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Derivation of the external magnetic field of a

homogeneously magnetized cylinder Here, we derive the external magnetic field of a homogeneously magnetized cylinder of infinite length. The cylinder is magnetized due to an constant magnetic field perpendic- ular to it. The applied field is high, so remanent magnetizationcan be neglected, and no currents are present. Then, the internal and external magnetic fields,H, and inductions, B, are found by solving the Laplace equation of the magnetic scalar potentialφm[1].

2φm= 0 (B.1)

with:

H=-?φm(B.2)

and:

B=μ0(-?φm+M) (B.3)

withμ0the permeability of free space. Because the cylinder is perpendicular to the applied field, the potential is independent of the coordinatey, with the ˆy-direction in the length direction of the cylinder. The general solutions of the Laplace equation are the so-called cylindrical harmonics:

φm=A+Bln(r) +∞?

n=0{Cnrncos(nφ) +Dnr-ncos(nφ) +Enrnsin(nφ) +Fnr-nsin(nφ)} (B.4) Now, the constantA,B,C,D,E,Fhave to be found for the regions inside (φm1) and outside (φm2) the cylinder. This is done by applying several boundary conditions.

1. The far field is parallel to cos(φ)ˆrand is not influenced by the disturbing cylinder,

which means: lim

2. The cylinder is homogeneously magnetized, which means:

M

1=M1(H1)cos(φ)ˆr(B.6)

91

92Appendix B.

3. The tangential component ofHis continuous across any surface:

H

1φ=H2φ(B.7)

withHiφ=Hi·ˆφ.

4. The normal component ofBis continuous across the surface of the cylinder:

B

1r=B2r(B.8)

withBir=Bi·ˆr. Because the magnetic potential is uniform and cannot get infinite at any point and by applying boundary condition 1, the external potential is written as: m2=∞? n=0{D2nr-ncos(nφ) +F2nr-nsin(nφ)} -B0

μ0μ2rcos(φ) (B.9)

The internal potential is only restricted by its uniformity: m1=∞? n=0{C1nrncos(nφ) +E1nrnsin(nφ)}(B.10) Now, the boundary conditions 2, 3 and 4 are applied at the cylinder surfacer=a. Only then= 1 components have to be taken into account, because of the cos(φ) term.

Boundary condition 3 becomes:

m1 or: -C11asin(φ) +E11acos(φ) =-D21a-1sin(φ) +F21a-1cos(φ) +B0

μ0μ2asin(φ) (B.12)

Boundary condition 4 becomes:

0(-δφm1

δr|r=a+M1cos(φ)) =-μ0μ2δφ

m2δr|r=a(B.13) or: μ0(-C11cos(φ)-E11sin(φ)+M1cos(φ)) =μoμ2(D21a-2cosφ-F21a-2sin(φ))+B0cos(φ) (B.14)

The coefficients are found to be:

C 11=M1

1 +μ2-2B0μ0(1 +μ2)(B.15)

D

21=M1a2

1 +μ2+B0(1-μ2)a2μ0(1 +μ2)μ2(B.16)

Magnetic field of a homogeneously magnetized cylinder93 E

21= 0 (B.17)

F

21= 0 (B.18)

The magnetic potential equations become:

m1=μ0M1-2B0

μ0(1 +μ2)rcos(φ) (B.19)

m2=μ0μ2M1+B0(1-μ2) μ0μ2(1 +μ2)cos(φ)a2r-B0μ0μ2rcos(φ) (B.20) Now, a change it made to cartesian coordinates, withz=rcos(φ) andx=rsin(φ): m1=μ0M1-2B0

μ0(1 +μ2)z(B.21)

m2=μ0μ2M1+B0(1-μ2)

μ0μ2(1 +μ2)a

2x2+z2z-B0μ0μ2z(B.22)

The magnetic field equations become:

H

1=-δφm1

δzˆz=-μ0M1-2B0μ0(1 +μ2)ˆz(B.23)

H

2=-δφm2

δxˆx-δφm2δzˆz(B.24)

μ0μ2M1+B0(1-μ2)

μ0μ2(1 +μ2)a2(z2-x2(x2+z2)2ˆz+2xz(x2+z2)2ˆx) +B0μ0μ2ˆz and the magnetic induction: B

1=-μ0δφ

m1 δzˆz+μ0M1ˆz=-μ0μ2M1-2B01 +μ2ˆz(B.25) B

2=μ0μ2(-δφm2

δxˆx-δφm2δzˆz)(B.26)

0μ2M1+B0(1-μ2)

1 +μ2a2(z2-x2(x2+z2)2ˆz+2xz(x2+z2)2ˆx) +B0ˆz

In the MR-scanner, the applied field around the iso-center is constant. During scan- ning, only very small, negligible variations around the mainfield are present. In that case, we may writeM1as proportional toB0using the dimensionless, field dependent parameterξ(B0): M

1=ξ(B0)B0

μ0μ2(B.27)

Now, equation B.26 can be written as:

B

2=ξ-χ2

π(2 +χ2)B0A?z2-x2(x2+z2)2ˆz+2xz(x2+z2)2ˆx? +B0ˆz(B.28)

94Appendix B.

withχ2=μ2-1 the magnetic susceptibility of the environment and A the cross-sectional area of the cylinder. Forχ2<<1, it becomes: B

2≈ξ-χ2

+B0ˆz(B.29) In MRI small frequency differences around the Larmor frequencyare analyzed. The frequency differences depend on the field variations induced bythe disturbing cylinder Δf= 2πγΔB, with ΔB=|B2| - |B0|andγthe gyromagnetic ratio. |B2|=B0?

1 +(ξ-χ2)A

2πz

2-x2(x2+z2)2?

2+?(ξ-χ2)A2π2xz(x2+z2)2?

2?1

2(B.30)

=B0?

1 +(ξ-χ2)A

πz

2-x2(x2+z2)2+ ((ξ-χ2)A2π)21(x2+z2)2?

1

2(B.31)

?B0?

1 +(ξ-χ2)A

πz

2-x2(x2+z2)2?

1

2(B.32)

?B0?

1 +(ξ-χ2)A

2πz

2-x2(x2+z2)2?

(B.33) Above, at the first approximation, it is assumed that the cross-sectional area of the cylinder is small compared to the field-of-view of the imaging slice, which implies that frequency differences close to the cylinder have negligible effect on the total distortion.

At the second, a Taylor-expansion of⎷

1 +u≈1 +u2-u28...is used. ΔBbecomes:

ΔB(x,z) =B0(ξ-χ2)A

2πz

2-x2(x2+z2)2(B.34)

References

1. Reitz JR, Milford FJ, Christy RW. Foundations of Electromagnetic Theory. Reading, MA

USA: Addison-Wesley Publishing Company, 4

thedition, 1993.quotesdbs_dbs19.pdfusesText_25