Question 3.5.3: Consider the circuit of Fig. 3.30. Derive an expression for ...

Apply Eq. (3.138) to a path around the circuit and obtain

Eq. (3.138): \oint_{c}H· dl = I_{tot},

\oint_{c}H· dl = H_{m}l_{m}+ H_{core}l_{c} +H_{g}l_{g} = 0,        (3.171)

or

H_{core}l_{c} +H_{g}l_{g} =  – H_{m}l_{m},         (3.172)

where H_{m},  H_{core}, and H_{g} are the fields in the magnet, core, and gap, respectively, and l_{m},  l_{c}, and l_{g} are the lengths of the paths in the respective sections. Here, I_{tot} = 0 and all of the flux is supplied by the magnet. In fact, the magnet can be replaced by an equivalent source equal to – H_{m}l_{m} (because the magnet operates in the second quadrant, H_{m} is negative and – H_{m}l_{m} is positive). We compare Eq. (3.172) with Eq. (3.155) and find that – H_{m}l_{m} plays the role of ni. However, H_{m} is not known a priori. Instead, it depends on the operating point of the magnet.

Eq. (3.155): H_{c}l_{c}  +  H_{g}l_{g} = ni

Operating point: The operating point ( H_{m},  B_{m},) is determined as follows. First, as the core has a high permeability, its field H_{core} is negligible compared to H_{m} and H_{g}. Thus, the second term on the right-hand side of Eq (3.171) can be ignored, which gives

H_{g}l_{g} = – H_{m}l_{m}.            (3.173)

It follows that

B_{g} = – μ_{0} H_{m} \frac{l_{m}}{l_{g}},            (3.174)

where we have used B_{g} = μ_{0} H_{g}. Because there is no flux leakage, Φ_{m}=  Φ_{m} or

B_{m}A_{m} = B_{g}A_{g}.                    (3.175)

Combining Eqs. (3.174) and (3.175) we obtain

B_{m} = \underbrace{- μ_{0} \frac{A_{g}}{A_{m}}  \frac{l_{m}}{l_{g}} H_{m}}_{slope  of  load  line}        (load line).            (3.176)

Equation (3.176) defines the magnet’s load line. This line has a slope equal to – μ_{0} (A_{g}/A_{m})(l_{m}/l_{g}) as shown in Fig. 3.31. The operating point is the intersection of the load line with the second quadrant demagnetization curve (3.170)

\underbrace{B_{r}+ μ_{m} H_{m}}_{second  quadrant \\ demagnetization  \\ curve} = \underbrace{- μ_{0} \frac{A_{g}}{A_{m}}  \frac{l_{m}}{l_{g}} H_{m}}_{load  line} .            (3.177)

This gives

H_{m} = \frac{-B_{r}}{(B_{r}/H_{c}+μ_{0} (A_{g}/A_{m})(l_{m}/l_{g}))}.              (3.178)

Notice that as the gap goes to zero, the H-field in the magnet goes to zero, that is, lim_{l_{g}→0}H_{m} = 0. As the gap goes to infinity, H_{m} approaches the coercivity lim_{l_{g}→∞}H_{m} = -H_{c}. From Eq. (3.178) we determine the equivalent source term

– H_{m}l_{m} = \frac{l_{m}B_{r}}{(B_{r}/H_{c}+μ_{0} (A_{g}/A_{m})(l_{m}/l_{g}))}             (equivalent source).

Gap field: To determine the field in the gap, substitute Eq. (3.178) into Eq (3.174) which gives

B_{g} = \frac{μ_{0}B_{r}}{(B_{r}/H_{c}+μ_{0} (A_{g}/A_{m})(l_{m}/l_{g}))} \frac{l_{m}}{l_{g}},

or

B_{g} = \frac{μ_{0}B_{r}}{((l_{g}/l_{m})(B_{r}/H_{c})+μ_{0}(A_{g}/A_{m}))}.              (3.179)

Gap energy: The magnetostatic energy density in the gap is given by Eq. (3.68),

Eq. (3.68): ω_{m} = \frac{1}{2} B ⋅ H  (J/m^{3}).

ω_{m} = \frac{1}{2} B_{g} ⋅ H_{g}.

Because B_{g} and H_{g} are colinear and constant throughout the gap, the total energy is

W_{m} = \frac{1}{2} B_{g}H_{g}l_{g}A_{g},        (3.180)

where l_{g}A_{g} is the volume of the gap. It is instructive to represent W_{m} in terms of the magnet’s field. To this end, substitute Eqs. (3.173) and (3.175) into Eq. (3.180) and obtain

W_{m} = -\frac{1}{2} B_{m}H_{m}l_{m}A_{m}

= -\frac{1}{2} B_{m}H_{m} × (volume of magnet).         (3.181)

Notice that W_{m} > 0 because H_{m} < 0. Thus, for a fixed volume of magnet (volume = l_{m}A_{m}), the gap energy is maximized when the magnet’s energy product B_{m}H_{m} is a maximum (B_{m}H_{m})_{max}