 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2018/paper-1/section-b/b7/b/solution

(x, 0, 0)

[d ℓ^{'} \times (x - x^{'}, - y^{'}, 0)]_{z} = - x d y^{'} + x^{'} d y^{'} - y^{'} d x^{'} .

The leading denominator variation must be kept:

\frac{1}{[( x - x ^{'} ) ^{2} + y ^{'2} ] ^{3/2}} = \frac{1}{∣ x ∣ ^{3}} (1 + \frac{3 x ^{'}}{x} + \dots) .

- x d y^{'}

is zero, but its product with

3 x^{'} / x

is not. Keeping all terms of the leading nonzero order gives

B_{z} ≃ \frac{μ _{0} I}{4 π ∣ x ∣ ^{3}} \oint [- 2 x^{'} d y^{'} - y^{'} d x^{'}] .

For the counterclockwise rectangle,

\oint x^{'} d y^{'} = Δ x Δ y

and

\oint y^{'} d x^{'} = - Δ x Δ y

. Hence

B (x, 0, 0) ≃ - \frac{μ _{0} m}{4 π ∣ x ∣ ^{3}} .

This transverse field is opposite to the magnetic dipole moment. Replacing every denominator by a constant before examining the term proportional to

x

would give the wrong answer.

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