Question

The vector potential for a uniform magnetostatic field is$\mathit{A}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\left(r\times B\right)$ (Prob. 5.25). Show that $\frac{\mathbf{d}\mathbf{A}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\left(v\times B\right)$, in this case, and confirm that Eq. 10.20 yields the correct equation of motion.

Short answer

It is proved that$\frac{dA}{dt}=-\frac{1}{2}\left(v\times B\right)$ and the equation 10.20 yields the correct equation of motion.

Step-by-step solution

Expression for the vector potential in a uniform magnetic field:

Write the expression for the vector potential in a uniform magnetic field.

$\mathit{A}\left(r\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\left(r\times B\right)$ …… (1)

Here, r is the distance vector, and B is the magnetic field.

Show that dAdt=-12(v×B) :

Take the first derivative of equation (1).

$$\begin{gathered} \frac{dA\left(r\right)}{dt}=\frac{\partial A}{\partial t}+\left(v·\nabla \right)A \\ \frac{dA\left(r\right)}{dt}=\left(v·\nabla \right)\left[-\frac{1}{2}\left(r\times B\right)\right] \\ \frac{dA\left(r\right)}{dt}=-\frac{1}{2}\left[v_x\frac{\partial }{\partial x}+v_y\frac{\partial }{\partial y}+v_z\frac{\partial }{\partial z}\right]\left[\left(y{B}_z-z{B}_y\right)\hat{x}+\left(z{B}_x-x{B}_z\right)\hat{y}+\left(x{B}_y-y{B}_x\right)\hat{z}\right] \\ \frac{dA\left(r\right)}{dt}=-\frac{1}{2}\left[v_x\left(-B_z\hat{y}+B_y\hat{z}\right)+v_y\left(B_z\hat{x}-B_x\hat{z}\right)+v_z\left(-B_y\hat{x}+B_x\hat{y}\right)\right] \end{gathered}$$

On further solving, the above equation becomes,

$$\begin{gathered} \frac{dA\left(r\right)}{dt}=-\frac{1}{2}\left[\left(v_y{B}_z-v_z{B}_y\right)\hat{x}+\left(v_z{B}_x-v_x{B}_z\right)\hat{y}+\left(v_x{B}_y-v_y{B}_x\right)\hat{z}\right] \\ \frac{dA\left(r\right)}{dt}=-\frac{1}{2}\left(v\times B\right) \end{gathered}$$

Confirm that the Equation 10.20 yields the correct equation of motion:

Write the expression of the equation 10.20 .

$\frac{d}{dt}\left(p+qA\right)=-\nabla \left[q\left(V-v·A\right)\right]$

Solve the L.H.S of the above equation.

$\begin{array}{rcl}\frac{d}{dt}\left(p+qA\right)& =& \frac{dp}{dt}-\frac{q}{2}\left(v\times B\right)\\ & =& q\nabla \left(v·A\right)\\ & =& -\frac{q}{2}\nabla \left[r·\left(B\times v\right)\right]\end{array}$

Now, for a vector c, i.e. independent of position, write the suitable equation.

$$\begin{gathered} \nabla \left(r·c\right)=\left(\hat{x}\frac{\partial }{\partial x}+\hat{y}\frac{\partial }{\partial y}+\hat{z}\frac{\partial }{\partial x}\right)\left(x{c}_x+y{c}_y+z{c}_z\right) \\ \nabla \left(r·c\right)=c_x\hat{x}+c_y\hat{y}+c_z\hat{z}=c \end{gathered}$$

Here, $c=\left(B\times v\right)=-\left(v\times B\right)$.

Solve as,

$$\begin{gathered} \frac{dp}{dt}=\frac{q}{2}\left(v\times B\right)+\frac{q}{2}\left(v\times B\right) \\ \frac{dp}{dt}=q\left(v\times B\right) \end{gathered}$$

Hence, the above equation yields the correct equation of motion.

Therefore, it is proved that$\frac{dA}{dt}=-\frac{1}{2}\left(v\times B\right)$ and the equation 10.20 yields the correct equation of motion.