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J.D. Jackson Problem 5 - joshorndorff.com · J.D. Jackson Problem 5.19 Josh Orndor admin@joshorndor...
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J.D. Jackson Problem 5.19 Josh Orndorff admin@joshorndorff.com May 1, 2012 1 Finding H and B using magnetic scalar potential Because there is no current in this problem, we are able to apply use the magnetic scalar potential method that Jackson derives in section 5.9. the scalar potential is given by equation (5.100). Φ M (r, θ, z)= - 1 4π Z V ∇ 0 · M(x 0 ) |x - x 0 | dV 0 + 1 4π I S ˆ n 0 · M(x 0 ) |x - x 0 | da 0 (1) In this case, the magnetization is uniform and pointed along the z-axis, M = M 0 ˆ z, so the first integral is zero, as is the sidewall’s contribution to the second integral. Further, by the cylindrical symmetry of the problem, we can see that the H-field at points on the z-axis will be oriented in the ˆ z direction. So rather than expanding |x-x 0 | -1 according to equation (3.70) and finding the scalar potential everywhere, finding it along the z-axis will suffice. Φ M (0, 0,z)= 1 4π Z top ˆ z · M 0 ˆ z |x - x 0 | da 0 + 1 4π Z bottom -ˆ z · M 0 ˆ z |x - x 0 | da 0 (2) = M 0 4π Z 2π 0 Z a 0 " 1 p (z - L) 2 + r 2 - 1 √ z 2 + r 2 # r dr dφ (3) = M 0 2 Z a 0 " r p (z - L) 2 + r 2 - r √ z 2 + r 2 # dr (4) = M 0 2 h p (z - L) 2 + r 2 - p z 2 + r 2 i a 0 (5) = M 0 2 h p (z - L) 2 + a 2 - p z 2 + a 2 - p (z - L) 2 + √ z 2 i (6) In order to simplify the latter two terms, we must break this potential into a piecewise function. The latter two terms must always be positive because they represent lengths. Φ M (0, 0,z< 0) = M 0 2 h p (z - L) 2 + a 2 - p z 2 + a 2 - L i (7) Φ M (0, 0, 0 <z<L)= M 0 2 h p (z - L) 2 + a 2 - p z 2 + a 2 - L +2z i (8) Φ M (0, 0,z>L)= M 0 2 h p (z - L) 2 + a 2 - p z 2 + a 2 + L i (9) The usefulness of defining the scalar potential is that we can now find H = -∇Φ M = - ∂Φ M ∂z . H inside = M 0 2 " L - z p (z - L) 2 + a 2 + z √ z 2 + a 2 - 2 # ˆ z (10) H outside = M 0 2 " L - z p (z - L) 2 + a 2 + z √ z 2 + a 2 # ˆ z (11) 1
Transcript of J.D. Jackson Problem 5 - joshorndorff.com · J.D. Jackson Problem 5.19 Josh Orndor admin@joshorndor...
J.D. Jackson Problem 5.19
Josh [email protected]
May 1, 2012
1 Finding H and B using magnetic scalar potential
Because there is no current in this problem, we are able to apply use the magnetic scalar potentialmethod that Jackson derives in section 5.9. the scalar potential is given by equation (5.100).
ΦM (r, θ, z) = − 1
4π
∫V
∇′ ·M(x′)
|x− x′|dV ′ +
1
4π
∮S
n′ ·M(x′)
|x− x′|da′ (1)
In this case, the magnetization is uniform and pointed along the z-axis, M = M0z, so the first integralis zero, as is the sidewall’s contribution to the second integral. Further, by the cylindrical symmetry ofthe problem, we can see that the H-field at points on the z-axis will be oriented in the z direction. Sorather than expanding |x−x′|−1 according to equation (3.70) and finding the scalar potential everywhere,finding it along the z-axis will suffice.
ΦM (0, 0, z) =1
4π
∫top
z ·M0z
|x− x′|da′ +
1
4π
∫bottom
−z ·M0z
|x− x′|da′ (2)
=M0
4π
∫ 2π
0
∫ a
0
[1√
(z − L)2 + r2− 1√
z2 + r2
]r dr dφ (3)
=M0
2
∫ a
0
[r√
(z − L)2 + r2− r√
z2 + r2
]dr (4)
=M0
2
[√(z − L)2 + r2 −
√z2 + r2
]a0
(5)
=M0
2
[√(z − L)2 + a2 −
√z2 + a2 −
√(z − L)2 +
√z2]
(6)
In order to simplify the latter two terms, we must break this potential into a piecewise function. Thelatter two terms must always be positive because they represent lengths.
ΦM (0, 0, z < 0) =M0
2
[√(z − L)2 + a2 −
√z2 + a2 − L
](7)
ΦM (0, 0, 0 < z < L) =M0
2
[√(z − L)2 + a2 −
√z2 + a2 − L+ 2z
](8)
ΦM (0, 0, z > L) =M0
2
[√(z − L)2 + a2 −
√z2 + a2 + L
](9)
The usefulness of defining the scalar potential is that we can now find H = −∇ΦM = −∂ΦM
∂z .
Hinside =M0
2
[L− z√
(z − L)2 + a2+
z√z2 + a2
− 2
]z (10)
Houtside =M0
2
[L− z√
(z − L)2 + a2+
z√z2 + a2
]z (11)
1
The B-field is now given by B = µ0(H + M). Magnetization is only non-zero inside the magnet, sowhen we calculate B there, it cancels. Now the B-field is continuous everywhere and satisfies ∇ ·B = 0.
B =µ0M0
2
[L− z√
(z − L)2 + a2+
z√z2 + a2
]z (12)
2 Plots of H and B
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Field
Str
eng
th
z/a
B and H fields along the z-axis
B/u0M0H/M0
2
