Chapter 7: Electrodynamics

Question: A rare case in which the electrostatic field E for a circuit can actually be calculated is the following: Imagine an infinitely long cylindrical sheet, of uniform resistivity and radius a . A slot (corresponding to the battery) is maintained at $\mathbf{\pm }\raisebox{1ex}{${\mathbf{V}}_{\mathbf{0}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.\mathbf{}\mathbf{}\mathit{a}\mathit{t}\mathbf{}\mathit{\varphi }\mathbf{=}\mathbf{\pm }\mathbf{\pi }$, and a steady current flows over the surface, as indicated in Fig. 7.51. According to Ohm's law, then,

$\mathbf{V}\left(a,\varphi \right)\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{0}}\mathbf{\varphi }}{\mathbf{2}\mathbf{\pi }}\mathbf{,}\left(-\pi <\varphi <+\pi \right)$

Figure 7.51

(a) Use separation of variables in cylindrical coordinates to determine $\mathbf{V}\left(s,\varphi \right)$ inside and outside the cylinder.

(b) Find the surface charge density on the cylinder.

The magnetic field outside a long straight wire carrying a steady current I is

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{2}\mathbf{\pi }}\frac{\mathbf{I}}{\mathbf{s}}\hat{\mathbf{\varphi }}$

The electric field inside the wire is uniform:

$\mathit{E}\mathbf{=}\frac{\mathbf{I}\mathbf{\rho }}{\mathbf{\pi }{\mathbf{a}}^{\mathbf{2}}}\hat{\mathbf{z}}$,

Where $\mathit{\rho }$is the resistivity and a is the radius (see Exs. 7.1 and 7 .3). Question: What is the electric field outside the wire? 29 The answer depends on how you complete the circuit. Suppose the current returns along a perfectly conducting grounded coaxial cylinder of radius b (Fig. 7.52). In the region a < s < b, the potential V (s, z) satisfies Laplace's equation, with the boundary conditions

(i) $\mathit{V}\mathbf{(}\mathit{a}\mathbf{,}\mathit{z}\mathbf{)}\mathbf{=}\frac{\mathbf{I}\mathbf{\rho }\mathbf{z}}{\mathbf{\pi }{\mathbf{a}}^{\mathbf{2}}}$ ; (ii) $\mathit{V}\mathbf{(}\mathit{b}\mathbf{,}\mathit{z}\mathbf{)}\mathbf{=}\mathbf{0}$

Figure 7.52

This does not suffice to determine the answer-we still need to specify boundary conditions at the two ends (though for a long wire it shouldn't matter much). In the literature, it is customary to sweep this ambiguity under the rug by simply stipulating that V (s,z) is proportional to V (s,z) = zf (s) . On this assumption:

(a) Determine (s).

(b) E (s,z).

(c) Calculate the surface charge density $\mathit{\sigma }\mathbf{\left(}\mathit{z}\mathbf{\right)}$on the wire.

[Answer: $\mathit{V}\mathbf{=}\left(-Iz\rho /\pi a^2\right)$ This is a peculiar result, since E_s and $\mathit{\sigma }\mathbf{\left(}\mathit{z}\mathbf{\right)}$are not independent of localid="1658816847863" $\mathit{z}\mathbf{\to }\mathit{a}\mathit{s}$ one would certainly expect for a truly infinite wire.]

If a magnetic dipole levitating above an infinite superconducting plane (Pro b. 7 .45) is free to rotate, what orientation will it adopt, and how high above the surface will it float?

Refer to Prob. 7.11 (and use the result of Prob. 5.42): How long does is take a falling circular ring (radius $a$, mass $m$, resistance $R$) to cross the bottom of the magnetic field $B$, at its (changing) terminal velocity?

(a) Referring to Prob. 5.52(a) and Eq. 7.18, show that

$E=-\frac{\partial A}{\partial t}$ (7.66) for Faraday-induced electric fields. Check this result by taking the divergence and curl of both sides.

(b) A spherical shell of radius$R$ carries a uniform surface charge $\sigma$. It spins about a fixed axis at an angular velocity $\omega \left(t\right)$that changes slowly with time. Find the electric field inside and outside the sphere. [Hint: There are two contributions here: the Coulomb field due to the charge, and the Faraday field due to the changing $\mathit{B}$. Refer to Ex. 5.11.]

Suppose the conductivity of the material separating the cylinders in Ex. 7.2 is not uniform; specifically, $\mathbf{\sigma }\left(s\right)\mathbf{=}\mathbf{k}\mathbf{/}\mathbf{s}$, for some constant . Find the resistance between the cylinders. [Hint: Because a is a function of position, Eq. 7.5 does not hold, the charge density is not zero in the resistive medium, and E does not go like 1/s. But we do know that for steady currents is the same across each cylindrical surface. Take it from there.]

Electrons undergoing cyclotron motion can be sped up by increasing the magnetic field; the accompanying electric field will impart tangential acceleration. This is the principle of the betatron. One would like to keep the radius of the orbit constant during the process. Show that this can be achieved by designing a magnet such that the average field over the area of the orbit is twice the field at the circumference (Fig. 7.53). Assume the electrons start from rest in zero field, and that the apparatus is symmetric about the center of the orbit. (Assume also that the electron velocity remains well below the speed of light, so that nonrelativistic mechanics applies.) [Hint: Differentiate Eq. 5.3 with respect to time, and use .$F=ma=qE$]

Question: An infinite wire carrying a constant current in the direction is moving in the direction at a constant speed . Find the electric field, in the quasistatic approximation, at the instant the wire coincides with the axis (Fig. 7.54).

Question; An atomic electron (charge q ) circles about the nucleus (charge Q) in an orbit of radius r ; the centripetal acceleration is provided, of course, by the Coulomb attraction of opposite charges. Now a small magnetic field dB is slowly turned on, perpendicular to the plane of the orbit. Show that the increase in kinetic energy, dT , imparted by the induced electric field, is just right to sustain circular motion at the same radius r. (That's why, in my discussion of diamagnetism, I assumed the radius is fixed. See Sect. 6.1.3 and the references cited there.)

The current in a long solenoid is increasing linearly with time, so the flux is proportional $t$:.$\varphi =\alpha t$Two voltmeters are connected to diametrically opposite points (A and B), together with resistors ( $R_1$and $R_2$), as shown in Fig. 7.55. What is the reading on each voltmeter? Assume that these are ideal voltmeters that draw negligible current (they have huge internal resistance), and that a voltmeter register -$-{\int }_a^{b}E\times dl$between the terminals and through the meter. [Answer: $V_1=\alpha R_1/(R_1+R_2)$. Notice that $V_1\ne V_2$, even though they are connected to the same points]