Question: In classical electromagnetism, the simplest magnetic dipole is a circular current loop, which behaves in a magnetic field just as an electric dipole does in an electric field. Both experience torques and thus have orientation energies -p.Eand$\mathbf{-}\mathit{\mu }\mathbf{·}\mathit{B}\mathbf{.}$(a) The designation "orientation energy" can be misleading. Of the four cases shown in Figure 8.4 in which would work have to be done to move the dipole horizontally without reorienting it? Briefly explain. (b) In the magnetic case, using B and u for the magnitudes of the field and the dipole moment, respectively, how much work would be required to move the dipole a distance dx to the left? (c) Having shown that a rate of change of the "orientation energy'' can give a force, now consider equation (8-4). Assuming that B and are general, write$\mathbf{-}\mathit{\mu }\mathbf{·}\mathit{B}\mathbf{.}$in component form. Then, noting thatis not a function of position, take the negative gradient. (d) Now referring to the specific magnetic field pictured in Figure 8.3 which term of your part (c) result can be discarded immediately? (e) Assuming thatandvary periodically at a high rate due to precession about the z-axis what else may be discarded as averaging to 0? (f) Finally, argue that what you have left reduces to equation (8-5).

Electromagnetism is defined as an attraction between particles which is defined as created by electricity. An example of electromagnetism is the force which is the basic operation of an electric motor.

Part (a) of this question wants us to determine which of the four cases require work to be done to move the dipole$\mu$ horizontally without reorientation.

Part (b) wants us to derive a formula for the work done to move a magnetic dipole a distanceto the left in a magnetic field .

Part (c) asks us to write$-\mu \times B$ in component form, and then take its negative gradient, keeping in mind thatis not a function of position.

Referring to Figure 3, part (d) asks us which terms of the result in (c) can be removed.

Part (e) assumes that${\mu }_x$ and${\mu }_y$ change at a very high rate due to precession about the z-axis, and asks -accordingly- which term averages to zero.

Lastly, part (f) requires us to argue that our result indeed reduces to equation (5).

To move the dipole horizontally along the x-axis, the work needed is

_{$W={\int }_{x_1}^{x_2} F.dx=\pm F_x\Delta x$………………….(1)}

According to Figure 8-4, in the uniform field case, $F_{net}$exerted by the field equal zero$\left(F_{net}=0\right)$. Hence, in the uniform case, no work is needed to move the dipole horizontally$\left(W_{uniform}=0\right)$.

However, in the nonuniform field case, $F_{net}$exerted by the field does not equal zero$F_{net}\ne 0$. Hence, the field does work on the dipole to move it horizontally in the direction of$\left(W_{uniform}\ne 0\right)$$\left(W_{uniform}\ne 0\right)$.

If we want to move the dipole in the opposite direction, we need to exert work equivalent to that done by the field but in the opposite direction (in other words, a force $\left(-F_{net}\right)$). This is analogous to lifting a ball straight up off the ground at constant speed, where we would need force .

(b)

As we stated in the previous part, for the uniform magnetic field case, no work is needed to move the dipole horizontally.

_{$W_{uniform}=0$…………………..(2)}

In a non-uniform magnetic field, there is already work done on the dipole due to causing the dipole to move to the right. This work is equivalent to the difference in the orientation energy between a point $x_0$and a point$x_0+dx$.

$$\begin{gathered} W_{by field}=\Delta U=-\mu ·B_f+\mu ·B_i \\ =-\mu B\left(x+dx\right)+\mu B\left(x\right) \\ =-\mu \left[\left(x+dx\right)-B\left(x\right)\right] \\ =-\mu \frac{\partial B}{\partial x}dx \end{gathered}$$

where is the dipole moment magnitude and -according to the figure- completely points in the –x-direction.

Accordingly, the work needed to move the dipole to the left is

_{$W_{nonuniform}=-W_{byfield}=\mu \frac{\partial B}{\partial x}dx$……………………..(3)}

Taking $\mathrm{B}$and$\mu$ to be general vectors, then

_{$-\mu .\mathrm{B}=-{\mu }_x{B}_x-{\mu }_y{B}_y-{\mu }_z{B}_z$………………….(4)}

Taking the gradient of the above expression, keeping in mind that is not a function of position, we get

_{$-\nabla \left(-\mu .B\right)={\mu }_x\frac{\partial B_x}{\partial x}\hat{x}+{\mu }_y\frac{\partial B_y}{\partial y}\hat{y}+{\mu }_z\frac{\partial B_z}{\partial z}\hat{z}$……………………….(5)}

Noting that

$$\begin{gathered} \frac{\partial B_y}{\partial x}=\frac{\partial B_z}{\partial x}=0 \\ \frac{\partial B_x}{\partial y}=\frac{\partial B_z}{\partial y}=0 \\ \frac{\partial B_x}{\partial z}=\frac{\partial B_y}{\partial z}=0 \end{gathered}$$

Referring to Figure 8-4, we can see that there is no$B_x$, so the first term in equation (5) equals zero and it becomes

$-\nabla \left(-\mu .B\right)={\mu }_y\frac{\partial B_y}{\partial y}\hat{y}+{\mu }_z\frac{\partial B_z}{\partial z}\hat{z}$………………………..(6)

If and vary periodically at a very high rate, both components can be averaged to zero, and therefore the first term in equation (6) equals zero and it becomes

_{$-\nabla \left(-\mu .B\right)={\mu }_z\frac{\partial B_z}{\partial z}\hat{z}$…………………………….(7)}

From equation (4),

$F=-\nabla \left(-\mu .B\right)$

Therefore, substituting from equation (6), we finally get,

$F={\mu }_z\frac{\partial B_z}{\partial z}\hat{z}$

Returning to part (b) for another method, from equations (5), (6), and (7), we get

$F_x={\mu }_x\frac{\partial B_x}{\partial x}=\mu \frac{\partial B_x}{\partial x}$

From our results in part (a) and equation (1), we get

${\text{W}}_{nonuniform}=-W_{by field}=-{\int }_{x1}^{x2}F·dx$

$$\begin{gathered} -{\int }_{x1}^{x2}F·dx=-F_x dx \cos \left(180°\right) \\ =\mu \frac{\partial B_x}{\partial x}dx \end{gathered}$$

Which is the same as equation (3). This makes sense intuitively because in a uniform field,$\frac{\partial B_x}{\partial x}=0$ , and we get the familiar result,${\text{W}}_{uniform}=0$