Question 3.7.4: Compute the field above a rectangular bar magnet of width 2a...
Compute the field above a rectangular bar magnet of width 2a, depth 2b and length L, which is polarized along its axis with uniform magnetization M = M_{s}\hat{z}. The magnet is resting on an infinite plate that has an infinite permeability μ ≈ ∞ (Fig. 3.47a).
We replace the plate by an image magnet as shown in Fig. 3.47b. The image magnet is identical to the source magnet, and from symmetry we know that the field due to the two magnets will satisfy the boundary conditions (3.301) at the interface. The resulting magnet structure is equivalent to a single magnet of length 2L as shown in Fig. 3.47c. The field for this structure is given by Eq. (3.114) with L replaced by 2L,
Eq. (3.114): B_{z}(z)= \frac{μ_{0}M_{s}}{π} [tan^{-1} (\frac{(z+L)\sqrt{a^{2}+b^{2}+(z+L)^{2}}}{ab}) – tan^{-1} (\frac{z\sqrt{a^{2}+b^{2}+z^{2}}}{ab})].
B_{z}(z)= \frac{μ_{0}M_{s}}{π} [tan^{-1}(\frac{(z+2L)\sqrt{a^{2}+b^{2}+(z+2L)^{2}}}{ab}) – tan^{-1}(\frac{z\sqrt{a^{2}+b^{2}+z^{2}}}{ab})]. (3.305)