 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2021/paper-1/section-c/c9/a/solution

For a steady filamentary electric current

I

B (r) = \frac{μ _{0} I}{4 π} \int \frac{d ℓ ^{'} \times ( r - r ^{'} )}{∣ r - r ^{'} ∣ ^{3}} .

As an application, an infinitely long straight wire has circular symmetry about its axis. At perpendicular distance

s

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

Its direction follows the right-hand rule. The following part applies the same law to a circular current loop.

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