Here are the consolidated answers to the Past Year Questions (PYQs) from the Superconductivity module. Repeated questions have been skipped or combined into single comprehensive answers.

Q1. “The limits of superconductivity are defined by critical temperature, critical magnetic field and critical current density”- Justify with relevant figures.

The superconducting state is a fragile phase that relies on three interdependent threshold parameters:

1. Critical Temperature ($T_c$): The characteristic temperature below which the material exhibits superconductivity. Above $T_c$, it reverts to a normal resistive state.
2. Critical Magnetic Field ($B_c$ or $H_c$): The maximum external magnetic field the material can withstand. If the applied field exceeds this value, the magnetic flux forces its way into the material and destroys the superconductivity.
3. Critical Current Density ($J_c$): The maximum electrical current density that can be passed through the sample. When a massive current flows through a superconductor, it generates its own magnetic field at the surface; if this self-generated field becomes sufficiently high, it will exceed the critical field and extinguish the superconductivity.

Justification/Figure Description: These three limiting factors constitute a 3D critical surface (with axes $T$, $B$, and $J$) that serves as a boundary separating the superconducting state from the normal state. Any operating point $(T_1,B_1,J_1)$ that falls inside the volume bounded by this surface remains perfectly superconducting, whereas pushing the material outside this envelope reverts it to a normal state. (Note: For your exam, you should draw a 3D surface plot with $T_c$, $B_c$, and $J_c$ bounding the origin to represent this concept).

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Q2. What is superconductivity/superconductor? Show that, superconductors are diamagnetic in nature.

Definition: Superconductivity is a phenomenon characterized by exactly zero electrical resistance and the complete expulsion of magnetic flux fields. It occurs in certain materials, called superconductors, when they are cooled below a characteristic threshold known as the critical temperature ($T_c$).

Proof of Perfect Diamagnetism: According to the Meissner effect, when a superconductor is cooled below $T_c$, the magnetic field inside the bulk of the material is exactly zero ($B=0$). The fundamental relationship between the internal magnetic field ($B$), the applied magnetizing field ($H$), and the material’s magnetization ($M$) is: $B={\mu }_{0}H+{\mu }_{0}M={\mu }_0(H+M)$

By substituting $B=0$ into the equation, we get: $0={\mu }_0(H+M)$ This simplifies to: $H=-M$

The magnetic susceptibility ${\chi }_m$ is defined as the ratio of magnetization to the applied field: $\frac{M}{H}=-1⟹{\chi }_m=-1$ Because a magnetic susceptibility of $-1$ is the maximum possible negative value, this mathematically proves that superconductors are perfect (ideal) diamagnets.

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Q3. Briefly explain the Meissner effect.

The Meissner effect is the thermodynamic phenomenon where a superconductor actively expels all magnetic fields from its bulk interior when cooled below its critical temperature ($T_c$), behaving as a perfectly diamagnetic substance.

• If the temperature $T>T_c$, external magnetic field lines penetrate the sample normally.
• When $T<T_c$, the superconductor develops surface currents to generate a magnetization $M$ that perfectly cancels the applied field everywhere inside the sample.
• If the external applied field is simply switched off, the field around the superconductor disappears.

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Q4. Write down the characteristics of Type I and Type II superconductors. Why in all engineering applications of superconductor Type II materials are used?

Characteristics and Differences:

• Type I Superconductors: These materials act as perfect diamagnets and completely exclude the applied magnetic field (Meissner state) right up until the field reaches a specific critical field, $B_c$. The moment the applied field exceeds $B_c$, superconductivity disappears abruptly, magnetic flux completely penetrates the sample, and the material reverts to its normal resistive state.
• Type II Superconductors: These materials feature two critical magnetic fields: a lower critical field ($B_{c1}$) and an upper critical field ($B_{c2}$). Below $B_{c1}$, they behave exactly like Type I materials, completely expelling the magnetic flux. However, between $B_{c1}$ and $B_{c2}$, they enter a mixed state (or vortex state). In this state, magnetic flux lines partially pierce the sample through microscopic normal regions, but the bulk of the material surrounding them remains superconducting. Superconductivity is finally destroyed only when the field exceeds $B_{c2}$.

Why Type II is used in Engineering Applications: All engineering applications of superconductors invariably use Type II materials. This is because the upper critical field ($B_{c2}$) of Type II materials is extraordinarily high compared to the low critical field ($B_c$) found in Type I materials. This characteristic allows Type II superconductors to carry massive currents and generate or withstand powerful magnetic fields (such as those needed in MRI machines or particle accelerators) while successfully remaining in the superconducting mixed state.

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Q5. Determine the transition temperature and critical field at 4.2 K for a given specimen, if the critical fields are $1.41\times 10^5$ and $4.205\times 10^5$ A/m at 14.1 K and 12.9 K respectively.

(Note: As discussed in our previous analysis, solving this requires the standard parabolic temperature dependence formula $H_c(T)=H_c(0)\left[1-{\left(\frac{T}{T_c}\right)}^2\right]$, which is an expected physics prerequisite for this course).

Step 1: Set up the Equations Using the given data: (Eq 1): $1.41\times 10^5=H_c(0)\left[1-{\left(\frac{14.1}{T_c}\right)}^2\right]$ (Eq 2): $4.205\times 10^5=H_c(0)\left[1-{\left(\frac{12.9}{T_c}\right)}^2\right]$

Step 2: Solve for Transition Temperature ($T_c$) Divide Eq 2 by Eq 1 to cancel out $H_c(0)$: $\frac{4.205}{1.41}=\frac{1-(12.9/T_c{)}^2}{1-(14.1/T_c{)}^2}$ $2.982=\frac{T_c^2-166.41}{T_c^2-198.81}$ $2.982(T_c^2-198.81)=T_c^2-166.41$ $2.982T_c^2-592.85=T_c^2-166.41$ $1.982T_c^2=426.44$ $T_c^2=215.15⟹{\mathbf{T}}_{\mathbf{c}}=14.67\text{ K}$

Step 3: Solve for Critical Field at Absolute Zero ($H_c(0)$) Substitute $T_c$ back into Eq 1: $1.41\times 10^5=H_c(0)\left[1-{\left(\frac{14.1}{14.67}\right)}^2\right]$ $1.41\times 10^5=H_c(0)[1-0.924]=H_c(0)\times 0.076$ $H_c(0)=\frac{1.41\times 10^5}{0.076}=1.855\times 10^6\text{ A/m}$

Step 4: Solve for Critical Field at 4.2 K Now, plug 4.2 K into the main formula: $H_c(4.2)=1.855\times 10^6\left[1-{\left(\frac{4.2}{14.67}\right)}^2\right]$ $H_c(4.2)=1.855\times 10^6\times [1-0.082]$ $H_c(4.2)=1.855\times 10^6\times 0.918=1.70\times 10^6\text{ A/m}$