Question 3.6.1: Figure 3.36 shows a cross-sectional view of a magnetic circu...

From symmetry we know that there is no variation in the field in the x or z directions. Therefore, the problem reduces to a 1D Dirichlet boundary-value problem. Because there is no applied current, we use the magnetic scalar potential formulation (Section 3.4). Specifically, we use Eqs. (3.101) and (3.102), which reduce to

H = – \frac{dφ_{m}}{dy} ,            (3.201)

and

\frac{d^{2}φ_{m}}{dy^{2}}  =  \frac{dM_{y}}{dy}.            (3.202)

There are two regions to consider, the magnet (region 1), and the gap (region 2). In these regions we have

M = \left\{\begin{matrix} M_{s}\hat{y} \quad  (region  1)\\  0  \quad (region  2).\end{matrix} \right.                  (3.203)

We find that the volume charge density is zero in both regions

ρ_{m}  =  – \frac{dM_{y}}{dy} = 0.

Therefore, Eq. (3.202) reduces to

\frac{d^{2}φ_{m}}{dy^{2}} = 0.

The general solutions to this equation is given by Eq. (3.196) with k_{y} = 0, that is,

Eq. (3.196): Y(y) = \left\{\begin{matrix} B_{1} y + B_{2} \qquad (k_{y}^{2}  =  0)\\ B_{3}sin(k_{y}y)+B_{4}cos(k_{y}y)\\ or \qquad (k_{y}^{2}  >  0)\\ B_{3}e^{jk_{y}t}+B_{4}e^{-jk_{y}t} \\B_{3}sinh(ky)+B_{4}cosh(ky)\\or \qquad (k_{y}^{2}  <  0  with  k= \sqrt{|k_{y}^{2} |}) \\B_{3}e^{ky}+B_{4}e^{-ky} \end{matrix} \right.

φ_{m}^{(1)}(y) = a_{1}y + b_{1}    (region 1),              (3.204)

and

φ_{m}^{(2)}(y) = a_{2}y + b_{2}    (region 2),              (3.205)

The coefficients are determined from the boundary conditions.

Magnet-plate interface (y = 0): We assume that the flux plates have infinite permeability ( μ ≈ ∞). Therefore, the H-field in them is negligible, and the potential φ_{plate} is constant. As the plates are connected (at infinity), they are both at the same potential. Without loss of generality, we set this potential to zero, that is, φ_{plate} = 0. As the potential is continuous at the magnet-plate interface, we have

φ_{m}^{(1)}(0) = φ_{plate}
= 0.

From this we find that b_{1} = 0.

Magnet-gap interface (y = l_{m}): At the magnet-gap interface the potential and normal component of B are continuous (B_{n} = μ_{0} (H_{y}+M_{s} )). These conditions give

φ_{m}^{(1)}(l_{m}) = φ_{m}^{(2)}(l_{m})    (continuity of φ_{m} ),

and

μ_{0}(-\frac{dφ_{m}^{(1)}}{dy}+M_{s})|_{y=l_{m}}  =  -μ_{0}\frac{dφ_{m}^{(2)}}{dy}|_{y=l_{m}}       (continuity of B_{n}).

From these we find that

a_{1}l_{m} = a_{2}l_{m}+B_{2},

or

B_{2} = (a_{1}-a_{2})l_{m}              (3.206)

and

–  a_{1}+ M_{s} = –  a_{2},

or

M_{s}= a_{1}-a_{2}.              (3.207)

Combining Eqs. (3.206) and (3.207) we obtain

b_{2}  =  M_{s}l_{m}.                  (3.208)

Gap-plate interface (y = g = l_{m}): At the gap-plate interface the potential is continuous, that is,

φ_{m}^{(2)}(l_{m}+g) = φ_{plate}
= 0,

which gives

a_{2}(l_{m}+g) + b_{2}= 0,

or

a_{2} = – \frac{b_{2}}{(l_{m}+g)}.              (3.209)

Finally, substitute Eq. (3.208) into Eq. (3.209) and obtain

a_{2} = – \frac{M_{s}l_{m}}{(l_{m}+g)}.                      (3.210)

Therefore, the field in the gap is

H_{g} = – \frac{dφ_{m}^{(2)}}{dy}

= M_{s}\frac{l_{m}}{(l_{m}+g)},

or

B_{g} = μ_{0} M_{s}\frac{l_{m}}{(l_{m}+g)}      (gap).          (3.211)

We can also determine the field in the magnet. Combining Eqs. (3.207) and
(3.210) we have

a_{1} = M_{s} (1- \frac{l_{m}}{(l_{m}+g)})

= M_{s}\frac{g}{(l_{m}+g)}.

Therefore,

H_{mag} = – \frac{dφ_{m}^{(1)}}{dy}

= – M_{s}\frac{g}{(l_{m}+g)} (magnet).             (3.212)

Notice that H_{mag} is in the opposite direction to the magnetization. This is the demagnetization field. Notice that this field goes to zero when the gap is reduced to zero, that is, lim_{g→0} H_{mag} = 0.