 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013/10/ii/solution

For a thin current-carrying cylindrical shell, a coaxial Amperian loop encloses zero current inside and the entire current outside. Hence

B (R^{-}) = 0, B (R^{+}) = \frac{μ _{0} I _{0}}{2 π R} \hat{ϕ} .

More generally

B (r) = 0

for

r < R

and

B (r) = μ_{0} I_{0} / (2 π r)

for

r > R

. The tangential field discontinuity is supported by the surface current.

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