91Ó°ÊÓ

Expansion of the Taylor series,

$\mathit{B}\mathbf{=}\mathit{B}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{∈}\mathbf{,}\mathit{z}\mathbf{)}\mathbf{â‰\dots }\mathit{B}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathit{z}\mathbf{)}\mathbf{+}\mathbf{∈}{\overline{)\frac{\mathbf{∂}\mathbf{B}}{\mathbf{∂}\mathbf{y}}}}_{\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{z}}$

Side of the square$=∈$

Current $=I$

Magnetic field=B

Force along each side =F

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

The force on the element of the core,

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

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

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

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

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

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

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

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

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

From the Taylor series,

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

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

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

The magnetic moment of the loop,

$m=I{∈}^2$

Substitute for into equation (2).

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

Noe solves the above equation,

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

From the gauss law of the magnetization,

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

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

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

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

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

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