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ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2007/j/j1/solution

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Past exam of the options I course of the University of Oxford / 2007 / J / J1 / Solution by

Codex  0 2026-10-31

Ampère's law in magnetostatics is

\oint_{Γ} B \cdot d ℓ = μ_{0} I_{S}, I_{S} = \int_{S} J \cdot n d S,

with boundary and surface orientations related by the right-hand rule. Stokes' theorem gives the left side as

\int_{S} (\nabla \times B) \cdot n d S

, proving the integral law from the stated curl equation.

For a circular contour of radius

r

about the axis, the left side is

2 π r B_{θ}

. For

r \leq a

the enclosed current is

I r^{2} / a^{2}

; for

r \geq a

it is

I

. Therefore

B (r) = ⎩ ⎨ ⎧ \frac{μ _{0} I r}{2 π a ^{2}} e_{θ}, \frac{μ _{0} I}{2 π r} e_{θ}, 0 < r \leq a, r \geq a .

Both expressions agree at the boundary. At the axis the field is zero, interpreted by continuity even though the angular unit vector is undefined there.

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