Extension-Torsion-Inflation Coupling in Compressible Magnetoelastomeric Thin Tubes with Helical Magnetic Anisotropy

Abstract

An axisymmetric and axially homogenous variational formulation is presented for coupled extension-torsion-inflation deformation in compressible magnetoelastomeric tubes in the presence of azimuthal and axial magnetic fields. The tube’s material is assumed to have a preferred magnetization direction which lie in the radial plane but at an angle from the tube’s axial direction - this imparts helical magnetic anisotropy to the tube. The governing differential equations necessary to solve the above deformation problem are obtained which are shown to reduce to a set of nonlinear algebraic equations in the thin tube limit. This allows us to obtain analytical expressions in terms of the applied magnetic field, preferred magnetization direction and magnetoelastic constants which tell us how these parameters can be tuned to generate unusual coupled deformations such as negative Poisson’s effect. The study can be useful in designing magnetoelastic soft tubular actuators.

Appendices

Appendix A: Constitutive Relations

In this section, the material models used in this study are discussed.

1.1 A.1 Saint Venant-Kirchhoff Model

For linearly magnetoelastic and orthotropic solid, the constitutive relation can be written as follows in the coordinate system of principal material directions (see Pan and Heyliger [24] for details):

$$ \begin{aligned} \begin{bmatrix} \stackrel{\circ}{S}_{11} \\ \stackrel{\circ}{S}_{22} \\ \stackrel{\circ}{S}_{33} \\ \stackrel{\circ}{S}_{23} \\ \stackrel{\circ}{S}_{13} \\ \stackrel{\circ}{S}_{12} \\ \stackrel{\circ}{B}_{1} \\ \stackrel{\circ}{B}_{2} \\ \stackrel{\circ}{B}_{3} \end{bmatrix}= \begin{bmatrix} a_{11} & a_{12} & a_{13} & 0 & 0 & 0&0 & 0 & -m_{31} \\ a_{12} & a_{22}& a_{23} & 0 & 0 & 0&0 & 0 & -m_{32} \\ a_{13} &a_{23} & a_{33} & 0 & 0 & 0&0 & 0 & -m_{33} \\ 0 & 0 & 0 & G_{23} & 0 & 0&0 & -m_{24}& 0 \\ 0 & 0 & 0 & 0 & G_{13} & 0&-m_{15}& 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & G_{12}&0 &0 & 0 \\ 0 & 0 & 0 & 0 & m_{15} & 0&\mu _{11} & 0& 0 \\ 0 & 0 & 0 & m_{24} &0 & 0&0& \mu _{22} & 0 \\ m_{31} & m_{32} & m_{33} & 0 & 0 &0&0 &0 & \mu _{33} \end{bmatrix} \begin{bmatrix} \stackrel{\circ}{\epsilon} _{11} \\ \stackrel{\circ}{\epsilon} _{22} \\ \stackrel{\circ}{\epsilon} _{33} \\ 2\stackrel{\circ}{\epsilon} _{23} \\ 2\stackrel{\circ}{\epsilon} _{13} \\ 2\stackrel{\circ}{\epsilon} _{12} \\ \stackrel{\circ}{H}_{1} \\ \stackrel{\circ}{H}_{2} \\ \stackrel{\circ}{H}_{3} \end{bmatrix}. \end{aligned} $$

(70)

The elastic constants satisfy the following relations:

$$\begin{aligned} a_{ii} =&E_{i}\frac{1-\nu _{jk}\nu _{kj}}{\Delta }, ~a_{ij}=a_{ji}=E_{i} \frac{\nu _{ij}+\nu _{kj}\nu _{ik}}{\Delta },~ \left [i, j = 1,2,3;~i \ne j\ne k\ne i\right ], \\ \Delta =&1-\nu _{12}\nu _{21}-\nu _{23}\nu _{32}-\nu _{31}\nu _{13}-2 \nu _{21}\nu _{32}\nu _{13}, \\ E_{i} =& \mbox{Young's moduli} \ \left [i = 1,2,3\right ], \\ G_{ij} =&\mbox{Shear moduli}\ \left [i, j = 1,2,3;~i< j\right ], \\ \nu _{ij} =&\nu _{ji}\frac{E_{i}}{E_{j}} = \mbox{Poisson's ratios}\ \left [i, j = 1,2,3;~i\ne j\right ]. \end{aligned}$$

(71)

There are a total of 17 independent constants in the compressible case (3 Young’s moduli, 3 shear moduli, 3 Poisson’s ratios and the remaining 8 are magnetic poling constants). Due to axisymmetric deformation, $\stackrel{\circ}{\epsilon} _{12}=\stackrel{\circ}{\epsilon} _{13}= \stackrel{\circ}{B}_{1}=0$ which implies that the constants $(G_{12},G_{13},m_{15},\mu _{11})$ do not show up in the formulation. In this paper, we further confine ourselves to transversely isotropic tubes having its principal material direction along $\mathbf{i}_{3}$ (see Fig. 2). The material constants satisfy the following constraints for the case of transverse isotropy:

$$\begin{aligned} E_{1}=E_{2},\quad & G_{13}=G_{23},\quad \nu _{31}=\nu _{32},\quad E_{1} =2G_{12}\ \left [1+\nu _{12}\right ], \\ &m_{31}=m_{32},\quad m_{15}=m_{24},\quad \mu _{11}= \mu _{22}. \end{aligned}$$

(72)

Thus, $(E_{2},E_{3},G_{23},\nu _{12},\nu _{32},m_{32},m_{33},m_{24},\mu _{22}, \mu _{33})$ are the only independent constants that show up in the formulation. The elastic constants $(E_{2},\nu _{32},G_{23})$ are represented alternatively using the following parameters:

$$ \lambda =E_{2}/E_{3},\quad \nu =\nu _{32},\quad \nu _{iso}= \frac{E_{3}}{2G_{23}}-1. $$

(73)

Here, the notation $\nu _{iso}$ derives its name from the fact that it would equal $\nu _{32}$ if the tube were isotropic.

1.2 A.2 Compressible Nonlinear Material Model

To account for compressibility, the incompressible and transversely isotropic energy model of Bustamante [3] is extended to the compressible case in the following way:

$$\begin{aligned} \Omega &= \Big[\frac{g_{0}+g_{1} I_{4}}{2}\Big] \Bigg[{ \gamma }\Big[I_{1}-1-\frac{1}{{\nu }}+\frac{1-2{\nu }}{{\nu }}I_{3}^{ \frac{-{\nu }}{1-2{\nu }}}\Big] \\ &\quad +[1-\gamma ]\Big[\frac{I_{2}}{I_{3}}-1- \frac{1}{{\nu }}+\frac{1-2{\nu }}{{\nu }}I_{3}^{ \frac{{\nu }}{1-2{\nu }}}\Big]\Bigg] \\ & -m_{0} m_{1}~ \text{ln}\Bigg[\text{cosh}\Big( \frac{\sqrt{I_{4}}}{m_{1}}\Big)\Bigg]-\frac{\zeta _{0}\,I_{4}}{2}+ K_{5} I_{5} \\ &\quad +\Big[h_{0}+h_{1} ln(I_{7})-\frac{h_{1}}{m}I^{m}_{7}\Big]\Big[ \omega _{0}+\omega _{1} I_{9}^{2}+\omega _{2} I_{10}^{2}+\omega _{3} I_{9}I_{10} \Big] \\ & +m_{0}~ \text{ln}\Bigg[ \frac{\cosh ^{m_{2}}\Big(\frac{I_{9}}{m_{2}}\Big)}{\cosh ^{m_{1}}\Big(\frac{I_{9}}{m_{1}}\Big)} \Bigg]+[\zeta _{0}-\zeta _{1}-\mu _{0}]\frac{I_{9}^{2}}{2}. \end{aligned}$$

(74)

Here $I_{1}: I_{10}$ are the invariants which are defined as

$$\begin{aligned} I_{1}&=tr\,{\mathbf{C}},\quad I_{2}=\frac{1}{2}[(tr\, \mathbf{C})^{2}-tr\, {\mathbf{C}}^{2}],\quad I_{3}=\text{det}\, \,{\mathbf{C}},\quad I_{4}=\mathbf{H}_{L}\cdot \mathbf{H}_{L}, \quad I_{5}=\,{\mathbf{C}}\mathbf{H}_{L}\cdot \mathbf{H}_{L}, \\ & I_{6}=\,{\mathbf{C}}^{2}\mathbf{H}_{L}\cdot \mathbf{H}_{L}, \quad I_{7}=\,{\mathbf{C}}\mathbf{a}_{0}\cdot \mathbf{a}_{0}, \quad I_{8}=\,{\mathbf{C}}^{2}\mathbf{a}_{0}\cdot \mathbf{a}_{0},\quad I_{9}=\mathbf{H}_{L}\cdot \mathbf{a}_{0},\quad I_{10}=\,{\mathbf{C}}\mathbf{H}_{L}\cdot \mathbf{a}_{0} \end{aligned}$$

(75)

and $\mathbf{a}_{0}$ is the preferred magnetization direction in the reference configuration. The corresponding incompressible model can be obtained from equation (74) in the limit of ${\nu }\rightarrow 0.5$. Accordingly, $\nu$ has the physical meaning of the Poisson’s ratio. It is to be noted that for an isotropic material model, only the invariants $I_{1}:I_{6}$ would be active in equation (74). Two special cases of the above model correspond to the parameter $\gamma$ being zero and unity. We call the case of $\gamma =1$ to be the Neo-Hookean model (NH) while $\gamma =0$ to be the Mooney-Rivlin model (MR). The various constants in equation (74) are taken from Bustamante [3] which are:

$g_{0}$	100 kPa	$\omega _{0}$	1334.296 kPa
$g_{1}$	$-0.001\ kPa/(kA/m)^{2}$	$\omega _{1}$	$-0.063748\ kPa/(kA/m)^{2}$
$m_{0}$	0.4998 N/(A.m)	$\omega _{2}$	$0.198605\ kPa/(kA/m)^{2}$
$m_{1}$	309.3395 kA/m	$\omega _{3}$	$-0.228273\ kPa/(kA/m)^{2}$
$m_{2}$	199.1828 kA/m	m	−10
$\mu _{0}$	$1.2566\times 10^{-6}\ N/A^{2}$	$\zeta _{0}$	$0.5\times \mu _{0}\ N/A^{2}$
$h_{0}$	0.002881	$\zeta _{1}$	0
$h_{1}$	0.012741	$K_{5}$	$\frac{\mu _{0}}{2}$

It is also noteworthy to mention the work of Shariff et al. [32] who propose a material model for transversely isotropic magnetoelastomers in terms of a different set of spectral invariants. It is noted there that the new set of invariants have better physical meaning and that the experimental characterization of a magnetoelastomer in terms of these new invariants are easier compared to the classical invariants in equation (75).

Appendix B: First Derivatives of $\tilde{\Phi }$ with Respect to $(\epsilon ,\kappa ,\beta ,\zeta ,\eta )$

We now obtain the expressions for first derivatives of $\tilde{\Phi }$ with respect to imposed parameters. Let us suppose $m$ denotes any of the imposed parameters $(\epsilon ,\kappa ,\beta ,\zeta , \eta )$. We then have

$$\begin{aligned} \frac{\partial {\tilde{\Phi }}}{\partial {m}}&= 2\pi \int _{R_{1}}^{R_{2}}\Big[ \frac{\partial \Omega }{\partial {\mathbf{F}}}: \frac{\partial {\mathbf{F}}}{\partial {m}}+ \frac{\partial \Omega }{\partial {\mathbf{H}_{L}}}\cdot \frac{\partial {\mathbf{H}_{L}}}{\partial {m}}\Big] R dR +\mathcal{C} \\ & = 2 \pi \int _{R_{1}}^{R_{2}}\Big[{\mathbf{T}}:\boldsymbol{\nabla }\left [\frac {\partial \mathbf {x}}{\partial m}\right ]- \mathbf{B}_{L}\cdot \frac{\partial {\mathbf{H}_{L}}}{\partial {m}}\Big] R dR+\mathcal{C} \\ & =2\pi \int _{R_{1}}^{R_{2}} \Bigg[\boldsymbol{\nabla }\cdot \left [{\mathbf{T}}^{T} \frac {\partial \mathbf {x}}{\partial m}\right ]-\left [\boldsymbol{\nabla }\cdot {\mathbf{T}}\right ]\cdot \frac {\partial \mathbf {x}}{\partial m}\Bigg] R\,dR \\ &\quad -2\pi \int _{R_{1}}^{R_{2}} \Bigg[B_{L \Theta } \frac{\partial {{H}}_{L\Theta }}{\partial {m}}+B_{L Z} \frac{\partial {{H}}_{L Z}}{\partial {m}}\Bigg] R dR+\mathcal{C} \\ &=2\pi \int _{R_{1}}^{R_{2}} \boldsymbol{\nabla }_{\alpha } \cdot \left [{\mathbf{T}}^{T} \frac {\partial \mathbf {x}}{\partial m}\right ] R\,dR ~+2\pi \int _{R_{1}}^{R_{2}} \frac {\partial }{\partial Z}\Big[{\mathbf{T}}^{T} \frac {\partial \mathbf {x}}{\partial m} \Big] \cdot {\mathbf{e}}_{Z} ~R\,dR \\ &\quad -2\pi \int _{R_{1}}^{R_{2}} \Bigg[B_{L \Theta } \frac{\partial {{H}}_{L\Theta }}{\partial {m}}+B_{L Z} \frac{\partial {{H}}_{L Z}}{\partial {m}}\Bigg] R dR+\mathcal{C} \quad \bigg[\because \boldsymbol{\nabla }\cdot {\mathbf{T}}=\mathbf{0},\quad {B}_{LR}={0} \bigg] \\ &=2\pi \left [R~{\mathbf{T}}{\mathbf{e}}_{R} \cdot \frac {\partial \mathbf {x}}{\partial m}\right ] \bigg\vert _{R_{1}}^{R_{2}} \\ &\quad +2\pi \int _{R_{1}}^{R_{2}} \Bigg[\frac {\partial }{\partial Z}\bigg[{\mathbf{T}}{\mathbf{e}}_{Z} \cdot \frac {\partial \mathbf {x}}{\partial m}\bigg]-B_{L \Theta }\frac{\partial {{H}}_{L\Theta }}{\partial {m}}-B_{L Z} \frac{\partial {{H}}_{L Z}}{\partial {m}}\Bigg] ~R\,dR +\mathcal{C}. \end{aligned}$$

(76)

Here $\boldsymbol{\nabla }$ denotes the Lagrangian gradient operator, $\boldsymbol{\nabla }\cdot$ the Lagrangian divergence operator and $\boldsymbol{\nabla }_{\alpha }\cdot$ the Lagrangian divergence operator in the cross-sectional plane. The last equality above was obtained after applying Gauss-divergence theorem in the cross-sectional plane and further using axisymmetry. The quantity $\mathcal{C}$ in the above equation is given by

$$\begin{aligned} \mathcal{C}&=\Bigg[\pi \mu _{0} [1+\epsilon ]{\mathbf{H}^{*}}\cdot { \mathbf{H}^{*}}~r\frac{\partial \,r}{\partial \,m} \Bigg]\Bigg\vert _{r_{1}}^{r_{2}}+ \pi \mu _{0}\int _{r_{1}}^{r_{2}}\frac{\partial }{\partial m} \Big[ [1+\epsilon ]{\mathbf{H}^{*}}\cdot {\mathbf{H}^{*}}\Big]r\,dr \\ &=-\Bigg[2\pi T_{{rR}}^{m^{*}}R^{2} \frac{\partial \tilde{b}}{\partial m}\Bigg]\Bigg\vert _{R_{1}}^{R_{2}}+ \pi \mu _{0}\int _{r_{1}}^{r_{2}}\frac{\partial }{\partial m} \Big[ [1+\epsilon ]{\mathbf{H}^{*}}\cdot {\mathbf{H}^{*}}\Big]r\,dr. \end{aligned}$$

(77)

The first term arises here due to change in the domain of integration while the second term is due to change in integrand itself.

Let us now set $m=\epsilon$. We then have $\frac {\partial \mathbf {x}}{\partial \epsilon }=Z\mathbf{e}_{z} +R \frac {\partial \tilde{b}}{\partial \epsilon }{\mathbf{e}}_{r}$ and $\frac {\partial \mathbf {H}_{L}}{\partial \epsilon }=\eta {\mathbf{e}}_{Z}$. Substituting this in (76) leads to

$$\begin{aligned} \frac{\partial {\tilde{\Phi }}}{\partial {\epsilon }}&=2\pi \,\Big[T_{rR}R^{2} \frac{\partial \tilde{b}}{\partial \epsilon }\Big]\Big\vert _{R_{1}}^{R_{2}}+2 \pi \int _{R_{1}}^{R_{2}}\frac{\partial }{\partial Z}[Z T_{zZ}] RdR-2\pi \int _{R_{1}}^{R_{2}}\eta B_{LZ}\,RdR \\ &\quad -2\pi \,\Big[T_{rR}^{m^{*}}R^{2} \frac{\partial \tilde{b}}{\partial \epsilon }\Big]\Big\vert _{R_{1}}^{R_{2}}+{ \pi \,\mu _{0}} \int _{r_{1}}^{r_{2}}{\mathbf{H}^{*}}\cdot { \mathbf{H}^{*}} ~r\,dr \\ &= 2\pi \,\Big[\left (T_{rR}-T_{rR}^{m^{*}}\right )R^{2} \frac{\partial \tilde{b}}{\partial \epsilon }\Big]\Big\vert _{R_{2}}+2 \pi \int _{R_{1}}^{R_{2}}\,T_{zZ} R\,dR \\ &\quad -2\pi \int _{R_{1}}^{R_{2}} \eta B_{LZ} R\,dR +{\pi \,\mu _{0}}\int _{r_{1}}^{r_{2}}{ \mathbf{H}^{*}}\cdot {\mathbf{H}^{*}} ~r\,dr \\ & =2\pi \int _{R_{1}}^{R_{2}} \tilde{T}\,^{p}_{{zZ}}\,RdR. \end{aligned}$$

(78)

In (78a), the first and the fourth terms evaluated at the inner radius vanish as $\frac {\partial \tilde{b}_{1}}{\partial \epsilon }=0$. Also these two terms when evaluated at the outer radius cancel each other, after using ${T_{rR_{2}}=T^{m*}_{rR_{2}}}$. The quantity $T_{zZ}$ in the second term of (78b) has also been decomposed into its corresponding mechanical and Maxwell contributions in order to obtain the final expression as shown above.

For $m=\kappa$, we have $\frac {\partial \mathbf {x}}{\partial \kappa }=R\frac {\partial \tilde{b}}{\partial \kappa }{\mathbf{e}}_{r} + \left [1+b\right ]RZ\mathbf{e}_{\theta }$ and $\frac {\partial \mathbf {H}_{L}}{\partial \kappa }=\zeta {\mathbf{e}}_{Z}$. Substituting this in (76) leads to

$$\begin{aligned} \frac{\partial \tilde{\Phi }}{\partial \kappa }&=2\pi \, \Big[T_{rR}R^{2}\frac{\partial \tilde{b}}{\partial \kappa }+[1+b]R^{2} \,Z\,T_{\theta \,R}\Big] \Big\vert _{R_{1}}^{R_{2}} \\ &\quad +2\pi \int _{R_{1}}^{R_{2}}\,\frac{\partial }{\partial Z}\Big[R \frac{\partial \tilde{b}}{\partial \kappa }T_{rZ}+[1+b]R\,Z\,T_{ \theta \,Z}\Big]RdR \\ &\quad -2\pi \int _{R_{1}}^{R_{2}}\zeta \,B_{LZ} RdR-2\pi \Bigg[T_{{rR}}^{m^{*}} R^{2} \frac{\partial \tilde{b}}{\partial \kappa }\Bigg]\Bigg\vert _{R_{1}}^{R_{2}} \\ & =2\pi \int _{R_{1}}^{R_{2}}\,[1+b]\quad {\,T_{{ \theta \,Z}}}R^{2}dR-2\pi \int _{R_{1}}^{R_{2}} \frac{\zeta }{r} B_{LZ}[1+b]R^{2}\,dR \\ & =2\pi \int _{R_{1}}^{R_{2}} R^{2} \big[1+{b}\big]{T}\,^{p}_{{\theta Z}} \,dR. \end{aligned}$$

(79)

The boundary terms vanish here for the same reason as in (78) together with the fact that $T_{\theta R}$ vanishes at the boundaries. Further, the term involving $T_{rZ}$ becomes zero when axial homogeneity is imposed. We have also decomposed $T_{\theta Z}$ in the first term of (79b) into its corresponding mechanical and Maxwell contributions.

For m= $\beta$, we then have $\displaystyle{ \frac {\partial \mathbf {x}}{\partial \beta }=R\frac {\partial \tilde{b}}{\partial \beta }{\mathbf{e}}_{r}}$ and $\displaystyle{ \frac {\partial \mathbf {H}_{L}}{\partial \beta }=0}$ which implies

$$\begin{aligned} \frac{\partial \tilde{\Phi }}{\partial \beta } &=2\pi \,\Big[T_{rR}R^{2} \frac{\partial \tilde{b}}{\partial \beta }\Big]\Big\vert _{R_{1}}^{R_{2}} -2\pi \Big[T_{{rR}}^{m^{*}} R^{2} \frac{\partial b}{\partial \beta } \Big]\Big\vert _{R_{1}}^{R_{2}} \\ & =\displaystyle {-2\pi R_{1}^{2}\left [T_{rR_{1}}-T_{rR_{1}}^{m^{*}} \right ]} =-2\pi {R_{1}}^{2}\,T^{p^{*}}_{{rR_{1}}}. \end{aligned}$$

(80)

The boundary term at inner radius survives here since $\tilde{b}_{1}=\beta$ implies $\frac {\partial \tilde{b}_{1}}{\partial \beta }=1$ whereas the boundary term at outer radius vanishes for the same reason as earlier.

For $m=\zeta$, we have $\displaystyle{ \frac {\partial \mathbf{x}}{\partial \zeta }=R\, \frac{\partial \tilde{b}}{\partial \zeta }{\mathbf{e}}_{r}}$ and $\displaystyle{ \frac {\partial \mathbf {H}_{L}}{\partial \zeta }=\frac{1}{R} {\mathbf{e}}_{\Theta }+ \kappa {\mathbf{e}}_{Z}}$. Substituting this in (76), we obtain:

$$\begin{aligned} \displaystyle {\frac{\partial \tilde{\Phi }}{\partial \zeta }}&=2\pi \Big[T_{rR}R^{2}\frac{\partial \tilde{b}}{\partial \zeta }\Big] \Big\vert _{R_{1}}^{R_{2}}+2\pi \int _{R_{1}}^{R_{2}}\, \frac{\partial }{\partial Z}\Big[R \frac{\partial \tilde{b}}{\partial \zeta }T_{rZ}\Big]R\,dR-2\pi \int _{R_{1}}^{R_{2}}\left [B_{L\,\Theta }+\kappa \,B_{L\,Z}R \right ]dR \\ &\quad -2\pi \Big[T_{{rR}}^{m^{*}}R^{2} \frac{\partial b}{\partial \zeta } \Big]\Big\vert _{R_{1}}^{R_{2}}+{2\pi \,\mu _{0}[1+\epsilon ]}\int _{r_{1}}^{r_{2}}H^{*}_{\theta }\,dr \\ & = -2\pi [1+\epsilon ]\int _{r_{1}}^{r_{2}}B_{ \theta }dr+{2\pi \,\mu _{0}[1+\epsilon ]}\int _{r_{1}}^{r_{2}}H_{ \theta }^{*}\,dr \\ & =-2\pi [1+\epsilon ]\int _{r_{1}}^{r_{2}} \left [B_{\theta }-\mu _{0} H_{\theta }^{*}\right ] dr. \end{aligned}$$

(81)

Again, the boundary terms vanish here for the same reason as in (78). Finally, for $m=\eta$, we have $\displaystyle{ \frac {\partial \mathbf{x}}{\partial \eta }=R\, \frac{\partial \tilde{b}}{\partial \eta }{\mathbf{e}}_{r}}$ and $\displaystyle{ \frac {\partial \mathbf {H}_{L}}{\partial \eta }=[1+\epsilon ] {\mathbf{e}}_{Z}}$. Substituting this in (76), we obtain:

$$\begin{aligned} \frac{\partial \tilde{\Phi }}{\partial \eta }&=2\pi \, \Big[T_{rR}R^{2}\frac{\partial \tilde{b}}{\partial \eta }\Big] \Big\vert _{R_{1}}^{R_{2}}+2\pi \int _{R_{1}}^{R_{2}}\, \frac{\partial }{\partial Z}\Big[R \frac{\partial \tilde{b}}{\partial \eta }T_{rZ}\Big]RdR-2\pi \,[1+ \epsilon ]\int _{R_{1}}^{R_{2}}B_{LZ}RdR \\ &\quad -2\pi \Bigg[ T_{{rR}}^{m^{*}} R^{2} \frac{\partial b}{\partial \eta } \Bigg]\Bigg\vert _{R_{1}}^{R_{2}}+{2\pi \,\mu _{0}[1+\epsilon ]}\int _{r_{1}}^{r_{2}}H^{*}_{z}~r\,dr \\ &= -2\pi [1+\epsilon ]\int _{r_{1}}^{r_{2}}\,B_{z}~rdr+2 \pi \,\mu _{0}[1+\epsilon ]\int _{r_{1}}^{r_{2}}H^{*}_{z}~r \,dr \\ & = -2\pi [1+\epsilon ]\int _{r_{1}}^{r_{2}} \left [B_{z}-\mu _{0} H_{z}^{*}\right ]r dr. \end{aligned}$$

(82)

Here again, the boundary terms vanish for the same reason as in (78).