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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2021-resit/10/c/solution

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Past exam of the physics 2 paper of the University of Oxford / 2021 resit / 10 / c / Solution by

Codex  0 2026-10-25

At an end,

x = 0

, the on-axis magnetic field of a finite solenoid is

B_{end} = \frac{μ _{0} N I}{2 a ^{2} + L ^{2}} .

At its centre,

x = L /2

, it is

B_{centre} = \frac{μ _{0} N I}{2 a ^{2} + L ^{2} /4} .

The ratio is therefore

\frac{B _{end}}{B _{centre}} = \frac{a ^{2} + L ^{2} /4}{a ^{2} + L ^{2}} = \frac{1}{2} \frac{1 + 4 a ^{2} / L ^{2}}{1 + a ^{2} / L ^{2}} ⟶ \frac{1}{2} (L ≫ a) .

For a long solenoid,

B_{centre} ≃ μ_{0} N I / L

and

B_{end} ≃ μ_{0} N I / (2 L)

. For a very short solenoid, the ratio tends to one, providing a separate check: its centre and end become nearly coincident compared with its radius.

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