 Solution

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

Let the wire lie along

z

r

. In the Biot-Savart law, the azimuthal contributions from line elements add with magnitude

B (r) = \frac{μ _{0} I _{0}}{4 π} \int_{- \infty}^{\infty} \frac{r d z}{( r ^{2} + z ^{2} ) ^{3/2}} = \frac{μ _{0} I _{0}}{2 π r} .

Independently, cylindrical symmetry makes the field constant and azimuthal on a circular Amperian loop, so Ampère's law gives

2 π r B = μ_{0} I_{0}

. Both yield

B = \frac{μ _{0} I _{0}}{2 π r} \hat{ϕ},

with direction fixed by the right-hand rule.

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