91影视

A piece of wire bent into a loop, as shown in Fig. 10.5, carries a current that increases linearly with time:

$\mathit{I}\left(t\right)\mathbf{=}\mathit{k}\mathit{t}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(-鈭�<t<鈭�\right)$

Calculate the retarded vector potential A at the center. Find the electric field at the center. Why does this (neutral) wire produce an electric field? (Why can鈥檛 you determine the magnetic field from this expression for A?)

Suppose the current density changes slowly enough that we can (to good approximation) ignore all higher derivatives in the Taylor expansion

$\mathit{J}\left(t_r\right)\mathbf{=}\mathit{J}\left(t\right)\mathbf{+}\left(t_r-t\right)\mathit{J}\left(t\right)\mathbf{+}\mathbf{鈥�}$

(for clarity, I suppress the r-dependence, which is not at issue). Show that a fortuitous cancellation in Eq. 10.38 yields

$\mathit{B}\left(r,t\right)\mathbf{=}\frac{{\mathbf{渭}}_{\mathbf{0}}}{\mathbf{4}\mathbf{蟺}}\mathbf{鈭�}\frac{\mathbf{J}\left(r\text{'},t\right)\mathbf{脳}\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}\mathit{d}\mathcal{b}\mathbf{\text{'}}$.

That is: the Biot-Savart law holds, with J evaluated at the non-retarded time. This means that the quasistatic approximation is actually much better than we had any right to expect: the two errors involved (neglecting retardation and dropping the second term in Eq. 10.38 ) cancel one another, to first order.

A particle of chargeq moves in a circle of radius a at constant angular velocity $\mathit{蝇}$. (Assume that the circle lies in thexy plane, centered at the origin, and at time$\mathit{t}\mathbf{=}\mathbf{0}$ the charge is at role="math" localid="1653885001176" $\left(\mathit{a}\mathbf{,}\mathbf{0}\right)$, on the positive x axis.) Find the Li茅nard-Wiechert potentials for points on the z-axis.

Show that the scalar potential of a point charge moving with constant velocity (Eq. 10.49) can be written more simply as

$\mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{蟺蔚}}_{\mathbf{0}}}\frac{\mathbf{q}}{\mathbf{R}\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{v}}^{\mathbf{2}}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{胃}}{{\mathbf{c}}^{\mathbf{2}}}}}}$ (10.51)

where$\mathit{R}\mathbf{鈮�}\mathit{r}\mathbf{-}\mathit{v}\mathit{t}$is the vector from the present (!) position of the particle to the field point r, and$\mathit{胃}$is the angle between R and v (Fig. 10.9). Note that for nonrelativistic velocities $(v^2鈮�c^2)$,

$\mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{鈮�}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{蟺蔚}}_{\mathbf{0}}}\frac{\mathbf{q}}{\mathbf{R}}$

Show that the differential equations for V and A (Eqs. 10.4 and 10.5) can be written in the more symmetrical form

$\begin{array}{r}\begin{array}{cc}{饾啅}^{2}V+\begin{array}{r}鈭�L\\ 鈭�t\end{array}=-\frac{1}{{蔚}_{鈭�}}p& \\ {饾啅}^{2}A-鈭�L=-{渭}_{鈭�}J& \end{array}\\ \end{array}\}$

Where

${\mathbf{饾啅}}^{\mathbf{2}}\mathbf{鈮�}{\mathbf{鈭�}}^{\mathbf{2}}\mathbf{-}{\mathit{渭}}_{\mathbf{鈭�}}{\mathit{蔚}}_{\mathbf{鈭�}}\frac{{\mathbf{鈭�}}^{\mathbf{2}}}{\mathbf{鈭�}{\mathbf{t}}^{\mathbf{2}}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{L}\mathbf{鈮�}\mathbf{鈭�}\mathbf{.}\mathit{A}\mathbf{+}{\mathit{渭}}_{\mathbf{鈭�}}{\mathit{蔚}}_{\mathbf{鈭�}}\frac{\mathbf{鈭�}\mathbf{V}}{\mathbf{鈭�}\mathbf{t}}$

For a point charge moving at constant velocity, calculate the flux integral$\mathbf{鈭�}\mathit{E}\mathbf{.}\mathit{d}\mathit{a}$ (using Eq. 10.75), over the surface of a sphere centered at the present location of the charge.

One particle, of charge ${\mathit{q}}_{\mathbf{1}}$, is held at rest at the origin. Another particle, of charge ${\mathit{q}}_{\mathbf{2}}$, approaches along the x axis, in hyperbolic motion:

$\mathit{x}\left(t\right)\mathbf{=}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{+}{\left(ct\right)}^{\mathbf{2}}}$

it reaches the closest point, b, at time $\mathit{t}\mathbf{=}\mathbf{0}$, and then returns out to infinity.

(a) What is the force ${\mathit{F}}_{\mathbf{2}}$on ${\mathit{q}}_{\mathbf{2}}$(due to ${\mathit{q}}_{\mathbf{1}}$ ) at time t?

(b) What total impulse $\left(I_2={鈭�}_{-鈭�}^{鈭�}{F}_{2}dt\right)$is delivered to ${\mathit{q}}_{\mathbf{2}}$by ${\mathit{q}}_{\mathbf{1}}$?

(c) What is the force ${\mathit{F}}_{\mathbf{1}}$on ${\mathit{q}}_{\mathbf{1}}$(due to ${\mathit{q}}_{\mathbf{2}}$ ) at time t?

(d) What total impulse $\left(I_1={鈭�}_{-鈭�}^{鈭�}{F}_{1}dt\right)$is delivered to ${\mathit{q}}_{\mathbf{1}}$by ${\mathit{q}}_{\mathbf{2}}$? [Hint: It might help to review Prob. 10.17 before doing this integral. Answer:${\mathit{I}}_{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathit{I}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{4}{\mathbf{蟺蔚}}_{\mathbf{0}}\mathbf{bc}}$ ]

A uniformly charged rod (length L, charge density $\mathit{位}$ ) slides out thex axis at constant speedv. At time t = 0 the back end passes the origin (so its position as a function of time is x = vt , while the front end is at x = vt + L ). Find the retarded scalar potential at the origin, as a function of time, for t > 0 . [First determine the retarded time t₁ for the back end, the retarded time t₂ for the front end, and the corresponding retarded positions x₁ and x₂ .] Is your answer consistent with the Li茅nard-Wiechert potential, in the point charge limit (L << vt , with $\mathit{位}\mathit{L}\mathbf{=}\mathit{q}$)? Do not assume v << c .

A particle of charge q is traveling at constant speed v along the x axis. Calculate the total power passing through the plane $\mathit{x}\mathbf{=}\mathit{a}$X, at the moment the particle itself is at the origin. [ Answer $\frac{{\mathbf{q}}^{\mathbf{2}}\mathbf{v}}{\mathbf{32}\mathbf{螤}{\mathbf{蔚}}_{\mathbf{0}}{\mathbf{a}}^{\mathbf{2}}}$]

(a) Find the fields, and the charge and current distributions, corresponding to

$\mathit{v}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{A}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}\mathbf{蟿}\mathbf{蟿}{\mathbf{蔚}}_{\mathbf{0}}}\frac{\mathbf{q}\mathbf{t}}{{\mathbf{r}}^{\mathbf{2}}}\overline{\mathbf{r}}$

(b) Use the gauge function $\mathit{位}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{4}\mathbf{蟿}\mathbf{蟿}{\mathbf{蔚}}_{\mathbf{0}}\mathbf{)}\mathbf{(}\mathbf{q}\mathbf{t}\mathbf{/}\mathbf{r}\mathbf{)}$to transform the potentials, and comment on the result.