Pinch effect in a current wire
An uniformly distributed current J flows along an infinite cylindrical
conductor of radius a. The current is due to electrons having number a
e
density ne and drift speed v (parallel to the axis of the cylinder).
v
Ions are considered to be fixed and having an uniform density ni
and charge Ze. The system is globally neutral.
a) Find the magnetic field generated by the current and the related
magnetic force on the electrons.
The electron displacement due to the magnetic force leads to the generation of a back-holding
electric field.
b) Find the electric field generated in order to compensate the magnetic force on the electrons, and
the related charge distribution.
c) Evaluate the effect in “standard” conditions for a good Ohmic conductor.
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Solution
a)
Let us fix a cylindrical coordinate frame with the axis
B
B
z along the axis of the cylinder. Assume v > 0, i.e. the
electrons flow in the positive direction of the z axis.
J
For symmetry reasons the magnetic field is azimuthal,
−ev × B
r
i.e. the only component is Bφ = Bφ (r). In order
to evaluate Bφ we apply Ampère’s theorem to a field
line, i.e. a circular path as shown in the figure. From 2πBφ = µ0 πr2 J we find Bφ = µ0 Jr/2 where
J = −ene v < 0. Thus Bφ < 0, i.e. it is oriented clockwise with respect to the z axis.
The magnetic force F = −e(v × B) is in the radial direction and it is directed towards the axis:
F = −µ0 e2 ne v 2 r/2 .
(1)
Thus, the magnetic force tends to push the electrons in inner region of the wire. Notice that the
dependence on e2 means that the force does not depend on the sign of the charge: in a flowing current
beam the self-generated magnetic field always acts to “pinch” the beam, i.e. to let it shrink around
the axis.
b) The Lorentz force is F = −e(E + v × B). Thus, we must have E = −v × B in order that F = 0,
i.e.
E = −µ0 ene v 2 r/2 .
(2)
From Gauss’s theorem we find that this field is generated by a charge density ρ which is uniform
over the cylinder, such that E = ρr/2ǫ0 . Thus ρ = −ǫ0 µ0 ene v 2 = −ene (v/c)2 . Since ρ = e(Zni − ne )
also holds, we obtain
ne =
Zni
.
1 − v 2 /c2
(3)
The electron density is uniform, but it exceeds the value ne0 = Zni that would ensure ρ = 0. This
means that there has been some accumulation of electrons inside the wire, rising the density by the
factor (1 − v 2 /c2 )−1 . The “missing” positive charge is located on the surface of the conductor.
c) For an electron in an Ohmic conductor usually v ≈ 1 cm/sec = 10−2 m/sec, thus (v/c)2 ≈ 10−21 ;
the “pinch” effect is extremely small and cannot be observed. The effect may be strong in high
density particle beams or plasma columns where v is not so small with respect to c.
To get further convinced that the effect is negligibly small in a standard conductor, let us assume that the piling up of the electrons in a central region of
d
radius a − d of a wire of radius a leaves a layer of ion density between r = a − d
and r = a. The thickness d can be estimated by the constraint of charge conservation, i.e. the ion charge δQ inside the outer layer, must compensate the excess
electron charge in the center. By considering the charge balance for a length ℓ we
have
a
.
δQ = Zeni π[a2 − (a − d)2 ]ℓ = −ρπ(a − d)2 ℓ,
(4)
2
which to order (v/c)2 ≪ 1 and assuming consistently d ≪ a, can be written as
Zeni (2πad) ≃ −ρ(πa2 ) ≃ Zeni (v/c)2 (πa2 ).
(5)
We thus get d ≃ (a/2)(v/c)2 , so that making d of the order of the crystal lattice spacing (≃ 10−10 m)
would require a ≃ 1011 m, a remarkably large radius!
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