The fields of a solenoid
Consider a infinite solenoid in the laboratory frame S. The solenoid
has n coils per unit length, in which the constant current I flows. At I
some instant a “test” particle of charge q is located at a position inside
the solenoid and has a velocity v perpendicular to the axis of the solenoid.
v
a) Find the electromagnetic fields and the force on the particle in S. (We
do not consider the fields generated by the particle itself!)
Now consider the system S 0 in which the particle is instantaneously at rest.
b) Find the electromagnetic fields and the force on the particle in S 0 .
c) In the limit (v/c)2 ¿ 1, i.e. to the lowest order in v/c, describe the sources of the fields in S 0 .
1
Solution
a) Let us assume v = vx x̂ and the axis of the solenoid along ẑ. The magnetic field of the solenoid is
B = Bẑ = µ0 nIẑ. The force on q is F = qv × B = −qvB ŷ.
b) Since v0 = 0 the force in S 0 must be due to an electric field. From the transofrmations of the
fields we obtain Ey0 = γ(Ey − vBz ) = −γvBz and Bz0 = γ(Bz − vEy /c2 ) = γBz . The force is
F0 = qEy0 ŷ = −qγvBz ŷ = γFy .
c) Since γ = (1 − v 2 /c2 )−1/2 ¿ 1 we can neglect the relativistic contraction of lengths; the section
of the solenoid remains circular in S 0 . As the electric field outside the solenoid is zero (we further
discuss this point below), the electric field component perpendicular to the surface of the walls is
discontinous, implying that on the walls of the solenoid a surface charge density σ 0 appears. Using
Gauss’s theorem σ 0 = ²0 E⊥0 = ²0 Ey0 sin φ = −²0 vB sin φ = −vnI/c2 sin φ, where φ is the azimuthal
angle, and φ = 0 corresponds to the x0 axis.
This result is in agreement with the transformation laws of the surface current on the walls of
the solenoid, ι = nI φ̂ = nI(− sin φ, cos φ). We obtain σ 0 = γ(σ − vιx /c2 ) ' −vιx /c2 = vnI sin φ/c2
The fact that a surface charge density varying as sin φ on the surface of a cylinder generates an
uniform electric field inside the cylinder may sound already familiar to the reader. In fact, if we
consider two identical, long parallel cylinders of opposite charge densities ±ρ, it is easy to see that a
small relative displacement of the two cylinders in the direction prependicular to the axis generates
such a surface density distribution and an uniform field in the overlap region, i.e. inside the cylinder.
But in electrostatics one also finds that such surface charge also generates a “dipole-like” field outside
the cylinder, which is at odd with the present problem, where since the field is zero in such region
in the S frame, one expects a null field in S 0 as well. Of course this is the case, and the reason is
that the field is not electrostatic in S 0 since the magnetic field is time-dependent due to the uniform
motion of the solenoid.
a
Looking at this issue in detail, we observe that an electric field
C
which is uniform inside the cylinder and null outside is not conservab
−v
tive, as the line integral of E along on a circuit across the surface of
φ
B
the cylinder will not vanish; in other words, the electrostatic condition
E
E1k = E2k is not satisfied. But this actually occurs since there is a
time-dependent flux of B across the surface of the circuit! As a matter
of fact consider the circuit C in the figure: the line integral of E0 equals
I
E0 · dl = Ek0 a = Ey0 a cos φ,
C
while the flux of the magnetic field is given by
Z
0
ΦC (B ) =
B0 · dS = Bz ab.
C
But the area ab over which the field is not zero varies in time as
d(ab)
db
= a = −av cos φ,
dt
dt
thus we obtain
dΦC
−
= Bz av cos φ = Ey0 a cos φ =
dt
2
I
E0 · dl.
C