 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/section-b/10/b/solution

Choose the electric current counterclockwise as viewed from

+ z

; reversing it reverses the magnetic field. Parameterize the circular wire by

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

d l \times s = (R z cos ϕ, R z sin ϕ, R^{2}) d ϕ, ∣ s ∣ = R^{2} + z^{2} .

Consequently the transverse terms integrate to zero over

0 \leq ϕ \leq 2 π

, while the axial term is constant:

B (0, 0, z) = \frac{μ _{0} I}{4 π ( R ^{2} + z ^{2} ) ^{3/2}} \int_{0}^{2 π} (R z cos ϕ, R z sin ϕ, R^{2}) d ϕ .

Thus

B (0, 0, z) = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + z ^{2} ) ^{3/2}} z .

The direction is

+ z

on both halves of the axis for this electric current orientation, including

z < 0

; it is not “away from the plane” on both sides. The right-hand rule gives the same direction.

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