 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2017/9/a/solution

I d ℓ

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

Take a straight wire along

z

, extending from

- a

a

, and observe its magnetic field at

(s, 0, 0)

. All contributions point in

+ \hat{ϕ}

for positive

I

along

+ z

, by the right-hand rule. Their magnitude integrates to

B_{ϕ} = \frac{μ _{0} I}{4 π} \int_{- a}^{a} \frac{s d z}{( s ^{2} + z ^{2} ) ^{3/2}} = \frac{μ _{0} I}{4 π} [\frac{z}{s s ^{2} + z ^{2}}]_{- a}^{a} = \frac{μ _{0} I a}{2 π s s ^{2} + a ^{2}} .

For

a / s \to \infty

this becomes

μ_{0} I / (2 π s)

, so

\oint B \cdot d ℓ = 2 π s B_{ϕ} = μ_{0} I

, exactly as Ampère's law requires for an infinite straight wire.

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