Download R H state will occur. [5 points] ∇ ·

7510: Homework # 8
Problem 13. Consider a cylindrical superconductor of radius R (and infinite length)
placed in the external magnetic field H which is directed perpendicular to the cylinder axis.
The critical magnetic field is Hc . Find the range of magnetic fields H for which intermediate
state will occur. [5 points]
Solution. Outside of the superconductor magnetic field satisfies ∇ · H=0, ∇ × H = 0.
The second condition is taken care of by H = ∇ψ, while the first one gives Laplace equation
∇2 ψ = 0. From the separation of variables it is found that any function rl cos (lθ) with
arbitrary l satisfies Laplace equation. The solution should be looked in the form,
ψ = Hr cos θ + C
cos θ
,
r
where the first term is the applied field, and the second term is the field created by the
cylinder (since it vanishes at r → ∞ only l < 0 are allowed, in addition any |l| 6= 1 will
not satisfy the boundary condition). The constant C is found from vanishing of the normal
component of H at the surface:
∂ψ
|r=R = 0,
∂r
which gives C = HR2 . Tangential component of magnetic field at the surface of a cylinder,
Hθ (R) =
1 ∂ψ
|r=R = −2H sin θ.
r ∂θ
The maximal value of |H(R)| is reached at θ = π/2 and is 2H. When this value exceeds Hc
the superconductor goes into intermediate state. Thus, for Hc /2 < H < Hc intermediate
state is realized.
Problem 14. Find surface density of the induced supercurrents as a function of the
position on a cylinder (consider a simple limit of small penetration depth λ/R → 0. Find the
total magnetic moment (per unit of length of the cylinder) created by these supercurrents.
[5 points]
Solution. The (volume) density of electric currents is found from the equation ∇ × H =
4πj/c. Writing z-component of this equation, we get,
1 ∂(rHθ ) 1 ∂Hr
4π
−
=
jz
r ∂r
r ∂θ
c
The second term on the left-hand side vanishes at the surface of the superconductor, while
the first term has a singularity. Integrating it from r = R − 0 to r = R + 0, we find the
density of surface currents
R+0
Z
jz (r)dr =
R−0
Hc
c
[Hθ (R + 0) − Hθ (R − 0)] = −
sin θ.
4π
2π