QB_PH_SEC7_020: Dielectric, Magnetic Properties & Superconductivity

The second London equation relating the magnetic induction $\vec{B}$ inside a superconductor to the supercurrent density $\vec{J}_s$ is

Correct Answer

### Step-by-Step Derivation & Verification: 1. **First London Equation (Lossless Acceleration)**: $m \frac{d\vec{v}_s}{dt} = -e \vec{E} \implies \frac{\partial \vec{J}_s}{\partial t} = \frac{n_s e^2}{m} \vec{E}$ 2. **Taking the Curl of Both Sides**: $\frac{\partial}{\partial t}(\nabla \times \vec{J}_s) = \frac{n_s e^2}{m} (\nabla \times \vec{E})$ 3. **Invoking Faraday's Law** ($\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$): $\frac{\partial}{\partial t}\left(\nabla \times \vec{J}_s + \frac{n_s e^2}{m}\vec{B}\right) = 0$ 4. **London Postulate (Meissner Effect)**: F. and H. London postulated that not only is the time derivative zero, but the bracketed quantity itself is identically zero in the superconducting state: $\nabla \times \vec{J}_s = -\frac{n_s e^2}{m} \vec{B}$ 5. **Conclusion**: Option (A) is correct.