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Comprehensive Notes on Superconductivity: Part 2

5. Type I and Type II Superconductors

While all superconductors exhibit zero resistance and the Meissner effect, they respond differently to strong applied magnetic fields. Based on this magnetic behavior, they are classified into two categories: Type I and Type II.

Magnetic Field Penetration Depth ($\lambda$)

Before distinguishing the two types, it is important to understand that the expulsion of the magnetic field (Meissner effect) is not perfectly abrupt at the surface. The external magnetic field actually penetrates the sample by a tiny amount, decreasing exponentially from the surface into the bulk: $B(x)=B_o\exp (-x/\lambda )$ Where $B_o$ is the field at the surface and $\lambda$ is the penetration depth.

Type I Superconductors

• Behavior: Type I superconductors act as perfect diamagnets (${\chi }_m=-1$) and completely exclude the magnetic field right up to a specific critical magnetic field, $B_c$.
• Transition: The moment the external field exceeds $B_c$, the superconducting state is abruptly destroyed, and the material instantly reverts to a normal, resistive state, allowing the magnetic field to fully penetrate.
• Drawback: Their critical field $B_c$ is typically very low (often less than 0.1 Tesla). Therefore, they cannot be used to generate strong magnetic fields because their own generated field would easily exceed $B_c$ and destroy their superconductivity.

Type II Superconductors

• Behavior: Type II superconductors have two critical magnetic fields: a lower critical field ($B_{c1}$) and an upper critical field ($B_{c2}$).

• The Vortex (Mixed) State:

  • Below $B_{c1}$, the material behaves exactly like a Type I superconductor (perfect Meissner effect).
  • Between $B_{c1}$ and $B_{c2}$, the material enters a mixed or vortex state. Instead of abruptly losing superconductivity, the material allows the magnetic field to partially penetrate the bulk in the form of microscopic, quantized tubes of magnetic flux called vortices. The regions inside these flux tubes become normal, but the bulk of the material surrounding them remains perfectly superconducting.
  • Above $B_{c2}$, the vortices overlap, and the material becomes completely normal.

• Conceptual Intuition (Answer to Slide Assignment): Why are Type II materials exclusively used in engineering applications? The upper critical field ($B_{c2}$) of a Type II superconductor can be extraordinarily high (tens of Teslas). Because they can remain superconducting in the vortex state even when subjected to massive magnetic fields, Type II materials (like Niobium-Titanium alloys) can be wound into coils to create incredibly powerful electromagnets, such as those required for MRI machines, particle accelerators (like the LHC), and maglev trains.

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6. Critical Current Density ($J_c$)

As defined by the 3D critical surface in Part 1, a material will only remain superconducting if the current flowing through it remains below a specific threshold known as the critical current density ($J_c$).

• Definition: $J_c$ is the maximum electrical current per unit cross-sectional area that a superconductor can carry without any electrical resistance. If $J>J_c$, the material reverts to its normal resistive state.
• Conceptual Intuition (Silsbee’s Rule): Why does an electrical current destroy superconductivity? According to Ampere’s Law, any electrical current generates a surrounding magnetic field. If a massive current is pumped through a superconducting wire, it generates a correspondingly massive magnetic field at the surface of the wire. If this self-generated magnetic field exceeds the material’s critical magnetic field ($B_c$ or $B_{c2}$), the superconductivity is destroyed. Therefore, the critical current is inherently linked to the critical magnetic field.

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7. Mathematical Problem: Example 8.10 (Kasap’s Textbook)

This problem demonstrates the massive amount of energy that can be stored in the magnetic field of a superconducting electromagnet (solenoid) without any Joule heating ($I^{2}R$) losses.

Problem Statement: Consider a superconducting solenoid (coil) that generates a massive internal magnetic field of $B=5\text{ T}$. Assume the solenoid has a cylindrical core with a length of $L=1\text{ m}$ and a radius of $r=0.05\text{ m}$. Calculate the total magnetostatic energy stored in the solenoid.

Step-by-Step Derivation: 1. Calculate the Energy Density ($E_{vol}$): The energy stored per unit volume in a magnetic field in free space (or air) is given by the formula: $E_{vol}=\frac{B^2}{2{\mu }_o}$ Where ${\mu }_o$ is the permeability of free space ($4\pi \times 10^{-7}\text{ T}\cdot \text{m/A}$). Plugging in the values: $E_{vol}=\frac{(5\text{ T}{)}^2}{2(4\pi \times 10^{-7}\text{ T}\cdot \text{m/A})}=9.95\times 10^6{\text{ J m}}^{-3}$

2. Calculate the Volume of the Solenoid ($V$): The volume of a cylinder is $V=\text{Length}\times \text{Cross-sectional Area}$. $V=L\times (\pi r^2)$ $V=(1\text{ m})\times (\pi \times (0.05\text{ m}{)}^2)=0.007854{\text{ m}}^3$

3. Calculate the Total Stored Energy ($E$): The total energy is the energy density multiplied by the volume. $E=E_{vol}\times V$ $E=(9.95\times 10^6{\text{ J m}}^{-3})\times (0.007854{\text{ m}}^3)$ $E=7.81\times 10^4\text{ J}=78.1\text{ kJ}$

Significance & Conceptual Intuition: The solenoid is storing 78.1 kJ of energy purely in the form of a magnetic field. Because the coil is made of superconducting wire ($\rho =0$), the current will flow forever without dissipating any heat. If this stored energy were extracted and converted entirely into work, it could power a standard 100 W lightbulb continuously for 13 minutes, or it could theoretically lift a massive 7,900 kg truck exactly 1 meter into the air! This demonstrates the incredible potential of superconducting Magnetic Energy Storage (SMES) systems for power grids.