 Solution

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

d B = \frac{μ _{0} I}{4 π} \frac{d ℓ \times s}{s ^{3}},

where

s

points from a current element to the field point. Choose the loop in the

yz

x x

. A parametrization consistent with positive axial magnetic field is

r^{'} (φ) = (0, a cos φ, a sin φ), d ℓ = (0, - a sin φ, a cos φ) d φ, s = (x, - a cos φ, - a sin φ) .

Then

s = a^{2} + x^{2}

and

d ℓ \times s = (a^{2}, a x cos φ, a x sin φ) d φ .

The transverse contributions cancel around the circle, leaving the on-axis magnetic field of a circular current loop

B (x) = \frac{μ _{0} I a ^{2}}{2 ( a ^{2} + x ^{2} ) ^{3/2}} x .

The right-hand rule fixes its direction; reversing the electric current reverses the magnetic field. At the centre this becomes

μ_{0} I / (2 a)

, and far away it decays as

∣ x ∣^{- 3}

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