Tunable microwave susceptibility of thin truncated conical permalloy nanodisks : a micromagnetic investigation

Abstract

Dynamic susceptibility of thin truncated conical permalloy nanodisk with height 20 nm is simulated using micromagnetic simulations. Starting from a relaxed magnetic vortex state, the effect of tapering on the resonant modes is analyzed by changing the bottom radius (R = 50 nm to 100 nm) and top radius (r = 10 nm to R − 10 nm) of the conical nanodisk. When an in-plane excitation field pulse is applied, the lowest resonant peak frequency increases with increase in r. This mode is originated due to vortex core translation and the variation of the resonant peak with r is well explained using Thiele’s equation. Other resonant modes appear at higher excitation frequencies which are identified as azimuthal modes. When an out-of-plane excitation field is applied, all the spin wave modes appear to be axially symmetric. Appearance of new resonant modes and shifting of the resonant frequency is recorded by changing either of the parameter r or R for both the excitation field directions. The spatial distribution of FFT of magnetization at each resonant mode links the area of nanodisk that has maximum contribution to each mode and combining this with average center of magnetization perturbation, shifting of resonance frequency is explained successfully. The findings from this study prove that tunable resonant frequency can be achieved by tapering of truncated conical nanodisks.

Appendix A: Calculation of vortex mode frequency

Using Feldtkeller model, [41] the magnetization of the vortex state is modelled as

$$ \vec{M}=M_{0}\sin\theta\hat{\varphi}+M_{0}\cos\theta\hat{z} $$

in which

$$ \cos\theta=e^{-2\rho^{2}\beta^{2}} $$

Now, the gyrovector is represented in components of gyrocoupling tensor as

$$ G=G_{xy}=\int d\tau g_{xy} $$

(A1)

The antisymmetric gyrocoupling density tensor is calculated as

$$ g_{xy}=\frac{M_{0}}{\gamma}\vec{m}\cdot\left( \partial_{x}\vec{m}\times\partial_{y}\vec{m}\right) $$

(A2)

$$ \begin{array}{@{}rcl@{}} \frac{\partial\vec{m}}{\partial x}&=&\hat{x}\Bigg[\frac{-4\beta^{2}\rho e^{-4\rho^{2}\beta^{2}} }{\sqrt{1-e^{-2\rho^{2}\beta^{2}}}}\tan\varphi+\frac{\sqrt{1-e^{-4\rho^{2}\beta^{2}}}}{\rho}\cot\varphi\Bigg] \\ &&+\hat{y}\Bigg[\frac{4\beta^{2}\rho e^{-4\rho^{2}\beta^{2}} }{\sqrt{1-e^{-4\rho^{2}\beta^{2}}}}+\frac{\sqrt{1-e^{-4\rho^{2}\beta^{2}}}}{\rho}\Bigg] \\ &&+\hat{z}\left[4\beta^{2}\rho e^{-2\rho^{2}\beta^{2}}\sec\varphi\right] \end{array} $$

$$ \begin{array}{@{}rcl@{}} \frac{\partial\vec{m}}{\partial y}&=&\hat{x}\Bigg[\frac{-4\beta^{2}\rho e^{-4\rho^{2}\beta^{2}} }{\sqrt{1-e^{-4\rho^{2}\beta^{2}}}}-\frac{\sqrt{1-e^{-4\rho^{2}\beta^{2}}}}{\rho}\Bigg] \\ &&+\hat{y}\Bigg[\frac{4\beta^{2}\rho e^{-4\rho^{2}\beta^{2}} }{\sqrt{1-e^{-4\rho^{2}\beta^{2}}}}\cot\varphi-\frac{\sqrt{1-e^{-4\rho^{2}\beta^{2}}}}{\rho}\tan\varphi\Bigg] \\ &&+\hat{z}\left[-4\beta^{2}\rho e^{-4\rho^{2}\beta^{2}}\text{cosec}\varphi\right] \end{array} $$

$$ \begin{array}{@{}rcl@{}} \left( \partial_{x}\vec{m}\times\partial_{y}\vec{m}\right)_{x}&=&-\hat{x}4\beta^{2}\rho e^{-2\rho^{2}\beta^{2}}\sqrt{1-e^{-4\rho^{2}\beta^{2}}}\text{cosec} \varphi\sec^{2}\varphi\\ (\partial_{x}\vec{m}\times\partial_{y}\vec{m})_{y}&=&\hat{y}4\beta^{2}\rho e^{-2\rho^{2}\beta^{2}}\sqrt{1-e^{-4\rho^{2}\beta^{2}}}\text{cosec}^{2} \varphi\sec\varphi\\ (\partial_{x}\vec{m}\times\partial_{y}\vec{m})_{z}&=&\hat{z}4\beta^{2}\rho e^{-4\rho^{2}\beta^{2}}\text{cosec}^{2} \varphi\sec^{2}\varphi \end{array} $$

(A3) (A4) (A5)

Substitute the above equations in Eq. A2 and by solving, we get

$$ g_{xy}=\frac{M_{0}}{\gamma}\left( 1-e^{-4\rho^{2}\beta^{2}}+4\beta^{2}e^{-6\rho^{2}\beta^{2}}\right)\text{cosec}^{2} \varphi\sec^{2}\varphi $$

(A6)

Therefore, the gyrovector is obtained by substituting Eqs. A6 in A1

$$ \begin{array}{@{}rcl@{}} G&=&16 \frac{M_{0}}{\gamma} {{\int}_{0}^{R}} d\rho {{\int}_{0}^{h}} dz \rho \Big(1-e^{-4\rho^{2}\beta^{2}}\\&&+4\beta^{2} e^{-6\rho^{2}\beta^{2}}\Big) \Bigg(2\pi+\sum\limits_{k=1}^{\infty}\frac{(2k-1)!!}{(2k)!!}\pi \Bigg) \end{array} $$

(A7)

Consider the vortex is shifted by a distance X along the x-axis, then the new centre is at (ϱ,φ) where as the original centre is at (ρ,ϕ) such that

$$ \begin{array}{@{}rcl@{}} {\varrho}^{2}=\rho^{2}+X^{2}-2\rho X \cos\varphi \\ \phi=\cos^{-1}\Bigg(\frac{\rho\cos\varphi-X}{{\varrho}}\Bigg) \end{array} $$

The magnetization is rewritten as

$$ \delta\vec{m}=\sqrt{1-e^{-4\rho^{2}\beta^{2}}}\Bigg(\frac{-X\sin\varphi}{\sqrt{\rho^{2}+X^{2}-2\rho X \cos\varphi}}\Bigg)\hat{\rho} $$

(A8)

Now, the exchange energy can be calculated by using

$$ \delta E_{ex}=A{\int}_{V} \left[(\nabla\delta m_{x})^{2}+(\nabla\delta m_{y})^{2}+(\nabla\delta m_{z})^{2}\right]dV $$

(A9)

Using Eq. A8 and solving

$$ \begin{array}{@{}rcl@{}} \nabla\delta m_{x}&=&\Bigg[\frac{4\rho\beta^{2} e^{-4\rho^{2}\beta^{2}}}{\sqrt{1-e^{-4\rho^{2}\beta^{2}}}}\Bigg(\frac{-X\sin\varphi}{\sqrt{\rho^{2}+X^{2}-2\rho X\cos\varphi}}\Bigg)\cos\varphi \\ &&-\sqrt{1-e^{-4\rho^{2}\beta^{2}}}\Bigg(\frac{-\rho X\sin\varphi}{(\rho^{2}+X^{2}-2\rho X\cos\varphi)^{3/2}}\Bigg)\cos\varphi\Bigg]\hat{\rho} \\ &&+\Bigg[\sqrt{1-e^{-4\rho^{2}\beta^{2}}}\Bigg(\frac{-X\cos 2\varphi}{\sqrt{\rho^{2}+X^{2}-2\rho X\cos\varphi}}\Bigg)\Bigg]\hat{\varphi} \end{array} $$

(A10)

$$ \begin{array}{@{}rcl@{}} \nabla\delta m_{y}&=&\Bigg[\frac{4\rho\beta^{2} e^{-4\rho^{2}\beta^{2}}}{\sqrt{1-e^{-4\rho^{2}\beta^{2}}}}\Bigg(\frac{-X\sin^{2}\varphi}{\sqrt{\rho^{2}+X^{2}-2\rho X\cos\varphi}}\Bigg)\\ &&-\sqrt{1-e^{-4\rho^{2}\beta^{2}}}\Bigg(\frac{-\rho X\sin^{2}\varphi}{(\rho^{2}+X^{2}-2\rho X\cos\varphi)^{3/2}}\Bigg)\Bigg]\hat{\rho} \\ &&+\Bigg[\sqrt{1-e^{-4\rho^{2}\beta^{2}}}\Bigg(\frac{-X\sin 2\varphi}{\sqrt{\rho^{2}+X^{2}-2\rho X\cos\varphi}}\Bigg)\Bigg]\hat{\varphi}\\ \nabla\delta m_{z}&=&0 \end{array} $$

(A11) (A12)

By substituting the above equations in Eq. A9, we have

$$ \begin{array}{@{}rcl@{}} \delta E_{ex}&=&A {\int}_{V} \Bigg[\frac{16\rho^{2}\beta^{4} e^{-8\rho^{2}\beta^{2}}}{1-e^{-4\rho^{2}\beta^{2}}}\Bigg(\frac{X^{2}\sin^{2}\varphi}{\rho^{2}+X^{2}-2\rho X\cos\varphi}\Bigg) \\ &&+\left( 1-e^{-4\rho^{2}\beta^{2}}\right)\Bigg(\frac{\rho^{2} X^{2}\sin^{2}\varphi}{\left( \rho^{2}+X^{2}-2\rho X\cos\varphi\right)^{3}}\Bigg) \\ &&-4 \rho^{2}\beta^{2} e^{-4\rho^{2}\beta^{2}}\Bigg(\frac{X^{2}\sin^{2}\varphi}{\left( \rho^{2}+X^{2}-2\rho X\cos\varphi\right)^{2}}\Bigg) \\ &&+\left( 1-e^{-4\rho^{2}\beta^{2}}\right)\Bigg(\frac{X^{2}}{\rho^{2}+X^{2}-2\rho X\cos\varphi}\Bigg)\Bigg]dV \end{array} $$

(A13)

Let the integrand be considered as

$$ \begin{array}{@{}rcl@{}} E_{0}&=&\Bigg[\frac{16\rho^{2}\beta^{4} e^{-8\rho^{2}\beta^{2}}}{1-e^{-4\rho^{2}\beta^{2}}} \Bigg(\pi\frac{X^{2}}{\rho^{2}+X^{2}}\bigg(1+\frac{\rho^{2} X^{2}}{(\rho^{2}+X^{2})^{2}}\bigg)\Bigg) \\ &&+\left( 1-e^{-4\rho^{2}\beta^{2}}\right)\Bigg(\pi\frac{\rho^{2} X^{2}}{\left( \rho^{2}+X^{2}\right)^{3}}\bigg(1+\frac{6\rho^{2} X^{2}}{\left( \rho^{2}+X^{2}\right)^{2}}\bigg)\Bigg) \\ &&-4 \rho^{2}\beta^{2} e^{-4\rho^{2}\beta^{2}}\Bigg(\pi\frac{X^{2}}{(\rho^{2}+X^{2})^{2}}\bigg(1+\frac{3\rho^{2} X^{2}}{(\rho^{2}+X^{2})^{2}}\bigg)\Bigg) \\ &&+\left( 1-e^{-4\rho^{2}\beta^{2}}\right)\Bigg(2\pi\frac{ X^{2}}{\rho^{2}+X^{2}}\bigg(1+\frac{2\rho^{2} X^{2}}{(\rho^{2}+X^{2})^{2}}\bigg)\Bigg)\Bigg] \end{array} $$

Splitting the limits to consider the curved part of the truncated conical nanodisk

$$ \begin{array}{@{}rcl@{}} \delta E_{ex1}&=&A {{\int}_{0}^{r}} \rho d{\rho{\int}_{0}^{h}} dz E_{0}\\ \delta E_{ex2}&=&A {{\int}_{r}^{R}} \rho d\rho {\int}_{0}^{\big(\frac{R-\rho}{R-r}\big)h} dz E_{0} \end{array} $$

(A14) (A15)

Thus, the exchange energy can be calculated by the summation of Eqs. A14 and A15.

The dipolar energy is given as

$$ E_{dip}=\frac{\mu_{0}}{2}\int \vec{M}(r).\vec{\nabla}U(r)d\nu $$

(A16)

where the potential is

$$ U(r)=-\frac{1}{4\pi}{\int}_{V} \frac{\vec{\nabla}.\vec{M}(r^{\prime})}{\lvert r-r^{\prime}\rvert}dV^{\prime}+\frac{1}{4\pi}{\int}_{S} \frac{\hat{n^{\prime}}.\vec{M}(r^{\prime})}{\lvert r-r^{\prime} \rvert}ds^{\prime} $$

(A17)

The demagnetization energy has no volume contribution, so we calculate the surface demagnetization energy and only the effect of curved surface is considered.

$$ \begin{array}{@{}rcl@{}} \hat{n^{\prime}}.\delta\vec{M}(r^{\prime})&=&M_{0}\frac{h}{\sqrt{h^{2}+(R-r)^{2}}}\sqrt{1-e^{-4\rho^{\prime^{2}}\beta^{2}}}\\ &&\times\Bigg(\frac{-X \sin\varphi^{\prime}}{\sqrt{\rho^{\prime^{2}}+X^{2}-2\rho^{\prime} X \cos\varphi^{\prime}}}\Bigg)\\ U(r)&=&\frac{1}{4\pi}M_{0}\frac{h}{\sqrt{h^{2}+(R-r)^{2}}}{\int}_{S} \frac{\sqrt{1-e^{-4\rho^{\prime^{2}}\beta^{2}}}}{\lvert r-r^{\prime} \rvert}\\ &&\times\Bigg(\frac{-X \sin\varphi^{\prime}}{\sqrt{\rho^{\prime^{2}}+X^{2}-2\rho^{\prime} X \cos\varphi^{\prime}}}\Bigg)ds^{\prime} \end{array} $$

Projecting the area integral onto the x-y plane

$$ \begin{array}{@{}rcl@{}} U(r)&=&\frac{1}{4\pi}M_{0}\frac{h}{\sqrt{h^{2}+(R-r)^{2}}}{\int}_{S} dx^{\prime} dy^{\prime} \frac{\sqrt{1-e^{-4\rho^{\prime^{2}}\beta^{2}}}}{\lvert r-r^{\prime} \rvert} \sec\gamma \\ &&\times \Bigg(\frac{-X \sin\varphi^{\prime}}{\sqrt{\rho^{\prime^{2}}+X^{2}-2\rho^{\prime} X \cos\varphi^{\prime}}}\Bigg) \end{array} $$

By approximating

$$ U(r)\approx-\frac{1}{4\pi}M_{0}\frac{h}{R-r}{\int}_{S} d\rho^{\prime} d\varphi^{\prime} \frac{\sqrt{1-e^{-4\rho^{\prime^{2}}\beta^{2}}}}{\lvert r-r^{\prime} \lvert}(X \sin\varphi^{\prime}) $$

(A18)

The radial part can be expressed as

$$ \frac{1}{\lvert r-r^{\prime} \lvert}=\sum\limits_{p=-\infty}^{\infty}e^{ip(\varphi - \varphi \prime)}{\int}_{0}^{\infty}J_{p}(k\rho)J_{p}(k\rho^{\prime})e^{k(z_<-z_>)}dk $$

where J_p are the first kind Bessel functions and using

$$ {\int}_{0}^{2\pi}e^{ip\varphi \prime} \sin \varphi^{\prime} d\varphi^{\prime} =\pm\frac{\pi}{i}\delta_{p,\pm1} $$

But $\delta \vec {M}(r^{\prime })$ has only radial component

$$ \begin{array}{@{}rcl@{}} \nabla U(r)_{\rho}\approx-\frac{1}{2}M_{0}\frac{h}{R-r}X {{\int}_{r}^{R}} d\rho^{\prime} \sqrt{1-e^{-4\rho^{\prime^{2}}\beta^{2}}} \sin\varphi \\ {\int}_{0}^{\infty}\Big[\frac{-1}{\rho}J_{1}(k\rho)+k J_{0}(k\rho)\Big]J_{1}(k\rho^{\prime})e^{k(z_<-z_>)} dk \end{array} $$

(A19)

Substitute in Eq. A16 to calculate dipolar energy

$$ \begin{array}{@{}rcl@{}} &&\delta E_{dip}=\frac{{M_{0}^{2}}\mu_{0}X^{2}h}{4(R-r)}{\int}_{V} \rho d\rho d\varphi dz \sqrt{1-e^{-4\rho^{2}\beta^{2}}} \\&& {{\int}_{r}^{R}} d\rho^{\prime} \sqrt{1-e^{-4\rho^{\prime^{2}}\beta^{2}}} \frac{\sin^{2}\varphi}{\sqrt{\rho^{2}+X^{2}-2\rho X\cos\varphi}} \\&& {\int}_{0}^{\infty} dk \Bigg[-\frac{1}{\rho}J_{1}(k\rho)+kJ_{0}(k\rho)\Bigg]J_{1}(k\rho^{\prime}) e^{k(z_<-z_>)} \end{array} $$

(A20)

We use the following equations to solve integration for z

$$ \begin{array}{@{}rcl@{}} {{\int}_{0}^{h}} e^{k(z_<-z_>)} dz&=&\frac{1}{k}\Bigg[2-e^{-kh\big(\frac{R-\rho^{\prime}}{R-r}\big)}-e^{-kh\big(\frac{\rho^{\prime}-r}{R-r}\big)}\Bigg]\\ {\int}_{0}^{h\big(\frac{R-\rho}{R-r}\big)}\! e^{k(z_<-z_>)} dz& = &\frac{1}{k}\Bigg[2 - e^{-kh\big(\frac{R-\rho^{\prime}}{R-r}\big)} - e^{-kh\big(\frac{\rho^{\prime}-\rho}{R-r}\big)}\!\Bigg] if \rho \!<\! \rho^{\prime}\\ {\int}_{0}^{h\big(\frac{R-\rho}{R-r}\big)} e^{k(z_<-z_>)} dz& = &\frac{1}{k}\Bigg[e^{-kh\big(\frac{\rho-\rho^{\prime}}{R-r}\big)} - e^{-kh\big(\frac{R-\rho^{\prime}}{R-r}\big)}\Bigg] if \rho \!>\! \rho^{\prime} \end{array} $$

(A21) (A22) (A23)

Let the total dipolar energy δE_dip = δE_dip1 + δE_dip2 + δE_dip3, where

$$ \begin{array}{@{}rcl@{}} \delta E_{dip1}&=&\frac{{M_{0}^{2}}\mu_{0}X^{2}h}{4(R-r)}{{\int}_{r}^{R}}d\rho^{\prime} \sqrt{1-e^{-4\rho^{\prime^{2}}\beta^{2}}}{{\int}_{0}^{r}}d\rho \sqrt{1-e^{-4\rho^{2}\beta^{2}}} \\ &&{\int}_{0}^{2 \pi} d\varphi \frac{\sin^{2}\varphi}{\sqrt{\rho^{2}+X^{2}-2\rho X\cos\varphi}} {\int}_{0}^{\infty}dk J_{1}(k\rho^{\prime}) \\ &&\times \Bigg[-\frac{1}{k}J_{1}(k\rho)+\rho J_{0}(k\rho)\Bigg] \Bigg[2-e^{kh\big(\frac{R-\rho^{\prime}}{R-r}\big)}-e^{kh\big(\frac{\rho^{\prime}-r}{R-r}\big)}\Bigg] \end{array} $$

(A24)

$$ \begin{array}{@{}rcl@{}} \delta E_{dip2}&=&\frac{{M_{0}^{2}}\mu_{0}X^{2}h}{4(R-r)}{{\int}_{r}^{R}}d\rho^{\prime} \sqrt{1-e^{-4\rho^{\prime^{2}}\beta^{2}}}{\int}_{r}^{\rho^{\prime}}d\rho \sqrt{1-e^{-4\rho^{2}\beta^{2}}} \\ &&{\int}_{0}^{2 \pi} d\varphi \frac{\sin^{2}\varphi}{\sqrt{\rho^{2}+X^{2}-2\rho X\cos\varphi}} \\ &&{\int}_{0}^{\infty}dk \Bigg[-\frac{1}{k}J_{1}(k\rho)+\rho J_{0}(k\rho)\Bigg]J_{1}(k\rho^{\prime})\\ &&\times \Bigg[2-e^{kh\big(\frac{R-\rho^{\prime}}{R-r}\big)}-e^{kh\big(\frac{\rho^{\prime}-\rho}{R-r}\big)}\Bigg] \end{array} $$

(A25)

$$ \begin{array}{@{}rcl@{}} \delta E_{dip3}&=&\frac{{M_{0}^{2}}\mu_{0}X^{2}h}{4(R-r)}{{\int}_{r}^{R}}d\rho^{\prime} \sqrt{1-e^{-4\rho^{\prime^{2}}\beta^{2}}}{\int}_{\rho^{\prime}}^{R}d\rho \sqrt{1-e^{-4\rho^{2}\beta^{2}}} \\ &&{\int}_{0}^{2 \pi} d\varphi \frac{\sin^{2}\varphi}{\sqrt{\rho^{2}+X^{2}-2\rho X\cos\varphi}} \\ &&{\int}_{0}^{\infty}dk \Bigg[-\frac{1}{k}J_{1}(k\rho)+\rho J_{0}(k\rho)\Bigg]J_{1}(k\rho^{\prime})\\ &&\times\Bigg[e^{kh\big(\frac{\rho-\rho^{\prime}}{R-r}\big)}-e^{kh\big(\frac{R-\rho^{\prime}}{R-r}\big)}\Bigg] \end{array} $$

(A26)

The summation of the above three equations gives the total demagnetization energy. Now, the total energy is calculated which is the algebraic sum of exchange energy and demagnetization energy that helps in finding the stiffness constant κ by utilizing W(X) = κX². Therefore, the vortex mode frequency can be calculated by $f=\frac {1}{2\pi }\frac {\kappa }{G}$.