Question

Derive Eq. 6.3. [Here's one way to do it: Assume the dipole is an infinitesimal square, of side E (if it's not, chop it up into squares, and apply the argument to each one). Choose axes as shown in Fig. 6.8, and calculate F = I J (dl x B) along each of the four sides. Expand B in a Taylor series-on the right side, for instance,

$\mathit{B}\mathbf{=}\mathit{B}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\in }\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{\cong }\mathit{B}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{Z}\mathbf{)}\mathbf{+}\mathbf{\in }{\overline{)\frac{\mathbf{\partial }\mathbf{B}}{\mathbf{\partial }\mathbf{y}}}}_{\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{z}}$

For a more sophisticated method, see Prob. 6.22.]

Short answer

Therefore, the equation$F=\nabla \left(m.B\right)$ is derived.

Step-by-step solution

Write the given data from the question.

Expansion of the Taylor series,

$\mathit{B}\mathbf{=}\mathit{B}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\in }\mathbf{,}\mathit{z}\mathbf{)}\mathbf{\cong }\mathit{B}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathit{z}\mathbf{)}\mathbf{+}\mathbf{\in }{\overline{)\frac{\mathbf{\partial }\mathbf{B}}{\mathbf{\partial }\mathbf{y}}}}_{\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{z}}$

Side of the square$=\in$

Current $=I$

Magnetic field=B

Force along each side =F

 Step 2: Derive the equation  F=∇m×B.

$d{F}_4$Consider the square of side $\in$in the plan.

The force on the element of the core,

$$\begin{gathered} d{F}_1=I\left(L\times B\right) \\ d{F}_1=I\left[dyy\times B\left(0,y,0\right)\right] \end{gathered}$$

The force on the element of the wire parallel to z axis,

$d{F}_2=I\left[dz\hat{z}\times B\left(0,\in ,z\right)\right]$

The force on the element of the wire parallel to $y$axis,

$d{F}_3=I\left[-dy\times B\left(0,y,\in \right)\right]$

The force on the element of the wire on $z$axis,

$d{F}_4=I\left[-dz\hat{z}\times B(0,0,z)\right]$

The total force on the element loop is sum of the forces along ach side.

$dF=d{F}_1+d{F}_{2}d{F}_3+d{F}_4$

Substitute $I\left[dyy\times B\left(0,y,0\right)\right]$for $d{F}_1,I\left[dz\hat{z}\times B\left(0,\in ,z\right)\right]$, for localid="1657711123898" $d{F}_2,I\left[-dy\times B\left(0,y,\in \right)\right]$, for $d{F}_3$and $I-\left[dz\hat{z}\times B\left(0,0,z\right)\right]$ for into above equation.

localid="1657711909865" $$\begin{gathered} dF=\mathrm{I}\left[dyy\times B\left(0,y,0\right)\right]+\mathrm{I}\left[dz\hat{z}\times B\left(0,y,0\in z\right)\right]+\mathrm{I}\left[-\mathrm{dy}\times \mathrm{Bo},\mathrm{y},\in \right]+\mathrm{I}\left[-\mathrm{dz}\hat{\mathrm{z}}\times \mathrm{B}0,0,\mathrm{z}\right] \\ \mathrm{dF}=\mathrm{I}\left\{-\mathrm{dyy}\times \left[\mathrm{B}\left(0,\mathrm{y},\in \right)-\mathrm{B}\left(0,\mathrm{y},0\right)\right]+\mathrm{dz}\hat{\mathrm{z}}\mathrm{B}\left(0,\in ,\mathrm{z}\right)-\mathrm{B}\left(0,0,\mathrm{z}\right)\right\} \end{gathered}$$……(1)

From the Taylor series,

localid="1657713074645" $$\begin{gathered} B\left(0,y,0\right)-B\left(O,y,0\right)=\in \frac{\partial B}{\partial z} \\ B\left(0,\in ,z\right)-B\left(O,0,Z\right)=\in \frac{\partial B}{\partial y} \end{gathered}$$

Substitute $\in \frac{\partial B}{\partial z}$for $B(o,y,\in )-b$and $\in \frac{\partial B}{\partial y}$for $B\left(0,\in ,z\right)-B\left(0,0,z\right)$into equation (1).

$$\begin{gathered} dF=I\left\{-dyy\times \left[\in \frac{\partial B}{\partial z}\right]+dz\hat{z}\left[\in \frac{\partial B}{\partial z}\right]\right\} \\ dF=I{\in }^2\left\{\hat{z}\times \left[\frac{\partial B}{\partial z}\right]-y\times \in \left[\frac{\partial B}{\partial z}\right]\right\} \end{gathered}$$ …… (2)

The magnetic moment of the loop,

$m=I{\in }^2$

Substitute for into equation (2).

$dF=m\left\{\hat{z}\times \left[\frac{\partial B}{\partial y}\right]-y\times \left[\frac{\partial B}{\partial z}\right]\right\}$

Noe solves the above equation,

localid="1657715061627" $$\begin{gathered} F=m\left|\hat{x}y\hat{z} \\ 001 \\ \frac{\partial B_x}{\partial y}\frac{\partial B_Y}{\partial Y}\frac{\partial B_Z}{\partial Y}\right|-\left|\hat{x}y\hat{z} \\ 001 \\ \frac{\partial B_x}{\partial z}\frac{\partial B_Y}{\partial z}\frac{\partial B_Z}{\partial z}\right| \\ F=m\left\{\hat{x}\left(-\frac{\partial B_y}{\partial y}\right)-y\left(-\frac{\partial B_x}{\partial y}\right)-\hat{z}\left(\frac{\partial B_Z}{\partial z}\right)\right\} \\ F=m\left\{\hat{x}\left[\frac{\partial B_y}{\partial y}+\frac{\partial B_Z}{\partial z}\right]+\left[\frac{\partial B_x}{\partial y}\right]+\hat{z}\left[\frac{\partial B_x}{\partial z}\right]\right\} \end{gathered}$$…… (3)

From the gauss law of the magnetization,

$$\begin{gathered} \left(\frac{\partial B_x}{\partial x}+\frac{\partial B_y}{\partial y}+\frac{\partial B_Z}{\partial z}\right)=0 \\ -\left(\frac{\partial B_y}{\partial y}+\frac{\partial B_Z}{\partial z}\right)=\frac{\partial B_x}{\partial x} \end{gathered}$$

Substitute $\frac{\partial B_x}{\partial x}$for$-\left(\frac{\partial B_y}{\partial y}+\frac{\partial B_Z}{\partial z}\right)$into equation (3).

$$\begin{gathered} F=m\left\{\hat{x}\left[\frac{\partial B_x}{\partial x}\right]+y\left[\frac{\partial B_x}{\partial y}\right]+z\left[\frac{\partial B_x}{\partial z}\right]\right\} \\ F=m\left(\nabla .B_X\right) \\ F=\nabla \left(m.B_x\right) \end{gathered}$$

In the general form the above equation can be written as,

localid="1657715013642" $F=\nabla \left(m.B\right)$

Hence, the equation$F=\nabla \left(m.B\right)$is obtained.