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Magnetic Fields Produced by Magnetic Dipoles 📂Electrodynamics

Magnetic Fields Produced by Magnetic Dipoles

Description

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The vector potential due to a magnetic dipole m\mathbf{m} is as given in magnetic dipole moment.

Adip(r)=μ04πm×r^r2=μ04πmsin⁡θr2ϕ^ \mathbf{A}_{\text{dip}}(\mathbf{r}) = \dfrac{\mu_{0}}{4 \pi} \dfrac{\mathbf{m} \times \hat{\mathbf{r}}}{r^2} = \dfrac{\mu_{0}}{4 \pi} \dfrac{m\sin\theta}{r^{2}} \hat{\boldsymbol{\phi}}

Now, let m\mathbf{m} be located at the origin and parallel to the zz axis, as shown in the figure above. Since the magnetic field is the curl of the vector potential, in spherical coordinates, it is as follows.

B=∇×Adip=1rsin⁡θ[∂(sin⁡θAϕ)∂θ−∂Aθ∂ϕ]r^+1r[1sin⁡θ∂Ar∂ϕ−∂(rAϕ)∂r]θ^+1r[∂(rAθ)∂r−∂Ar∂θ]ϕ^ \begin{align*} \mathbf{B} = \nabla \times \mathbf{A}_{\text{dip}} &= \frac{1}{r\sin\theta} \left[\frac{\partial (\sin\theta A_\phi)}{\partial \theta} - \frac{\partial A_\theta}{\partial \phi} \right]\mathbf{\hat{\mathbf{r}}} + \frac{1}{r}\left[\frac{1}{\sin\theta} \frac{\partial A_{r}}{\partial \phi} - \frac{\partial (rA_\phi)}{\partial r} \right] \boldsymbol{\hat{\boldsymbol{\theta}}} \\[1em] &\quad+ \frac{1}{r} \left[\frac{\partial (rA_\theta)}{\partial r}-\frac{\partial A_{r}}{\partial \theta} \right]\boldsymbol{\hat \phi} \end{align*}

Calculating each component gives the following. Since the Adip\mathbf{A}_{\text{dip}} component has only AϕA_{\phi},

Br=1rsin⁡θ∂(sin⁡θAϕ)∂θ=1rsin⁡θ∂∂θ(μ04πmsin⁡2θr2)=μ04π2mcos⁡θr3Bθ=−1r∂(rAϕ)∂r=−1r∂∂r(μ04πmsin⁡θr)=μ04πmsin⁡θr3Bϕ=0 \begin{align*} B_{r} &= \frac{1}{r\sin\theta} \frac{\partial (\sin\theta A_\phi)}{\partial \theta} = \frac{1}{r\sin\theta}\frac{\partial }{\partial \theta}\left( \dfrac{\mu_{0}}{4 \pi} \dfrac{m\sin^{2}\theta}{r^{2}} \right) = \dfrac{\mu_{0}}{4 \pi} \dfrac{2m\cos\theta}{r^{3}} \\[1em] B_{\theta} &= - \frac{1}{r} \frac{\partial (rA_\phi)}{\partial r} = - \frac{1}{r} \frac{\partial }{\partial r} \left( \dfrac{\mu_{0}}{4 \pi} \dfrac{m\sin\theta}{r} \right) = \dfrac{\mu_{0}}{4 \pi} \dfrac{m\sin\theta}{r^{3}} \\[1em] B_{\phi} &= 0 \end{align*}

Therefore, the magnetic field created by the magnetic dipole is as follows.

Bdip(r,θ)=μ04πmr3(2cos⁡θr^+sin⁡θθ^) \begin{equation} \mathbf{B}_{\text{dip}} (r,\theta)=\frac{\mu_{0} }{4 \pi }\frac{m}{r^3}(2\cos\theta \hat{\mathbf{r}} + \sin\theta \hat{\boldsymbol{\theta}}) \end{equation}

Interestingly, it is exactly the same formula as the electric field created by an electric dipole.

Formula

The formula above can be changed to be independent of the coordinate system as follows.

Bdip(r)=μ04π1r3[3(m⋅r^)r^−m] \mathbf{B}_{\text{dip}}(\mathbf{r}) = \frac{\mu_{0}}{4 \pi}\frac{1}{r^3}[3 (\mathbf{m} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - \mathbf{m}]

Derivation

First, if we express the unit vectors of spherical coordinates in Cartesian coordinates, it is as follows.

r^= cos⁡ϕsin⁡θx^+sin⁡ϕsin⁡θy^+cos⁡θz^θ^= cos⁡ϕcos⁡θx^+sin⁡ϕcos⁡θy^−sin⁡θz^ \begin{align*} \hat{\mathbf{r}} =&\ \cos\phi \sin\theta \hat{\mathbf{x}} + \sin\phi \sin\theta\hat{\mathbf{y}} + \cos\theta\hat{\mathbf{z}} \\ \hat{\boldsymbol{\theta}} =&\ \cos\phi \cos\theta \hat{\mathbf{x}} + \sin\phi \cos\theta\hat{\mathbf{y}} - \sin\theta\hat{\mathbf{z}} \end{align*}

Therefore, calculating the expression inside the brackets of (1)(1) gives the following.

2cos⁡θr^+sin⁡θθ^= 2cos⁡ϕsin⁡θcos⁡θx^+2sin⁡ϕsin⁡θcos⁡θy^+2cos⁡2θz^+cos⁡ϕsin⁡θcos⁡θx^+sin⁡ϕcos⁡θsin⁡θy^−sin⁡2θz^= 3cos⁡ϕsin⁡θcos⁡θx^+3sin⁡ϕsin⁡θcos⁡θy^+3cos⁡2θz^−(sin⁡2θ+cos⁡2θ)z^= 3cos⁡θ(cos⁡ϕsin⁡θx^+sin⁡ϕsin⁡θy^+cos⁡θz^)−z^= 3(m^⋅r^)r^−z^ \begin{align*} & 2\cos\theta \hat{\mathbf{r}} + \sin \theta \hat{\boldsymbol{\theta}} \\ =&\ 2 \cos\phi \sin\theta \cos\theta \hat{\mathbf{x}} + 2 \sin\phi \sin\theta \cos\theta \hat{\mathbf{y}} + 2 \cos^2 \theta \hat{\mathbf{z}} \\ & + \cos\phi \sin\theta \cos\theta \hat{\mathbf{x}} + \sin\phi \cos\theta \sin\theta \hat{\mathbf{y}} -\sin^2\theta \hat{\mathbf{z}} \\ =&\ 3\cos\phi \sin\theta \cos\theta \hat{\mathbf{x}} + 3 \sin\phi \sin\theta \cos\theta \hat{\mathbf{y}} + 3 \cos^2 \theta \hat{\mathbf{z}} -(\sin^2\theta + \cos^2\theta)\hat{\mathbf{z}} \\ =&\ 3 \cos\theta (\cos\phi \sin\theta \hat{\mathbf{x}} + \sin\phi \sin\theta\hat{\mathbf{y}} + \cos\theta\hat{\mathbf{z}}) - \hat{\mathbf{z}} \\ =&\ 3 (\hat{\mathbf{m}} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - \hat{\mathbf{z}} \end{align*}

The last equality holds because of cos⁡θ=m^⋅r^\cos\theta = \hat{\mathbf{m}} \cdot \hat{\mathbf{r}}. Now, the following result is obtained.

Bdip(r,θ)=μ04πmr3(2cos⁡θr^+sin⁡θθ^)=μ04πmr3[3(m^⋅r^)r^−z^]=μ04π1r3[3(m⋅r^)r^−mz^]=μ04π1r3[3(m⋅r^)r^−m]=Bdip(r)) \begin{align*} \mathbf{B}_{\text{dip}}(r,\theta) &=\frac{\mu_{0} }{4 \pi }\frac{m}{r^3}(2\cos\theta \hat{\mathbf{r}} + \sin\theta \hat{\boldsymbol{\theta}}) \\[1em] &= \frac{\mu_{0} }{4 \pi }\frac{m}{r^3}[3 (\hat{\mathbf{m}}\cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - \hat{\mathbf{z}}] \\[1em] &= \frac{\mu_{0} }{4 \pi }\frac{1}{r^3}[3 (\mathbf{m} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - m \hat{\mathbf{z}}] \\[1em] &= \frac{\mu_{0}}{4 \pi }\frac{1}{r^3}[3 (\mathbf{m} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - \mathbf{m}] \\[1em] &= \mathbf{B}_{\text{dip}}(\mathbf{r})) \end{align*}

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