= Solution
\b[Field law and centre value.] The <Biot-Savart law> for a steady thin wire is
$$
d\mathbf B=\frac{\mu_0I}{4\pi}
\frac{d\boldsymbol\ell'\times(\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}.
$$
Here $d\boldsymbol\ell'$ follows the conventional <electric current> and the displacement points from source to observation point. For a circular loop of radius $a$, every element at the centre has distance $a$ and contributes in the same axial direction. Choose $+z$ by the <right-hand rule>. Then
$$
\boxed{\mathbf B(0)=\frac{\mu_0I}{4\pi a^2}(2\pi a)\widehat{\mathbf z}
=\frac{\mu_0I}{2a}\widehat{\mathbf z}.}
$$
\b[On-axis integration.] At axial distance $z$, every wire element has distance $\sqrt{a^2+z^2}$. Azimuthal symmetry cancels transverse components. Parameterizing $\mathbf r'=(a\cos\varphi,a\sin\varphi,0)$ gives axial cross-product element $a^2\,d\varphi$, so the <on-axis magnetic field of a circular current loop> is
$$
\boxed{\mathbf B(0,z)=B_0(z)\widehat{\mathbf z},\qquad
B_0(z)=\frac{\mu_0Ia^2}{2(a^2+z^2)^{3/2}}.}
$$
It points the same way on the two halves of the axis; its strength is even in $z$.
\b[Directions near the axial point.] At a point on the positive axis the field is axial, but nearby lines spread outward because $B_0$ decreases as $z$ increases. On the negative-axis side they converge toward the axis while still progressing in the positive axial direction. There is no azimuthal field. The near-axis sketch shows this local spreading; the complete <magnetic field lines> return around the outside of the loop and form closed curves.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2017/paper-1-b8-near-axis-loop-field.png]
{title=Original calculated leading near-axis field-line geometry on the positive side of a circular current loop; the lines spread as the axial field weakens}
\b[Radial component from zero divergence.] Smooth axial symmetry gives $B_z(\rho,z)=B_0(z)+O(\rho^2)$ and $B_\rho=O(\rho)$. By <Gauss's law for magnetism>,
$$
\frac1\rho\partial_\rho(\rho B_\rho)+\partial_zB_z=0.
$$
Integrate radially from zero, excluding a singular $1/\rho$ term, to obtain the <near-axis radial magnetic field>, $B_\rho=-\rho B_0'(z)/2+O(\rho^3)$. Differentiating the axial field therefore gives
$$
\boxed{B_\rho\simeq
\frac{3\mu_0Ia^2z\rho}{4(a^2+z^2)^{5/2}}.}
$$
For $z>0$ it is outward; for $z<0$ it is inward. This supplies the direction reasoning used in the sketch.
\b[Force on the second loop.] Put the second loop at $z=d$ and radius $\rho=a$. For $d\gg a$, the radial field due to the first loop is $B_\rho\simeq3\mu_0Ia^3/(4d^4)$. The element $d\boldsymbol\ell=a\,d\varphi\,\widehat{\boldsymbol\varphi}$ experiences <Lorentz force>
$$
d\mathbf F=I a\,d\varphi\left(B_z\widehat{\boldsymbol\rho}-B_\rho\widehat{\mathbf z}\right).
$$
The radial contributions cancel as vectors around the loop. The axial contributions add, giving the <force between distant coaxial current loops>,
$$
\boxed{F_z=-2\pi aI B_\rho
\simeq-\frac{3\pi\mu_0I^2a^4}{2d^4}.}
$$
The sign follows $\widehat{\boldsymbol\varphi}\times\widehat{\boldsymbol\rho}=-\widehat{\mathbf z}$. Both the small radial coordinate relative to $d$ and the dipole approximation are justified by $a/d\ll1$.
\b[Force direction.] The second loop is pulled back toward the first: \b[the force is attractive] for the same current sense. Reversing one current reverses the <force>. The much larger axial field chiefly produces radial element forces; retaining the smaller radial field is essential for the net axial attraction.
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