= Solution
Set $\beta=\pi/\alpha$ and use the branch $\phi(z)=z^\beta$ with argument in $(-\alpha/2,\alpha/2)$. This is a <conformal map> from the wedge to the right half-plane, fixes the starting point $1$, and sends the outer circle of radius $r$ to that of radius
$$
\boxed{R=r^{\pi/\alpha}.}
$$
The two wedge sides map to the imaginary axis.
By <conformal invariance of planar Brownian motion>, the image of the stopped path is <planar Brownian motion> after the increasing <conformal Brownian clock> $A(t)=\int_0^t|\phi'(B_s)|^2\,ds$. This clock does not change which boundary portion is reached first. Localization away from the vertex justifies the map even when its derivative is unbounded there; the vertex is a <polar point for planar Brownian motion> and has zero hitting probability from $1$.
Consequently
$$
\mathbb P_1\{B[0,T(r)]\subset W_\alpha\}
=\mathbb P_1\{T(R)<S\},
$$
where the probability on the right is for the right half-plane. The <power-map reduction for Brownian exit from a wedge> also works at $\alpha=2\pi$, when the wedge is the plane slit along the negative real axis.
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