Its direction is again azimuthal, fixed by the right-hand rule. The inside and outside formulas join continuously at the maximum

B (a) = μ_{0} I / (2 πa)

. The outer branch decays as

1/ r

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