Electromagnetic torque
A plane, monochromatic, circularly polarized EM wave of wavelength λ = 2πc/ω and amplitude E0
impinges on a small metallic sphere of radius a ≪ λ. We assume to be in the low-frequency regime
such that the metal can be considered as an Ohmic conductor having a conducitivity σ equal to the
static value and, thus, independent on the frequency ω.
a) Find the induced dipole momentum in the sphere.
b) Show that the EM wave exerts a torque on the sphere.
1
Solution
a) We can write the EM wave as
E(z, t) = E0 [x̂ cos(kz − ωt) − ŷ sin(kz − ωt)]
= Re[E0 (x̂ + iŷ)ei(kz−ωt) ] ,
(1)
where k = ω/c = 2π/λ. Since a ≪ λ, we can neglect both the spatial variation of the field across
the volume of the sphere and magnetic induction effects, so that the sphere can be considered inside
an uniform rotating field
Ẽ0 = E0 (x̂ + iŷ) .
(2)
E0 (t) = Re Ẽ0 e−iωt ,
For harmonic fields a real conductivity σ is equivalent to a dielectric function
ǫr =
ǫ
iσ
=1+
,
ǫ0
ωǫ0
(3)
and we recognize that the problem is equivalent to that of a dielectric sphere in an uniform field.
Thus the amplitudes of the internal electric field and the dipole momentum are given by
3
3Ẽ0
3iωτ
Ẽ0 =
=−
Ẽ0 ,
ǫr + 2
3 + i(σ/ωǫ0 )
1 − 3iωτ
ǫr − 1
i(σ/ωǫ0 )
3ǫ0 V
p̃ = PV = ǫ0 χEint V = 3ǫ0 V
Ẽ0 = 3ǫ0 V
Ẽ0 =
Ẽ0 ,
ǫr + 2
3 + i(σ/ωǫ0 )
1 − 3iωτ
Ẽint =
where V = 4πa3 /3 and τ = ǫ0 /σ. By writing
p
1 − 3iωτ = 1 + (3ωτ )2 e−iφ ,
tan φ = 3ωτ ,
(4)
(5)
(6)
one obtains
3V ǫ0 Ẽ0
p̃ = p
eiφ ,
2
1 + (3ωτ )
i.e.
p = Re
3V ǫ0 E0
p
(x̂ + iŷ)e−i(ωt−φ)
2
1 + (3ωτ )
(7)
!
.
(8)
Thus the dipole momentum rotates with a phase delay φ with respect to the electric field of the
wave.
b) The torque on a dipole is given by M = p × E0 . Since the angle between p and E0 is constant
and equal to φ, we immediately find
3V ǫ0 E02
M = |p||E0 | sin φ = p
ẑ sin φ .
1 + (3ωτ )2
2
(9)
This is also given by the explicit calculation
where we used
1
3V ǫ0
hMi = hp × E0 i = Re(p̃ × Ẽ∗0 ) = p
Re(eiφ Ẽ0 × Ẽ∗0 )
2
2
2 1 + (3ωτ )
3V ǫ0 E02
3V ǫ0
Re(−2ieiφ E02 ẑ) = p
= p
ẑ sin φ ,
2 1 + (3ωτ )2
1 + (3ωτ )2
Ẽ0 × Ẽ∗0 = E02 (x̂ + iŷ) × (x̂ − iŷ) = E02 (−ix̂ × ŷ + iŷ × x̂) = −2iE02 ẑ .
3
(10)
(11)