Superconductor numericals

Question 1: Transition Temperature & Critical Field “Determine the transition temperature and critical field at 4.2 K for a given specimen of a semiconductor [Note: This is a typo in the exam paper; it should say ‘superconductor’], if the critical fields are $1.41\times 10^5\text{ A/m}$ and $4.205\times 10^5\text{ A/m}$ at 14.1 K and 12.9 K respectively.”

• Years of Appearance: 2018, 2022

Explanation: The critical magnetic field $H_c(T)$ of a superconductor at any temperature $T$ follows a parabolic law mathematically defined as: $H_c(T)=H_0\left[1-{\left(\frac{T}{T_c}\right)}^2\right]$ Where:

• $H_0$ is the critical field at absolute zero ($0\text{ K}$).
• $T_c$ is the transition (critical) temperature.

Step-by-Step Solution: Step 1: Set up the equations based on given data. Let $H_1=1.41\times 10^5\text{ A/m}$ at $T_1=14.1\text{ K}$ Let $H_2=4.205\times 10^5\text{ A/m}$ at $T_2=12.9\text{ K}$

1. $1.41\times 10^5=H_0\left[1-{\left(\frac{14.1}{T_c}\right)}^2\right]$
2. $4.205\times 10^5=H_0\left[1-{\left(\frac{12.9}{T_c}\right)}^2\right]$

Step 2: Solve for the Transition Temperature ($T_c$). Divide equation (1) by equation (2) to eliminate $H_0$: $\frac{1.41\times 10^5}{4.205\times 10^5}=\frac{1-\frac{198.81}{T_c^2}}{1-\frac{166.41}{T_c^2}}$ $0.3353=\frac{T_c^2-198.81}{T_c^2-166.41}$ Cross-multiply to solve for $T_c^2$: $0.3353(T_c^2-166.41)=T_c^2-198.81$ $0.3353T_c^2-55.80=T_c^2-198.81$ $198.81-55.80=T_c^2-0.3353T_c^2$ $143.01=0.6647T_c^2$ $T_c^2=215.15⟹{\mathbf{T}}_{\mathbf{c}}=14.67\text{ K}$ (This is the transition temperature).

Step 3: Solve for the critical field at absolute zero ($H_0$). Substitute $T_c$ back into equation (1): $1.41\times 10^5=H_0\left[1-{\left(\frac{14.1}{14.67}\right)}^2\right]$ $1.41\times 10^5=H_0[1-0.9238]$ $H_0=\frac{1.41\times 10^5}{0.0762}=18.50\times 10^5\text{ A/m}$

Step 4: Calculate the critical field at 4.2 K. $H_c(4.2)=18.50\times 10^5\left[1-{\left(\frac{4.2}{14.67}\right)}^2\right]$ $H_c(4.2)=18.50\times 10^5\left[1-0.0820\right]$ $H_c(4.2)=18.50\times 10^5\times 0.918=16.98\times 10^5\text{ A/m}\text{ (or }1.7\times 10^6\text{ A/m}\text{)}$

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Question 2: Superconducting Solenoids (Example 8.10) (This is the one we discussed previously, included here so your list is exhaustive). “Suppose that we have a superconducting solenoid that is 10 cm in diameter and 1 m in length and has 500 turns of $N{b}_{3}Sn$ wire. What is the current necessary to set up a field of 5 T at the center? What will be the corresponding current density? What is the approximate energy stored in the solenoid?”

Explanation: This evaluates the properties of a superconducting electromagnet. The field inside a long solenoid depends on the turn density and current. The energy stored is the magnetostatic energy density multiplied by the core volume.

Step-by-Step Solution:

• Current ($I$): Using $B=\frac{{\mu }_{0}NI}{\ell }$, we rearrange to $I=\frac{B\ell }{{\mu }_{0}N}$. $I=\frac{5\times 1}{(4\pi \times 10^{-7})\times 500}=7958\text{ A}$.
• Current Density ($J_{wire}$): The area of one wire is needed. 500 turns pack into 1 m, so the diameter of one wire is $1/500=0.002\text{ m}$ (radius $r=0.001\text{ m}$). Cross-sectional area $A=\pi (0.001{)}^2=3.14\times 10^{-6}{\text{ m}}^2$. $J_{wire}=\frac{I}{A}=\frac{7958}{\pi \times (0.001{)}^2}=2.53\times 10^9{\text{ A/m}}^2$.
• Energy ($E$): $E=\text{Volume}\times \text{Energy Density}$. Volume $=\ell \times \pi r_{solenoid}^2=1\times \pi (0.05{)}^2=0.00785{\text{ m}}^3$. Energy Density $E_{vol}=\frac{B^2}{2{\mu }_0}=\frac{5^2}{2(4\pi \times 10^{-7})}=9.95\times 10^6{\text{ J/m}}^3$. Total Energy $=(0.00785)\times (9.95\times 10^6)=7.81\times 10^4\text{ J}\text{ (or }78.1\text{ kJ)}$.

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Question 3: Critical Current Density in Wires (Problem 8.23) “Consider two superconducting wires, tin (Sn; Type I) and $N{b}_{3}Sn$ (Type II), each 1 mm in thickness (diameter). The magnetic field on the surface of a current-carrying conductor is given by $B=\frac{{\mu }_{0}I}{2\pi r}$. (a) Assuming Sn wire loses its superconductivity when the field at the surface reaches the critical field (0.2 T), calculate the maximum current and hence the critical current density near 0 K. (b) Calculate for $N{b}_{3}Sn$ taking $B_{c2}$ as 24.5 T.”

Explanation: A current flowing through a wire generates its own magnetic field around the surface of the wire. If this self-generated magnetic field exceeds the superconductor’s critical magnetic field ($B_c$ for Type I, $B_{c2}$ for Type II), the wire will lose its superconductivity. This is known as the Silsbee effect.

Step-by-Step Solution: Parameters: Diameter $=1\text{ mm}⟹\text{radius }r=0.5\text{ mm}=0.0005\text{ m}$. Cross-sectional Area $A=\pi r^2=\pi (0.0005{)}^2=7.854\times 10^{-7}{\text{ m}}^2$.

(a) For Tin (Sn; Type I): Critical Field $B_c=0.2\text{ T}$. The maximum current $I_{max}$ is reached when the surface field equals $B_c$: $B_c=\frac{{\mu }_0{I}_{max}}{2\pi r}$ $I_{max}=\frac{B_c(2\pi r)}{{\mu }_0}$ $I_{max}=\frac{0.2\times 2\pi \times 0.0005}{4\pi \times 10^{-7}}=\frac{0.0002}{2\times 10^{-7}}=1000\text{ A}$ The critical current density ($J_c$) is: $J_c=\frac{I_{max}}{A}=\frac{1000}{7.854\times 10^{-7}}=1.27\times 10^9{\text{ A/m}}^2$

(b) For $N{b}_{3}Sn$ (Type II): Upper Critical Field $B_{c2}=24.5\text{ T}$. $I_{max}=\frac{B_{c2}(2\pi r)}{{\mu }_0}$ $I_{max}=\frac{24.5\times 2\pi \times 0.0005}{4\pi \times 10^{-7}}=\frac{0.0245}{2\times 10^{-7}}=122,500\text{ A}$ The critical current density ($J_c$) is: $J_c=\frac{I_{max}}{A}=\frac{122,500}{7.854\times 10^{-7}}=1.56\times 10^{11}{\text{ A/m}}^2$ (Note: As the textbook highlights, Type II materials can carry phenomenally larger currents before self-induced magnetic fields destroy their superconducting state).