Q5P - Introduction to Electrodynamics

Question

A uniform current density $\mathbf{J}\mathbf{=}{\mathbf{J}}_{\mathbf{0}}\hat{\mathbf{z}}$ fills a slab straddling the $\mathbf{yz}$  plane, from $\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{a}\mathbf{ }\mathbf{to}\mathbf{ }\mathbf{x}\mathbf{=}\mathbf{+}\mathbf{a}$ to . A magnetic dipole $\mathbf{m}\mathbf{=}{\mathbf{m}}_{\mathbf{0}}\mathbf{x}$ is situated at the origin. 

(a) Find the force on the dipole, using Eq. 6.3. 

(b) Do the same for a dipole pointing in the direction: $\mathbf{m}\mathbf{=}{\mathbf{m}}_{\mathbf{0}}\mathbf{y}$ . 

(c) In the electrostatic case, the expressions $\mathbf{F}\mathbf{=}\mathbf{\nabla }\left(p.E\right)$ and $\mathbf{F}\mathbf{=}\left(p.\nabla \right)\mathbf{E}$are equivalent (prove it), but this is not the case for the magnetic analogs (explain why). As an example, calculate $\left(m.\nabla \right)$ for the configurations in (a) and (b).

Step-by-Step Solution

Answer

(a) The force on the dipole is zero.

(b) The force on the dipole is ${\mathrm{m}}_0{\mathrm{\mu }}_0{\mathrm{J}}_0\hat{\mathrm{x}}$ .

(c) The expressions $\mathrm{F}=\nabla \left(\mathrm{p}.\mathrm{E}\right)$ and $\mathrm{F}=\left(\mathrm{p}.\mathrm{E}\right)\nabla$are equivalent for electrostatic is proved and for magnetism $\nabla \left(\mathrm{m}.\mathrm{B}\right)=\mathrm{m}\times \left(\nabla \times \mathrm{B}\right)+\left(\mathrm{B}.\nabla \right)\mathrm{m}$. The calculation $\left(\mathrm{m}.\nabla \right)$ of for configuration (a) and (b) are $m_0{\mu }_0{J}_{0}y$  and respectively.

Step 1: Write the given data from the question:

Current density, $J=J_0\stackrel{⏜}{z}$fills a slab in plane fr to $x=+a$.

Magnetic dipole situated at origin,  $\mathrm{m}={\mathrm{m}}_{\mathrm{o}}\hat{\mathrm{x}}$ 

Step 2: Calculate the force on the dipole by using equation 6.3.

(a)

For the current density,

$$\begin{gathered} \mathbf{\oint }\mathbf{B}\mathbf{.}\mathbf{dl}\mathbf{=}\mathbf{Bl} \\ \mathbf{\oint }\mathbf{B}\mathbf{.}\mathbf{dl}\mathbf{=}{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{l}}_{\mathbf{enc}} \\ \mathbf{\oint }\mathbf{B}\mathbf{.}\mathbf{dl}\mathbf{=}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}\mathbf{\times }\mathbf{J} \end{gathered}$$ $$

Therefore, the magnitude $B={\mu }_0\times J$  .

By using the flaming’s right-hand rule,

$B={\mu }_0{J}_0\times y$

 

The expression of force acting on the dipole of magnetic moment, $F=\nabla \left(m.B\right)$

Substitute $m_0\hat{x}$ for and ${\mu }_0{J}_{0}xy$ for B into above equation.

$$\begin{gathered} F=\nabla \left(m_0\hat{x}.{\mu }_0{J}_{0}xy\right) \\ F=\nabla \left(m_0{\mu }_0{J}_{0}x\hat{x}.y\right) \\ F=\nabla \left(0\right) \\ F=0 \end{gathered}$$ 

Hence the force on the dipole is zero.

Step 3: Calculate the force on the dipole by using equation 6.3 dipole pointing in direction.

(b)

The dipole pointing in the direction, $\mathrm{m}={\mathrm{m}}_{\mathrm{o}}\hat{y}$

 The expression of force acting on the dipole of magnetic moment, $F=\nabla \left(m.B\right)$

Substitute$m_0\stackrel{⏜}{y}$ for m  and${\mu }_0{J}_{0}x\stackrel{⏜}{y}$for B into above equation.

$$\begin{gathered} F=\nabla \left(m_0\stackrel{⏜}{y}.{\mu }_0{J}_{0}x\stackrel{⏜}{y}\right) \\ F=\nabla \left(m_0{\mu }_0{J}_{0}x\stackrel{⏜}{y}.\stackrel{⏜}{y}\right) \\ F=\nabla \left(m_0{\mu }_0{J}_{0}x\right) \\ F=m_0{\mu }_0{J}_0\hat{x} \end{gathered}$$

 Hence the force on the dipole is $m_0{\mu }_0{J}_0\hat{x}$. 

Step 4: show that the expressions F = ∇ ( p ⋅ E ) and F = ( p ⋅ ∇ ) E are equivalent (prove it) in case of electrostatic but not for analods.

(c)

From the product rule,

$\nabla \left(A.B\right)=A\times \left(\nabla \times B\right)+B\times \left(\nabla \times A\right)+\left(A.\nabla \right)B+\left(B.\nabla \right)A$

By suing the above product rule,

  $\nabla \left(p.E\right)=p\times \left(\nabla \times E\right)+E\times \left(\nabla \times p\right)+\left(p.\nabla \right)E+\left(E.\nabla \right)p$….. (1)

Here, does not depends on x,y and z. Therefore, terms of the equation (1).

$$\begin{gathered} E\times \left(\nabla \times p\right)=0 \\ \left(E.\nabla \right)P=0 \end{gathered}$$

Since is irrational. So,$\nabla \times E=0$ .

 

Substitute 0 for $E\times \left(\nabla \times P\right),\left(E.\nabla \right)P$, and $\nabla \times E$ into equation (1).

  

$$\begin{gathered} \nabla \left(p.E\right)=0+0+\left(p.\nabla \right)E+0 \\ \nabla \left(p.E\right)=\left(p.\nabla \right)E \end{gathered}$$

Hence the expressions $F=\nabla \left(p.E\right)$ and $F=\left(p.E\right)$ are equivalent for electrostatic is proved.

 

For the magnetism,

$$\begin{gathered} \nabla (m\cdot B)=m\times (\nabla \times B)+B\times (\nabla \times m)+(m\cdot \nabla )B+(B\cdot \nabla )m \\ \nabla (m\cdot B)=m\times (\nabla \times B)+B(B\cdot \nabla )m \end{gathered}$$ 

 

For the configuration a,

$$\begin{gathered} (m\cdot \nabla )B_a=(m_0\stackrel{⏜}{x}\nabla ){\mu }_0{J}_{0}x\stackrel{⏜}{y} \\ (m\cdot \nabla )B_a=\left(m_0\frac{\partial }{\partial x}\right){\mu }_0{J}_{0}x\stackrel{⏜}{y} \\ (m\cdot \nabla )B_a=m_0{\mu }_0{J}_0\stackrel{⏜}{y} \end{gathered}$$

 

For the configuration b,

$$\begin{gathered} (m\cdot \nabla )B_b=(m_0\stackrel{⏜}{y}\nabla ){\mu }_0{J}_{0}x\stackrel{⏜}{y} \\ (m\cdot \nabla )B_b=\left(m_0\frac{\partial }{\partial y}\right){\mu }_0{J}_{0}x\stackrel{⏜}{y} \\ (m\cdot \nabla )B_b=0 \end{gathered}$$

Hence expressions $F=\nabla (p\cdot E)$ and $F=(p\cdot E)\nabla$ are equivalent for electrostatic is proved and for magnetism $\nabla (m\cdot B(\left)\right(\nabla \times B\left)\right(B\cdot \nabla \left)\right)$. The calculation of $(m\cdot \nabla )B$for configuration (a) and (b) are $m_0{\mu }_0{J}_0\stackrel{⏜}{y}$ and respectively.