= Solution
A highly conducting <solar wind> carries the <Sun>'s <magnetic field> outward by <magnetic flux freezing>. This can open <magnetic field lines> that would form closed loops in a static dipole model. Beyond an opening radius $r_o$, approximate the field as radial. Constancy of the <magnetic flux> in a fixed solid-angle <flux tube> gives the <radial solar-wind magnetic flux> law
$$
B_r(r)r^2\simeq\text{constant},\qquad B_r\propto r^{-2},
$$
which decays more slowly than the vacuum <magnetic dipole field>, proportional to $r^{-3}$.
For example, if the field becomes radial already near the solar radius $R_\odot$, normalizing both estimates to the same field there gives
$$
\frac{B_{\rm wind}(r_E)}{B_{\rm dip}(r_E)}\sim\frac{r_E}{R_\odot}
=\frac1{0.005}=200.
$$
Thus a dipole estimate $10^{-11}\ \mathrm T$ becomes of order $2\times10^{-9}\ \mathrm T$, comparable to the stated hundredfold enhancement. More generally, matching dipolar decay inside $r_o$ to radial decay outside gives
$$
\boxed{\frac{B_{\rm wind}(r_E)}{B_{\rm dip}(r_E)}\sim\frac{r_E}{r_o};
\quad r_o\sim2R_\odot\ \Rightarrow\ B(r_E)\sim10^{-9}\ \mathrm T.}
$$
This is a plausible scaling explanation, not a determination of the actual opening radius. The magnetic field need not be outward everywhere: opposite-polarity sectors can preserve zero net <magnetic flux> through a closed <surface>, consistent with <Gauss's law for magnetism>. The argument concerns the changed geometry of the field in a flowing <plasma>, rather than changing the dipole exponent by assumption.
Back to article page
