= Solution
For a <uniformly magnetized ring with a narrow gap>, take the local direction of <magnetization> as positive around the <magnetic circuit>. Neglect leakage and fringing in the narrow gap, and take the cross section as uniform. <Continuity> of <magnetic flux> gives $H_g=H_r+M$. With no free current, <Ampère's law> gives $lH_r+aH_g=0$. Hence
$$
\boxed{H_g=\frac{lM}{l+a}\simeq M,\qquad H_r=-\frac{aM}{l+a}\simeq-\frac al M.}
$$
The rod's field is a <demagnetizing field> opposite its <magnetization>; its magnitude is $aM/(l+a)$. The gap field points with the <magnetization> and $B_g=\mu_0H_g$.
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