= Solution
For <binary strip position resolution>[binary strip readout], report the centre of the hit strip. With uniform impact position across pitch $w$, the position error is uniformly distributed on $[-w/2,w/2]$, so
$$
\sigma_x^2=\frac1w\int_{-w/2}^{w/2}x^2dx=\frac{w^2}{12},\qquad
\boxed{\sigma_x=\frac{100\,\mu\mathrm m}{\sqrt{12}}=28.87\,\mu\mathrm m.}
$$
This assumes no analogue charge sharing between strips.
The <Lorentz force>[magnetic force] is $d\mathbf p/dt=q\mathbf v\times\mathbf B$. Since relativistic <momentum> is parallel to <velocity>, $d(p^2)/dt=2\mathbf p\cdot q(\mathbf v\times\mathbf B)=0$. A uniform transverse field therefore changes <momentum> direction but not magnitude. If the trajectory turns at angular <speed> $v/R$, then $|d\mathbf p/dt|=p_Tv/R$. Equating force magnitudes yields
$$
\boxed{R=\frac{p_T}{|q|B}.}
$$
This is the positive <radius>. The sign of $q$ determines the direction of <curvature>; the source's signed formula can describe an oriented <radius>. In practical tracking units,
$$
p_T[\mathrm{GeV}/c]=0.299792458\,|q/e|B[\mathrm T]R[\mathrm m].
$$
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