Physics 305
Problem set 4
Due: Fri, Feb. 17
[1] (a) Consider two very long charged wires, both with uniform charge per length +λ,
positioned parallel to the y axis at locations z = ±d in the y-z plane. Compute the force per unit
length on the upper wire by using the Maxwell stress tensor method: Consider the surface
surrounding the wire to be the x-y plane plus the upper hemisphere at r ! " (the latter
experiencing a vanishingly small field). First find the electric field in the x-y plane, and then
write down the elements of the stress tensor. Finally, integrate to find the force. (The total force
is infinite, but it can be computed for a length l of the wire by considering a strip of width l in the
x-y plane, and noting that by symmetry all such strips will give identical contributions.)
(b) In a similar way, compute the force per length between two long (uncharged) wires carrying
currents I in opposite directions, positioned parallel to the y axis at locations z = ±d in the y-z
plane, using a stress tensor method.
[2] In problem 7.29, you showed that an inductor L carrying current I o , when switched so that it
can “discharge” through a resistor R, will have a decaying current I = I o e ! Rt/L . For this problem,
suppose that the inductor is a long single-layer solenoid, with n turns per length, radius a, and
length l, and assume that it can be treated as an ideal solenoid, neglecting end effects.
(a) Find the magnetic field vs time, and the Faraday-induced electric field vs time, for points
inside the solenoid. (Neglect the small change in B field due to the time-dependent E field.)
(b) Calculate the time-dependent Poynting vector in the solenoid, and find its direction.
(c) Find the time-dependent rate of energy flow into the curved outer surface of the solenoid (at
a), by integrating the Poynting-vector flux. Show that the instantaneous total flux is equal to the
instantaneous ohmic power loss in the resistor.
(d) Compute the instantaneous total field energy contained in the solenoid (B field only; assume
that L/R sufficiently large that the E field contribution is small). Then, show that the Poynting
vector flux is also equal to the rate of change of the field energy inside.
[3] A straight section of coaxial cable has an inner conducting wire of radius R1, and a thin outer
conducting sheath of radius R2. The length of the cable is ! , and assume that the inner and outer
conductor have negligible resistance. A battery, Vo, is connected to one end, and a resistor, R, to
the other end. Between R1 and R2 is vacuum.
(a) Neglecting end effects in the cable, find the electric and magnetic fields between R1 and R2.
(b) Find the Poynting vector, with direction, and integrate the Poynting flux density to obtain the
total power flowing in at one end of the cable. Show that this power flows from the generator
toward the resistor, and that it is equal in magnitude to the rate of energy dissipation in the
resistor.
(c) From the Poynting vector find the total momentum “ pem ” belonging to the electromagnetic
fields in this case. Also evaluate pem numerically for the case Vo, = 1 V, R = 1 Ω, ! = 1 m.
[4] Starting with the definition of the stress tensor! components,
eqn. (8.19) in the text, work out
! "
explicitly the x-components of the divergence of T , (! " T )x . Show that they equal the
corresponding components in the expression for the force density, eqn. (8.18), with the exception
of the Poynting vector time-derivative term.
[5] Griffiths problem 8.6 (page 358).