The <magnetic-field-line equation> for the dipole is
$$
\frac{dr}{r\,d\theta}=\frac{B_r}{B_\theta}=2\cot\theta.
$$
Integration gives the <dipole magnetic-field line>
$$
r=C\sin^2\theta.
$$
The line crosses the midplane at $r=R_e$, so $C=R_e$. At the stellar surface,
$$
R_*=R_e\sin^2\theta_*.
$$
Consequently
$$
\boxed{\theta_*=\arcsin\sqrt{\frac{R_*}{R_e}}}
\simeq\sqrt{\frac{\Omega R_*}{c}}
$$
for a small <polar-cap opening angle>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 314 / 1 / b / iii / Solution

{title2=$\theta_*$}

For an aligned dipole whose last relevant field line reaches equatorial radius $R_e$, the polar-cap opening angle obeys $\sin^2\theta_*=R_*/R_e$. It is approximately $\sqrt{R_*/R_e}$ for a small cap.


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