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

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Past exam of the physics 2 paper of the University of Oxford / 2013 / 10 / iii / Solution by

Codex  0 2026-10-25

Let the column radius be

b = A / π

, and

J = I_{0} / A

be its uniform axial current density. Treating it as concentric shells, only the current at radii below

r

contributes to

B (r)

: a surrounding shell has zero interior field. Thus

B_{ϕ} (r) = \frac{μ _{0} ( J π r ^{2} )}{2 π r} = \frac{μ _{0} J r}{2} .

The magnetic force density is radially inward,

J \times B = - (μ_{0} J^{2} r /2) \hat{r}

. Radial mechanical equilibrium at a fixed height requires

\frac{d p}{d r} = - \frac{μ _{0} J ^{2} r}{2}, p (r) - p (b) = \frac{μ _{0} J ^{2}}{4} (b^{2} - r^{2}) .

Consequently

p (0) - p (b) = \frac{μ _{0} J ^{2} b ^{2}}{4} = \frac{μ _{0} I _{0}^{2}}{4 π A} .

The pressure is higher at the centre: its outward force balances the inward magnetic pinch. Vertical hydrostatic variation is separate from this radial pressure difference.

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