Problem 225

Solution:
(I) Since $2b\sin \left(C+ \frac {\pi}{6}\right)=a+c$,
$\therefore b( \sqrt {3}\sin C+\cos C)=a+c$, which means $\sqrt {3}b\sin C+b\cos C=a+c$,
$\therefore \sqrt {3}\sin B\sin C+\sin B\cos C=\sin A+\sin C=\sin (B+C)+\sin C=\sin B\cos C+\cos B\sin C+\sin C$
$\therefore \sqrt {3}\sin B\sin C=\cos B\sin C+\sin C$,
Since $\sin C\neq 0$,
$\therefore \sqrt {3}\sin B=\cos B+1$,
Squaring both sides gives: $3\sin ^{2}B=\cos ^{2}B+1+2\cos B$,
$\therefore 2\cos ^{2}B+\cos B-1=0$,
Solving this, we get $\cos B= \frac {1}{2}$ or $-1$,
Since $0 < B < \pi$,
$\therefore B= \frac {\pi}{3}$. So, the magnitude of angle $B$ is $\boxed{\frac {\pi}{3}}$.
(II) $BM=CM= \frac {a}{2}$,
In $\triangle ABC$, by the cosine rule, we have: $\cos B= \frac {a^{2}+c^{2}-b^{2}}{2ac}= \frac {1}{2}$, which means $\left( \frac {a^{2}+c^{2}-4}{2ac}= \frac {1}{2}\right)$, $\therefore a^{2}+c^{2}-4=ac$,
In $\triangle ABM$, by the cosine rule, we have: $\cos B= \frac {BM^{2}+AB^{2}-AM^{2}}{2BM\cdot AB}= \frac {1}{2}$, which means $\left( \frac { \frac {a^{2}}{4}+c^{2}-4}{ac}= \frac {1}{2}\right)$, $\therefore \frac {a^{2}}{4}+c^{2}-4= \frac {1}{2}ac$.
$\therefore$ By solving the system of equations $\begin{cases} a^{2}+c^{2}-4=ac \\ \frac {a^{2}}{4}+c^{2}-4= \frac {ac}{2}\end{cases}$, we get $a= \frac {4 \sqrt {7}}{7}$. So, the value of $a$ is $\boxed{\frac {4 \sqrt {7}}{7}}$.