Download Solid State 2 – Homework 9 Use the Maxwell equation

Solid State 2 – Homework 9
1) Comparison between perfect conductivity and perfect diamagnetism:
Use the Maxwell equation
∂B
∂t
Using the equation above, write an expression for B, for a perfect
conductor.
Suppose a material is a perfect conductor. Apply an external magnetic
field. What is the magnetic field in the material? Explain !
Suppose the material is a perfect conductor only at temperatures below
Tc. Start with a sample at T>Tc, apply an external magnetic field and then,
keeping the external field constant, decrease the temperature below Tc.
What is the magnetic field in the material? Explain !
Repeat parts (b) and (c) for a perfect diamagnet. Is there a difference in
the behaviour ? If so, what is it, and how does it happen ?
∇× E = −
a)
b)
c)
d)
2) The coexistence of the normal and superconducting states:
a) We can use the Helmholtz free energy F(B,T,N) for cases where the magnetic
field B inside a material is constant. But when we set the external magnetic field
constant, we need to minimize a different energy: X(H,T,N) . Write an expression
for X and identify it with a thermodynamic energy you know.
b) Now calculate the difference between the energy X in the normal and
superconducting states:
Xnormal – Xsuperconducting = ?
c) Using the definition of the critical field:
VH c 2
Fn − Fs =
8π
Find the value of the applied field Ha so that the normal and superconducting
states will be in equilibrium.
3) A perfect diamagnet pushes the field lines away from it, so we can expect a
higher field outside it:
S
If the superconducting sample is large enough, we can expect that the field
pushed sideways will be come higher than the critical field Hc thereby destroying
superconductivity in those regions.
a) For the geometry above, suggest a way for superconducting-normal (SN)
regions to arrange themselved.
b) What is the value of the magnetic field in the normal regions (where it has
destroyed superconductivity). Explain. (Hint: Q2).
c) We can call the transition region between a superconducting domain to a
normal one a ‘domain wall’. For a perfect diamagnet, what is the sign of the
energy required to make a domain wall ? What about a partial diamagnet ?
4) Abrikosov vortex energy:
Consider an isolated vortex (see figure below) which is called the Abrikosov
vortex line (Abrikosov, 1956). It has a core with a radius ξ in which the density ns
of SC electrons decays to zero in the center. The magnetic field is maximal in the
center while at r > λ it decays as a result of screening by circular currents.
a) Find the kinetic energy of the electrons in the vortex state for a ring of width dr.
b) Write the total energy (kinetic + magnetic) of a vortex as an integral. Ignore the
energy of the (tiny) normal core.
=
∫ξ dr (???)
r>
c) Derive the London equation (5.7 from lecture notes) from the condition of
minimal energy . Identify λL . Hint:
∫ rot (h)rot (δ h)dr = ∫ rot (rot (h))δ hdr
d) In the interior of the vortex, we need to replace the London equation by
something more complicated. (Remember we integrated only for r > ξ ). But
since the hard core has a very small radius, we can replace it with a 2D delta
function:
h + λ 2 rot (rot (h)) = φ0δ 2 (r )
By integrating the above equation show that φ0 is all the flux carried in the
vortex.
e) Now we define h along the z-axis. Find rot(h) and h(r) in the region ξ < r ≪ λ .
λ
Bring it to the form h ∝ ln   + const .
r
Hint: Use the same integral you used in part (d).
f) The exact solution for h(r) is
r
h( r ) ∝ K 0  
λ
K0 being the zero order Bessel function of an imaginary
argument. Luckily, the asymptotic solution is : the constant in part (e) vanishes,
and for large r:
h( r ≫ λ ) ∝ e
−r
λ
.
Now find the energy density per unit length (along z direction) of the vortex.
What happens when the flux is 2φ0 ?