Magnetization reversal and coercivity mechanism in truncated conical nanowires of permalloy

Abstract

Magnetization reversal mechanism and stable remanent states of truncated conical nanowire of high aspect ratio are examined using micromagnetic simulation tool. The length of nanowire is fixed at $1\,\upmu$m, while its base radius R is varied from 50 to 10 nm and top radius r is varied $R-$10 nm to $R-$5 nm. The axial magnetization reversal proceeds predominantly through domain wall motion which is hindered as the tapering of the conical nanowire is increased. At remanence, a homogeneous magnetization persists throughout the nanowire except at the ends. Vortex states appear at both the ends in case of nanowires close to cylindrical shape, whereas a flower state of magnetization is exhibited at the tapered end if r is as small as $\le 20$ nm. Control of magnetization reversal through propagation of domain wall can be observed as tapered end with flower configuration successfully restricts the movement of domain wall. Expressions for total energy density of both vortex-vortex state and vortex-flower state are constructed, and the obtained values are compared with homogeneous configuration to understand the extent of stability. The transition of remanent state is successfully obtained by energy calculations. Angular dependency of coercivity shows the existence of a critical angle above which the coercivity follows Stoner–Wohlfarth model. Below the critical angle, the coercivity mechanism is explained by Kondorsky model if R is large, whereas a modified Kondorsky model is applicable for small base radius.

Appendices

Appendix A Total energy density for vortex-vortex state

In order to include the effect of top and bottom surface, the continuum function is divided into two parts for each half of the wire.

$$\begin{aligned} \cos \theta (\rho ,z) = \left\{ \begin{array}{ c l } N_{0}e^{-2\rho ^n[\beta _{0}(1-\frac{2z}{h})^\eta ]^n}+(1-N_{0}) &{} \quad 0\le z\le \frac{h}{2} \\ N_{h}e^{-2\rho ^n[\beta _{h}(\frac{2z}{h}-1)^\lambda ]^n}+(1-N_{h}) &{} \quad \frac{h}{2} \le z\le h \end{array} \right. \end{aligned}$$

Here $\eta$ and $\lambda$ are parameters useful to control the effective length of the vortex on bottom and top surface, respectively.

Now the exchange energy can be calculated by using

$$\begin{aligned} E_{\mathrm{ex}}=A\int _V \left[ \left( \vec {\nabla } m_{x}\right) ^2+\left( \vec {\nabla } m_{y}\right) ^2+\left( \vec {\nabla } m_{z}\right) ^2\right] dV \end{aligned}$$

(A1)

On the curved surface, the variation of z is given as $z=\left( \frac{R-\rho }{R-r}\right) h$ (See Fig. 18). Now we use Eq. 3 and solve to give

$$\begin{aligned} E_{\mathrm{ex}}= & {} A\int _{0}^{R-\left( \frac{R-r}{h}\right) z} \rho \text {d}\rho \int _{0}^{2\pi } \text {d}\varphi \nonumber \\{} & {} \int _{0}^{h} \text {d}z \left[ \left( \frac{\partial \theta }{\partial \rho }\right) ^2+\frac{\sin ^2\theta }{\rho ^2}+\left( \frac{\partial \theta }{\partial z}\right) ^2\right] \end{aligned}$$

(A2)

For bottom half $0\le z\le \frac{h}{2}$

$$\begin{aligned} \left( \frac{\partial \theta }{\partial \rho }\right)= & {} \frac{2 n \rho ^{n-1}\beta _{0}^n (1-\frac{2z}{h})^{n\eta } N_{0}e^{-2\rho ^n\beta _{0}^n(1-\frac{2z}{h})^{n\eta }}}{\sqrt{1-\left[ N_{0}e^{-2\rho ^n[\beta _{0}(1-\frac{2z}{h})^\eta ]^n}+(1-N_{0})\right] ^2}} \end{aligned}$$

(A3)

$$\begin{aligned} \left( \frac{\partial \theta }{\partial z}\right)= & {} \frac{-\frac{4}{h} n \eta \rho ^n\beta _{0}^n \left( 1-\frac{2z}{h}\right) ^{n\eta -1} N_{0}e^{-2\rho ^n\beta _{0}^n\left( 1-\frac{2z}{h}\right) ^{n\eta }}}{\sqrt{1-\left[ N_{0}e^{-2\rho ^n\left[ \beta _{0}\left( 1-\frac{2z}{h}\right) ^\eta \right] ^n}+\left( 1-N_{0}\right) \right] ^2}} \end{aligned}$$

(A4)

For top half $\frac{h}{2} \le z\le h$

$$\begin{aligned} \left( \frac{\partial \theta }{\partial \rho }\right)= & {} \frac{2 n \rho ^{n-1} \beta _{h}^n \left( \frac{2z}{h}-1\right) ^{n \lambda } N_{h}e^{-2\rho ^n\beta _{h}^n\left( \frac{2z}{h}-1\right) ^{n\lambda }}}{\sqrt{1-\left[ N_{h}e^{-2\rho ^n\left[ \beta _{h}\left( \frac{2z}{h}-1\right) ^\lambda \right] ^n}+(1-N_{h})\right] ^2}} \end{aligned}$$

(A5)

$$\begin{aligned} \left( \frac{\partial \theta }{\partial z}\right)= & {} \frac{\frac{4}{h} n \lambda \rho ^{n} \beta _{h}^n \left( \frac{2z}{h}-1\right) ^{n \lambda -1} N_{h}e^{-2\rho ^n\beta _{h}^n\left( \frac{2z}{h}-1\right) ^{n\lambda }}}{\sqrt{1-\left[ N_{h}e^{-2\rho ^n\left[ \beta _{h}\left( \frac{2z}{h}-1\right) ^\lambda \right] ^n}+\left( 1-N_{h}\right) \right] ^2}} \end{aligned}$$

(A6)

By substituting in Eq. A2

$$\begin{aligned} E_{\mathrm{ex}}= & {} 2\pi A\int _{0}^{\rho _{\mathrm{lim}}} \rho \text {d}\rho \int _{0}^{\frac{h}{2}} \text {d}z \nonumber \\{} & {} \Bigg [\left( \frac{2n\rho ^{n-1}\beta _{0}^n \left( c_{1}\right) ^{n\eta }f_{1}}{\sqrt{1-f_{2}^2}}\right) ^2+\frac{1}{\rho ^2}\left( 1-f_{2}^2\right) \nonumber \\{} & {} +\left( \frac{-\frac{4}{h}n \eta \rho ^{n}\beta _{0}^n \left( c_{1}\right) ^{n\eta -1}f_{1}}{\sqrt{1-f_{2}^2}}\right) ^2\Bigg ] \nonumber \\{} & {} + 2\pi A\int _{0}^{\rho _{\mathrm{lim}}} \rho \text {d}\rho \int _{\frac{h}{2}}^{h} \text {d}z \nonumber \\{} & {} \Bigg [\left( \frac{2n\rho ^{n-1}\beta _{h}^n \left( c_{2}\right) ^{n\lambda }f_{3}}{\sqrt{1-f_{4}^2}}\right) ^2+\frac{1}{\rho ^2}\left( 1-f_{4}^2\right) \nonumber \\{} & {} +\left( \frac{\frac{4}{h}n \lambda \rho ^{n}\beta _{h}^n \left( c_{2}\right) ^{n\lambda -1}f_{3}}{\sqrt{1-f_{4}^2}}\right) ^2\Bigg ] \end{aligned}$$

(A7)

where $\rho _{\mathrm{lim}}=R-\left( \frac{R-r}{h}\right) z$, $c_{1}=1-\frac{2z}{h}$, $c_{2}=\frac{2z}{h}-1$, $f_{1}=N_{0}e^{-2\rho ^n\beta _{0}^n\left( c_1\right) ^{\eta n}}$, $f_{2}=N_{0}e^{-2\rho ^n\beta _{0}^n\left( c_1\right) ^{\eta n}}+\left( 1-N_{0}\right)$, $f_{3}= N_{h}e^{-2\rho ^n \beta _{h}^n\left( c_{2}\right) ^{\lambda n}}$ and $f_{4}= N_{h}e^{-2\rho ^n\beta _{h}^n\left( c_{2}\right) ^{\lambda n}}+\left( 1-N_{h}\right)$

Figure 18

Geometry of conical nanowire. Top radius r and bottom radius R in the x–y plane and height h along the z-direction

Volume of the truncated conical nanodisk is calculated using

$$\begin{aligned} V=\pi r^2 h + \int _{r}^{R} \rho \text {d}\rho \int _{0}^{2\pi } \text {d}\varphi \int _{0}^{\left( \frac{R-\rho }{R-r}\right) h} \text {d}z \end{aligned}$$

which is found to be

$$\begin{aligned} V=\pi r^2 h + \frac{2\pi h}{R-r} \left( \frac{R^3}{6}+r^2 \left( \frac{r}{3}-\frac{R}{2}\right) \right) \end{aligned}$$

(A8)

The dipolar interaction energy is given as

$$\begin{aligned} E_\mathrm{dip}=\frac{\mu _{0}}{2}\int \vec {M}(r)\cdot \vec {\nabla }U(r)\text {d}\nu \end{aligned}$$

(A9)

where the magnetic potential is

$$\begin{aligned} U(r)= & {} -\frac{1}{4\pi }\int _V \frac{\vec {\nabla }\cdot \vec {M}(r^\prime )}{|r-r^\prime |}dV^\prime \nonumber \\{} & {} +\frac{1}{4\pi }\int _S \frac{\hat{n^\prime }\cdot \vec {M}(r^\prime )}{|r-r^\prime |}ds^\prime \end{aligned}$$

(A10)

To calculate the volume magnetic potential, we first solve divergence of $\vec {M}$

for $0\le z\le \frac{h}{2}$

$$\begin{aligned} \vec {\nabla }\cdot \vec {M}=\frac{4 M_{0}}{h} \rho ^{\prime ^n} \beta _{0}^n n \eta \left( c_3\right) ^{n\eta -1} f_5 \end{aligned}$$

for $\frac{h}{2}\le z\le h$

$$\begin{aligned} \vec {\nabla }\cdot \vec {M}=-\frac{4 M_{0}}{h} \rho ^{\prime ^n} \beta _{h}^n n \lambda \left( c_4\right) ^{n\lambda -1} f_6 \end{aligned}$$

where $c_3=1-\frac{2z^\prime }{h}$, $c_4=\frac{2z^\prime }{h}-1$ and $f_5= N_{0} e^{-2\rho ^{\prime ^n}\beta _{0}^n \left( c_3\right) ^{\eta n}}$, $f_6= N_{h} e^{-2\rho ^{\prime ^n}\beta _{h}^n\left( c_4\right) ^{\lambda n}}$

Substitute in the first half of Eq. A10

$$\begin{aligned} U_{\nu }(r)=-\frac{1}{4\pi } \int _{0}^{R} \rho ^\prime \text {d}\rho ^\prime \int _{0}^{2\pi } \text {d}\varphi ^\prime \int _{0}^{h} \text {d}z^\prime \left( \vec {\nabla }\cdot \vec {M}\right) \frac{1}{|r-r^\prime |} \end{aligned}$$

The radial part can be expressed as

$$\begin{aligned} \frac{1}{|r-r^\prime |}=\sum _{p=-\infty }^{\infty }e^{ip(\varphi - \varphi \prime )}\int _{0}^{\infty }J_{p}(k\rho )J_{p}(k\rho ^\prime )e^{k(z_<-z_>)}\text {d}k \end{aligned}$$

(A11)

where $J_{p}$ are the first kind Bessel functions and using

$$\begin{aligned} \int _{0}^{2\pi }e^{-ip\varphi \prime } \text {d}\varphi ^\prime =2\pi \delta _{p,0} \end{aligned}$$

(A12)

After solving the radial part using above expression, we have

$$\begin{aligned} U_{\nu }(r)= & {} -\frac{1}{2} \int _{0}^{R} \rho ^\prime \text {d}\rho ^\prime \int _{0}^{h} \text {d}z^\prime \left( \vec {\nabla }\cdot \vec {M}\right) \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{k(z_<-z_>)}\text {d}k \end{aligned}$$

Substituting the divergence of $\vec {M}$ expressions appropriately

$$\begin{aligned} U_{\nu }(r)= & {} -\frac{M_0}{2} \int _{0}^{\rho ^\prime _{\mathrm{lim}}} \rho ^\prime \text {d}\rho ^\prime \int _{0}^{\frac{h}{2}} \text {d}z^\prime \frac{4}{h} \rho ^{\prime ^n} \beta _{0}^n n \eta \left( c_3\right) ^{n\eta -1} f_5\\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime ) \\{} & {} \times e^{k(z_<-z_>)}\text {d}k + \frac{M_0}{2} \int _{0}^{\rho ^\prime _{\mathrm{lim}}} \rho ^\prime \text {d}\rho ^\prime \\{} & {} \int _{\frac{h}{2}}^{h} \text {d}z^\prime \frac{4}{h} \rho ^{\prime ^n} \beta _{h}^n n \lambda \left( c_4\right) ^{n\lambda -1} f_6 \\{} & {} \times \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{k(z_<-z_>)}\text {d}k \end{aligned}$$

Here, we have $\rho ^\prime _{\mathrm{lim}}= R-\left( \frac{R-r}{h}\right) z^\prime$

$\vec {M}(r)$ has both $\varphi$ and z components. Thus, we need only these components of $\vec {\nabla }U_{\nu }(r)$. But $\vec {\nabla }U_{\nu }(r)_\varphi =0$. So we solve only the z-component to calculate the required dipolar energy.

$$\begin{aligned} \vec {\nabla }U_{\nu }(r)_z= & {} -\frac{M_0}{2} \int _{0}^{\rho ^\prime _{\mathrm{lim}}} \rho ^\prime \text {d}\rho ^\prime \\{} & {} \int _{0}^{\frac{h}{2}} \text {d}z^\prime \frac{4}{h} \rho ^{\prime ^n} \beta _{0}^n n \eta \left( c_3\right) ^{n\eta -1} f_5 J_{00}\\{} & {} + \frac{M_0}{2} \int _{0}^{\rho ^\prime _{\mathrm{lim}}} \rho ^\prime \text {d}\rho ^\prime \\{} & {} \int _{\frac{h}{2}}^{h} \text {d}z^\prime \frac{4}{h} \rho ^{\prime ^n} \beta _{h}^n n \lambda \left( c_4\right) ^{n\lambda -1} f_6 J_{00} \end{aligned}$$

where $J_{00}=\int _{0}^{\infty } ksgn(z-z^\prime ) J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{k(z_<-z_>)}\text {d}k$

Now, the dipolar energy is calculated by substituting the interacting potential

$$\begin{aligned} E_\mathrm{dip}^v= & {} \mu _{0}M_0 \pi \left[ \int _{0}^{\rho _{\mathrm{lim}}}\rho \text {d}\rho \int _{0}^{\frac{h}{2}} \text {d}z f_2 \right. \nonumber \\{} & {} \left. + \int _{0}^{\rho _{\mathrm{lim}}}\rho \text {d}\rho \int _{\frac{h}{2}}^{h} \text {d}z f_4 \right] \vec {\nabla }U_\nu (r)_z \end{aligned}$$

(A13)

By further simplification

$$\begin{aligned} E_\mathrm{dip}^v= & {} \frac{2n \pi \mu _0 M_0^2 }{h} \Bigg [ -\int _{0}^{\rho _{\mathrm{lim}}}\rho \text {d}\rho \int _{0}^{\frac{h}{2}} \text {d}z f_2\\{} & {} \int _{0}^{\rho ^\prime _{\mathrm{lim}}}\rho ^{\prime ^{n+1}} \text {d}\rho ^\prime \int _{0}^{\frac{h}{2}} \text {d}z^\prime \beta _{0}^n \eta (c_3)^{n\eta -1} f_5 J_{00} \\{} & {} -\int _{0}^{\rho _{\mathrm{lim}}}\rho \text {d}\rho \int _{\frac{h}{2}}^{h} \text {d}z f_4 \\{} & {} \int _{0}^{\rho ^\prime _{\mathrm{lim}}}\rho ^{\prime ^{n+1}} \text {d}\rho ^\prime \int _{0}^{\frac{h}{2}} \text {d}z^\prime \beta _{0}^n \eta (c_3)^{n\eta -1} f_5 J_{00} \\{} & {} + \int _{0}^{\rho _{\mathrm{lim}}}\rho \text {d}\rho \int _{0}^{\frac{h}{2}} \text {d}z f_2 \\{} & {} \int _{0}^{\rho ^\prime _{\mathrm{lim}}}\rho ^{\prime ^{n+1}} \text {d}\rho ^\prime \int _{\frac{h}{2}}^{h} \text {d}z^\prime \beta _{h}^n \lambda (c_4)^{n\lambda -1} f_6 J_{00} \\{} & {} +\int _{0}^{\rho _{\mathrm{lim}}}\rho \text {d}\rho \int _{\frac{h}{2}}^{h} \text {d}z f_4 \int _{0}^{\rho ^\prime _{\mathrm{lim}}}\rho ^{\prime ^{n+1}} \text {d}\rho ^\prime \\{} & {} \int _{\frac{h}{2}}^{h} \text {d}z^\prime \beta _{h}^n \lambda (c_4)^{n\lambda -1} f_6 J_{00} \Bigg ] \end{aligned}$$

The equation of curved surface is of the form

$$\begin{aligned} \phi =z-\left( \frac{R-\rho }{R-r}\right) h=0 \end{aligned}$$

Here, the unit vector is

$$\begin{aligned} \hat{n}=\frac{h}{\sqrt{h^2+(R-r)^2}}\hat{\rho }+\frac{R-r}{\sqrt{h^2+(R-r)^2}}\hat{z} \end{aligned}$$

(A14)

Dot product with magnetization yields

$$\begin{aligned} \hat{n^\prime }.\vec {M}(r^\prime )=M_0 \frac{R-r}{\sqrt{h^2+(R-r)^2}} \cos \theta (\rho ^\prime ,z^\prime ) \end{aligned}$$

On the top surface

$$\begin{aligned} \hat{n^\prime }.\vec {M}(r^\prime )= M_{0} \left[ N_{h}e^{-2\rho ^{\prime ^n}\beta _{h}^n}+\left( 1-N_{h}\right) \right] \end{aligned}$$

On the bottom surface

$$\begin{aligned} \hat{n^\prime }.\vec {M}(r^\prime )=- M_{0}\left[ N_{0} e^{-2\rho ^{\prime ^n}\beta _{0}^n}+\left( 1-N_{0}\right) \right] \end{aligned}$$

Along the curved surface, we have $ds^\prime =\rho ^\prime \text {d}\rho ^\prime \text {d}\varphi ^\prime \sec \gamma ,$ where $\sec \gamma = \frac{\sqrt{h^2+(R-r)^2}}{R-r}$

The surface interacting potential due to the top, bottom and curved surface is

$$\begin{aligned} U_s(r)= & {} \frac{M_0}{2}\int _{0}^{r} \rho ^\prime \text {d}\rho ^\prime f_8 \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{k(z-h)}\text {d}k \\{} & {} -\frac{M_0}{2}\int _{0}^{R} \rho ^\prime \text {d}\rho ^\prime f_7 \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{-kz}\text {d}k \\{} & {} + \frac{M_0}{2}\int _{r}^{\frac{R+r}{2}} \rho ^\prime \text {d}\rho ^\prime f_{10} \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{-k|z-\left( \frac{R-\rho ^\prime }{R-r}\right) h|}\text {d}k \\{} & {} + \frac{M_0}{2}\int _{\frac{R+r}{2}}^{R} \rho ^\prime \text {d}\rho ^\prime f_9 \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{-k|z-\left( \frac{R-\rho ^\prime }{R-r}\right) h|}\text {d}k \end{aligned}$$

where $f_7 =N_{0}e^{-2\rho ^{\prime ^n}\beta _{0}^n}+\left( 1-N_{0}\right)$, $f_8 = N_{h}e^{-2\rho ^{\prime ^n}\beta _{h}^n}+\left( 1-N_{h}\right)$, $f_9= N_{0} e^{-2\rho ^{\prime ^n}\beta _{0}^n \left( c_3\right) ^{\eta n}}+\left( 1-N_{0}\right)$, $f_{10}= N_{h} e^{-2\rho ^{\prime ^n}\beta _{h}^n\left( c_4\right) ^{\lambda n}}+\left( 1-N_{h}\right)$ and $z^\prime =\left( \frac{R-\rho ^\prime }{R-r}\right) h$

Again, only the z-component is required

$$\begin{aligned} \vec {\nabla }U_s(r)_z= & {} \frac{M_0}{2}\int _{0}^{r} \rho ^\prime \text {d}\rho ^\prime f_8\\{} & {} \int _{0}^{\infty }k J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{k(z-h)}\text {d}k\\{} & {} +\frac{M_0}{2}\int _{0}^{R} \rho ^\prime \text {d}\rho ^\prime f_7\\{} & {} \int _{0}^{\infty }k J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{-kz}\text {d}k \\{} & {} + \frac{M_0}{2}\int _{r}^{\frac{R+r}{2}} \rho ^\prime \text {d}\rho ^\prime f_{10} J_{00} \\{} & {} + \frac{M_0}{2}\int _{\frac{R+r}{2}}^{R} \rho ^\prime \text {d}\rho ^\prime f_9 J_{00} \end{aligned}$$

The dipolar surface energy is calculated as

$$\begin{aligned} E_\mathrm{dip}^s= & {} \mu _{0}M_0 \pi \left[ \int _{0}^{\rho _{\mathrm{lim}}}\rho \text {d}\rho \int _{0}^{\frac{h}{2}} \text {d}z f_2 \right. \nonumber \\{} & {} \left. + \int _{0}^{\rho _{\mathrm{lim}}}\rho \text {d}\rho \int _{\frac{h}{2}}^{h} \text {d}z f_4 \right] \vec {\nabla }U_s(r)_z \end{aligned}$$

(A15)

Appendix B Total energy density for vortex-flower state

Calculation of dipolar energy requires the divergence of $\vec {M}$

for $0\le z\le \frac{h}{2}$

$$\begin{aligned} \vec {\nabla }\cdot \vec {M}= & {} \frac{M_0}{\rho ^\prime }\left[ \frac{\partial }{\partial \varphi ^\prime }\left( \sin \theta \left( \rho ^\prime ,z^\prime \right) \right) +\frac{\partial }{\partial z^\prime }\left( \rho ^\prime \cos \theta \left( \rho ^\prime ,z^\prime \right) \right) \right] \nonumber \\ \vec {\nabla }\cdot \vec {M}_{_{0\le z\le \frac{h}{2}}}= & {} \frac{4 M_{0}}{h} \rho ^{\prime ^n} \beta _{0}^n n \eta \left( c_3\right) ^{n\eta -1} f_5 \end{aligned}$$

(B16)

which is similar to vortex-vortex state, as the vortex is still stabilized on the bottom half.

for $\frac{h}{2} \le z\le h$

$$\begin{aligned} \vec {\nabla }\cdot \vec {M}= & {} \frac{M_0}{\rho ^\prime }\left[ \frac{\partial }{\partial \rho ^\prime }\left( \rho ^\prime \sin \theta \left( \rho ^\prime ,z^\prime \right) \right) +\frac{\partial }{\partial z^\prime }\left( \rho ^\prime \cos \theta \left( \rho ^\prime ,z^\prime \right) \right) \right] \nonumber \\ \vec {\nabla }\cdot \vec {M}_{\frac{h}{2} \le z\le h}= & {} M_0 \left[ \frac{\sin \theta \left( \rho ^\prime ,z^\prime \right) }{\rho ^\prime }+T_1+T_2\right] \end{aligned}$$

(B17)

Here $T_1$ and $T_2$ are solved as

$$\begin{aligned} T_1= & {} \frac{\partial }{\partial \rho ^\prime }\left( \sin \theta \left( \rho ^\prime ,z^\prime \right) \right) \\= & {} 2n \lambda \rho ^{\prime ^{n-1}} \beta _{h}^n \cot \theta \left( \rho ^\prime ,z^\prime \right) \left( c_4\right) ^{n\lambda } f_6\\ T_2= & {} \frac{\partial }{\partial z^\prime }\left( \rho ^\prime \cos \theta \left( \rho ^\prime ,z^\prime \right) \right) \\= & {} -\frac{4}{h} n \lambda \rho ^{\prime ^n} \beta _{h}^n \left( c_4\right) ^{n\lambda -1} f_6 \end{aligned}$$

Magnetic potential is given as

$$\begin{aligned} U_{\nu }(r)= & {} -\frac{1}{2} \int _{0}^{\rho ^\prime _{\mathrm{lim}}} \rho ^\prime \text {d}\rho ^\prime \int _{0}^{\frac{h}{2}} \text {d}z^\prime \left( \vec {\nabla }\cdot \vec {M}\right) _{0 \le z\le \frac{h}{2}} \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime ) e^{k(z_<-z_>)}\text {d}k\\{} & {} -\frac{1}{2} \int _{0}^{\rho ^\prime _{\mathrm{lim}}} \rho ^\prime \text {d}\rho ^\prime \int _{\frac{h}{2}}^{h} \text {d}z^\prime \left( \vec {\nabla }\cdot \vec {M}\right) _{\frac{h}{2} \le z\le h} \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{k(z_<-z_>)}\text {d}k \end{aligned}$$

For the bottom half, we require only the z-component of $\vec {\nabla }U_{\nu }(r)$, as it exhibits vortex configuration. But for the top half, due to flower domain configuration $\vec {M}(r^\prime )$ has both $\rho$ and z component. So we need both of these components for $\vec {\nabla }U_{\nu }(r)$.

$$\begin{aligned} \vec {\nabla }U_{\nu }(r)_z= & {} -\frac{1}{2} \int _{0}^{\rho ^\prime _{\mathrm{lim}}} \rho ^\prime \text {d}\rho ^\prime \nonumber \\{} & {} \int _{0}^{\frac{h}{2}} \text {d}z^\prime \left( \vec {\nabla }\cdot \vec {M}\right) _{0 \le z\le \frac{h}{2}} J_{00} -\frac{1}{2} \int _{0}^{\rho ^\prime _{\mathrm{lim}}} \rho ^\prime \text {d}\rho ^\prime \nonumber \\{} & {} \times \int _{\frac{h}{2}}^{h} \text {d}z^\prime \left( \vec {\nabla }\cdot \vec {M}\right) _{\frac{h}{2} \le z\le h} J_{00} \end{aligned}$$

(B18)

$$\begin{aligned} \vec {\nabla }U_{\nu }(r)_\rho= & {} \frac{1}{2} \int _{0}^{\rho ^\prime _{\mathrm{lim}}} \rho ^\prime \text {d}\rho ^\prime \nonumber \\{} & {} \int _{0}^{\frac{h}{2}} \text {d}z^\prime \left( \vec {\nabla }\cdot \vec {M}\right) _{0 \le z\le \frac{h}{2}} J_{10}+\frac{1}{2} \int _{0}^{\rho ^\prime _{\mathrm{lim}}} \rho ^\prime \text {d}\rho ^\prime \nonumber \\{} & {} \times \int _{\frac{h}{2}}^{h} \text {d}z^\prime \left( \vec {\nabla }\cdot \vec {M}\right) _{\frac{h}{2} \le z\le h} J_{10} \end{aligned}$$

(B19)

Here $J_{10} =\int _{0}^{\infty }k J_{1}(k\rho )J_{0}(k\rho ^\prime )e^{k(z_<-z_>)}\text {d}k$

Using Eq. B18 and Eq. B19, we have

$$\begin{aligned} E_\mathrm{dip}^\nu= & {} \mu _{0}M_0 \pi \Bigg [ \int _{0}^{\rho _{\mathrm{lim}}}\rho \text {d}\rho \int _{0}^{\frac{h}{2}} \text {d}z f_2 \vec {\nabla }U_\nu (r)_z \nonumber \\{} & {} + \int _{0}^{\rho _{\mathrm{lim}}}\rho \text {d}\rho \int _{\frac{h}{2}}^{h} \text {d}z f_4 \vec {\nabla }U_\nu (r)_z \nonumber \\{} & {} + \int _{0}^{\rho _{\mathrm{lim}}}\rho \text {d}\rho \int _{\frac{h}{2}}^{h} \text {d}z \sqrt{1-f_4^2} \vec {\nabla }U_\nu (r)_\rho \Bigg ] \end{aligned}$$

(B20)

Now the dipolar contribution of bottom surface is calculated in a similar fashion to vortex-vortex state, but for the top half, i.e.

for $\frac{h}{2} \le z\le h$

$$\begin{aligned} \hat{n^\prime }.\vec {M}(r^\prime )= & {} M_0 \frac{h}{\sqrt{h^2+(R-r)^2}} \sin \theta \\{} & {} +M_0 \frac{R-r}{\sqrt{h^2+(R-r)^2}} \cos \theta \end{aligned}$$

The potential with contributions from top, bottom and curved surface is followed as

$$\begin{aligned} U_s(r)= & {} \frac{1}{4\pi }\int _S {\frac{M_0 \cos \theta }{|r-r^\prime |}ds^\prime }_{z^\prime =h}\nonumber \\{} & {} -\frac{1}{4\pi }\int _S {\frac{M_0 \cos \theta }{|r-r^\prime |}ds^\prime }_{z^\prime =0} \nonumber \\{} & {} + \frac{1}{4\pi } \frac{R-r}{\sqrt{h^2+(R-r)^2}} \int _S {\frac{M_0 \cos \theta }{|r-r^\prime |}ds^\prime }_{z^\prime < \frac{h}{2}}\nonumber \\{} & {} + \frac{1}{4\pi } \frac{1}{\sqrt{h^2+(R-r)^2}} \nonumber \\{} & {} \times \int _S {\frac{M_0 \left( h\sin \theta +\left( R-r\right) \cos \theta \right) }{|r-r^\prime |}ds^\prime }_{z^\prime > \frac{h}{2}} \end{aligned}$$

(B21)

Using the simplification procedure for radial component yields

$$\begin{aligned} U_s(r)= & {} \frac{M_0}{2}\int _{0}^{r} \rho ^\prime \text {d}\rho ^\prime f_8 \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{k(z-h)}\text {d}k\\{} & {} -\frac{M_0}{2}\int _{0}^{R} \rho ^\prime \text {d}\rho ^\prime f_7 \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{-kz}\text {d}k \\{} & {} + \frac{M_0}{2} \frac{h}{R-r}\int _{r}^{\frac{R+r}{2}} \rho ^\prime \text {d}\rho ^\prime \sqrt{1-f_{10}^2}\\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{-k|z-\left( \frac{R-\rho ^\prime }{R-r}\right) h|}\text {d}k \\{} & {} + \frac{M_0}{2}\int _{r}^{\frac{R+r}{2}} \rho ^\prime \text {d}\rho ^\prime f_{10} \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{-k|z-\left( \frac{R-\rho ^\prime }{R-r}\right) h|}\text {d}k \\{} & {} + \frac{M_0}{2}\int _{\frac{R+r}{2}}^{R} \rho ^\prime \text {d}\rho ^\prime f_9 \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{-k|z-\left( \frac{R-\rho ^\prime }{R-r}\right) h|}\text {d}k \end{aligned}$$

Again we require both z and $\rho$ components of $\vec {\nabla }U_s(r)$

$$\begin{aligned} \vec {\nabla }U_s(r)_z= & {} \frac{M_0}{2}\int _{0}^{r} \rho ^\prime \text {d}\rho ^\prime f_8 \\{} & {} \int _{0}^{\infty }k J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{k(z-h)}\text {d}k\\{} & {} +\frac{M_0}{2}\int _{0}^{R} \rho ^\prime \text {d}\rho ^\prime \\{} & {} \times f_7 \int _{0}^{\infty }k J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{-kz}\text {d}k \\{} & {} + \frac{M_0}{2}\frac{h}{R-r}\int _{r}^{\frac{R+r}{2}} \rho ^\prime \text {d}\rho ^\prime \sqrt{1-f_{10}^2} I_{00} \\{} & {} + \frac{M_0}{2}\int _{r}^{\frac{R+r}{2}} \rho ^\prime \text {d}\rho ^\prime f_{10} I_{00}\\{} & {} + \frac{M_0}{2}\int _{\frac{R+r}{2}}^{R} \rho ^\prime \text {d}\rho ^\prime f_9 I_{00} \end{aligned}$$

where $I_{00} = \int _{0}^{\infty }k sgn\left( z-\left( \frac{R-\rho ^\prime }{R-r}\right) h\right) J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{-k|z-\left( \frac{R-\rho ^\prime }{R-r}\right) h|}\text {d}k$

$$\begin{aligned} \vec {\nabla }U_s(r)_\rho= & {} -\frac{M_0}{2}\int _{0}^{r} \rho ^\prime \text {d}\rho ^\prime f_8 \int _{0}^{\infty }k J_{1}(k\rho )J_{0}(k\rho ^\prime )e^{k(z-h)}\text {d}k\\{} & {} +\frac{M_0}{2}\int _{0}^{R} \rho ^\prime \text {d}\rho ^\prime \\{} & {} \times f_7 \int _{0}^{\infty }k J_{1}(k\rho )J_{0}(k\rho ^\prime )e^{-kz}\text {d}k \\{} & {} - \frac{M_0}{2}\frac{h}{R-r}\int _{r}^{\frac{R+r}{2}} \rho ^\prime \text {d}\rho ^\prime \sqrt{1-f_{10}^2}J_{10}\\{} & {} - \frac{M_0}{2}\int _{r}^{\frac{R+r}{2}} \rho ^\prime \text {d}\rho ^\prime f_{10} J_{10} \\{} & {} - \frac{M_0}{2}\int _{\frac{R+r}{2}}^{R} \rho ^\prime \text {d}\rho ^\prime f_9 J_{10} \end{aligned}$$

Substitution of above equations gives the expression for surface dipolar energy as

$$\begin{aligned} E_\mathrm{dip}^s= & {} \mu _{0}M_0 \pi \Bigg [ \int _{0}^{\rho _{\mathrm{lim}}}\rho \text {d}\rho \int _{0}^{\frac{h}{2}} \text {d}z f_2 \vec {\nabla }U_s(r)_z \nonumber \\{} & {} + \int _{0}^{\rho _{\mathrm{lim}}}\rho \text {d}\rho \int _{\frac{h}{2}}^{h} \text {d}z f_4 \vec {\nabla }U_s(r)_z \nonumber \\{} & {} + \int _{0}^{\rho _{\mathrm{lim}}}\rho \text {d}\rho \int _{\frac{h}{2}}^{h} \text {d}z \sqrt{1-f_4^2} \vec {\nabla }U_s(r)_\rho \Bigg ] \end{aligned}$$

(B22)

Appendix C Angular variation of dipolar energy for homogeneous state

The magnetization of single-domain state along an arbitrary direction that makes an angle $\psi$ with the z-axis can be represented as

$$\begin{aligned} \vec {M}=M_{0} \cos \psi \ \hat{z}+M_{0} \sin \psi \ \hat{x} \end{aligned}$$

(C23)

Using Eq. A1, the exchange energy is found to be zero.

For the dipolar energy, there is no volume contribution as the divergence of $\vec {M}$ is again zero. We have only the surface contribution.

From Eq. A14,

$$\begin{aligned} \hat{n^\prime }\cdot \vec {M}(r^\prime )= & {} M_0 \frac{h}{\sqrt{h^2+(R-r)^2}} \cos \varphi ^{\prime } \sin \psi \\{} & {} +M_0 \frac{R-r}{\sqrt{h^2+(R-r)^2}} \cos \psi \end{aligned}$$

On the top surface

$$\begin{aligned} \hat{n^\prime }\cdot \vec {M}(r^\prime )=M_0 \cos \psi \end{aligned}$$

On the bottom surface

$$\begin{aligned} \hat{n^\prime }\cdot \vec {M}(r^\prime )=- M_0 \cos \psi \end{aligned}$$

The expression for surface potential after including the top, bottom and curved part becomes

$$\begin{aligned} U_s(r)= & {} \frac{1}{4\pi } \frac{h}{R-r} \sin \psi \nonumber \\{} & {} \int _S {\frac{M_0 cos\varphi ^{\prime }}{|r-r^\prime |}\rho ^\prime \text {d}\rho ^\prime \text {d}\varphi ^\prime }_{z^\prime = \left( \frac{R-\rho ^\prime }{R-r}\right) h}+ \frac{1}{4\pi } \cos \psi \nonumber \\{} & {} \times \int _S {\frac{M_0 }{|r-r^\prime |}\rho ^\prime \text {d}\rho ^\prime \text {d}\varphi ^\prime }_{z^\prime = \left( \frac{R-\rho ^\prime }{R-r}\right) h} \nonumber \\{} & {} +\frac{1}{4\pi } \cos \psi \int _S {\frac{M_0}{|r-r^\prime |}\rho ^\prime \text {d}\rho ^\prime \text {d}\varphi ^\prime }_{z^\prime =h}-\frac{1}{4\pi } \cos \psi \nonumber \\{} & {} \int _S {\frac{M_0}{|r-r^\prime |}\rho ^\prime \text {d}\rho ^\prime \text {d}\varphi ^\prime }_{z^\prime =0} \end{aligned}$$

(C24)

By using Eq. A11 and Eq. A12 along with $\int _{0}^{2\pi }e^{-ip\varphi \prime } cos \varphi ^\prime \text {d}\varphi ^\prime =\pi \delta _{p1,-1}$ to solve the radial part, we have

$$\begin{aligned} U_s(r)= & {} \frac{M_0 h \sin \psi }{4\pi (R-r)} \int _{r}^{R} 2 \pi \cos \varphi \rho ^\prime \text {d}\rho ^\prime \\{} & {} \int _{0}^{\infty }J_{1}(k\rho )J_{1}(k\rho ^\prime )e^{-k|z-h\left( \frac{R-\rho ^\prime }{R-r}\right) |}\text {d}k \\{} & {} +\frac{M_0}{4 \pi } \cos \psi \int _{r}^{R} 2 \pi \rho ^\prime \text {d}\rho ^\prime \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{-k|z-h\left( \frac{R-\rho ^\prime }{R-r}\right) |}\text {d}k \\{} & {} + \frac{M_0}{4 \pi } \cos \psi \int _{0}^{r} 2 \pi \rho ^\prime \text {d}\rho ^\prime \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{-k(h-z)}\text {d}k \\{} & {} -\frac{M_0}{4 \pi } \cos \psi \int _{0}^{R} 2 \pi \rho ^\prime \text {d}\rho ^\prime \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )e^{-kz}\text {d}k \end{aligned}$$

Using Eq. A9, the surface dipolar interaction energy energy is calculated as

$$\begin{aligned} E_\mathrm{dip}^s= & {} \frac{\mu _{0}M_0}{2} \int _{\rho =0}^{R} \int _{\varphi =0}^{2\pi } \int _{z=0}^{h} \nonumber \\{} & {} \Bigg [ \sin \psi \cos \varphi \frac{\partial U_s(r)}{\partial \rho } -\frac{1}{\rho }\sin \psi \sin \varphi \frac{\partial U_s(r)}{\partial \varphi } \nonumber \\{} & {} + \cos \psi \frac{\partial U_s(r)}{\partial z} \Bigg ] \rho \text {d}\rho \text {d}\varphi \text {d}z \end{aligned}$$

(C25)

$$\begin{aligned} \frac{\partial U_s(r)}{\partial \rho }= & {} \frac{M_0}{4\pi } \frac{h}{R-r} \sin \psi \int _{r}^{R} 2 \pi \cos \varphi \ \rho ^\prime \text {d}\rho ^\prime \\{} & {} \int _{0}^{\infty } \left[ \frac{-1}{\rho }J_{1}(k\rho )+kJ_{0}(k\rho )\right] \\{} & {} \times J_{1}(k\rho ^\prime )e^{-k|z-h\left( \frac{R-\rho ^\prime }{R-r}\right) |}\text {d}k \\{} & {} -\frac{M_0}{4 \pi } \cos \psi \int _{r}^{R} 2 \pi \rho ^\prime \text {d}\rho ^\prime \\{} & {} \int _{0}^{\infty }kJ_{1}(k\rho )J_{0}(k\rho ^\prime )e^{-k|z-h\left( \frac{R-\rho ^\prime }{R-r}\right) |}\text {d}k \\{} & {} - \frac{M_0}{4 \pi } \cos \psi \int _{0}^{r} 2 \pi \rho ^\prime \text {d}\rho ^\prime \\{} & {} \int _{0}^{\infty }kJ_{1}(k\rho )J_{0}(k\rho ^\prime )e^{-k(h-z)}\text {d}k \\{} & {} +\frac{M_0}{4 \pi } \cos \psi \int _{0}^{R} 2 \pi \rho ^\prime \text {d}\rho ^\prime \int _{0}^{\infty }kJ_{1}(k\rho )J_{0}(k\rho ^\prime )e^{-kz}\text {d}k\\ \frac{\partial U_s(r)}{\partial \varphi }= & {} -\frac{M_0 h \sin \psi }{4\pi (R-r)} \int _{r}^{R} 2 \pi \sin \varphi \rho ^\prime \text {d}\rho ^\prime \\{} & {} \int _{0}^{\infty }J_{1}(k\rho ) J_{1}(k\rho ^\prime )e^{-k|z-h\left( \frac{R-\rho ^\prime }{R-r}\right) |}\text {d}k\\ \frac{\partial U_s(r)}{\partial z}= & {} \frac{M_0}{4\pi } \frac{h}{R-r} \sin \psi \int _{r}^{R} 2 \pi \cos \varphi \rho ^\prime \text {d}\rho ^\prime K_{11}\\{} & {} +\frac{M_0}{4 \pi } \cos \psi \int _{r}^{R} 2 \pi \rho ^\prime \text {d}\rho ^\prime K_{00} \\{} & {} + \frac{M_0}{4 \pi } \cos \psi \int _{0}^{r} 2 \pi \rho ^\prime \text {d}\rho ^\prime \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )ke^{-k(h-z)}\text {d}k \\{} & {} +\frac{M_0}{4 \pi } \cos \psi \int _{0}^{R} 2 \pi \rho ^\prime \text {d}\rho ^\prime \\{} & {} \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )ke^{-kz}\text {d}k \end{aligned}$$

Here $K_{00}=\int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )k sgn\left( h\left( \frac{R-\rho ^\prime }{R-r}\right) -z\right) e^{-k|z-h\left( \frac{R-\rho ^\prime }{R-r}\right) |}\text {d}k$ and

$K_{11}=\int _{0}^{\infty }J_{1}(k\rho )J_{1}(k\rho ^\prime )k sgn\left( h\left( \frac{R-\rho ^\prime }{R-r}\right) -z\right) e^{-k|z-h\left( \frac{R-\rho ^\prime }{R-r}\right) |}\text {d}k$

$$\begin{aligned} E_\mathrm{dip}^s= & {} \frac{\mu _{0}M_0^2 \pi }{4} \frac{h}{R-r} \sin ^2 \psi \int _{0}^{R} \nonumber \\{} & {} \int _{0}^{h} \rho \text {d}\rho \text {d}z \int _{r}^{R} \rho ^\prime \text {d}\rho ^\prime K_{01} \nonumber \\{} & {} + \frac{\mu _{0}M_0^2 \pi }{2} \cos ^2 \psi \int _{0}^{R} \nonumber \\{} & {} \int _{0}^{h} \rho \text {d}\rho \text {d}z \int _{r}^{R} \rho ^\prime \text {d}\rho ^\prime K_{00} \nonumber \\{} & {} + \frac{\mu _{0}M_0^2 \pi }{2} \cos ^2 \psi \int _{0}^{R} \int _{0}^{h} \rho \text {d}\rho \text {d}z \nonumber \\{} & {} \int _{0}^{r} \rho ^\prime \text {d}\rho ^\prime \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )ke^{-k(h-z)}\text {d}k \nonumber \\{} & {} + \frac{\mu _{0}M_0^2 \pi }{2} \cos ^2 \psi \int _{0}^{R} \int _{0}^{h} \rho \text {d}\rho \text {d}z \nonumber \\{} & {} \int _{0}^{R} \rho ^\prime \text {d}\rho ^\prime \int _{0}^{\infty }J_{0}(k\rho )J_{0}(k\rho ^\prime )ke^{-kz}\text {d}k \end{aligned}$$

(C26)

Here $K_{01}=\int _{0}^{\infty } k J_{0}(k\rho )J_{1}(k\rho ^\prime )e^{-k|z-h\left( \frac{R-\rho ^\prime }{R-r}\right) |}\text {d}k$