 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 $l{H}_r+a{H}_g=0$. HenceThe rod's field is a demagnetizing field opposite its magnetization; its magnitude is $aM/(l+a)$. The gap field points with the magnetization and $B_g={\mu }_0{H}_g$.