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The London equation in a super conductor is:
Question

The London equation in a super conductor is:

A.

2B=B(mc2)/(4πnq2)\nabla^2 \mathrm{B}=\mathrm{B}\left(\mathrm{mc}^2\right) /\left(4 \pi \mathrm{nq}^2\right)​​

B.

2B=B(mc2)/(4πnq2)1/2\nabla^2 \mathrm{B}=\mathrm{B}\left(\mathrm{mc}^2\right) /\left(4 \pi \mathrm{nq}^2\right)^{1 / 2}​​

C.

2B=B(mc2)/(4πnq2)2\nabla^2 \mathrm{B}=\mathrm{B}\left(\mathrm{mc}^2\right) /\left(4 \pi \mathrm{nq}^2\right)^2​​

D.

ΔB=B(mc2)/(4πnq2)\Delta \mathrm{B}=\mathrm{B}\left(\mathrm{mc}^2\right) /\left(4 \pi \mathrm{nq}^2\right)​​

Correct option is A

The London equations describe the electromagnetic behavior of superconductors.2B=B mc24πnq2Where:B=Magnetic fieldm=Mass of the charge carrier (usually electron)q=Charge of the carriern=Number density of superconducting carriersc=Speed of light in vacuum\bullet \text{The \textbf{London equations} describe the electromagnetic behavior of superconductors.} \\\nabla^2 \vec{B} = \frac{B\, mc^2}{4\pi n q^2}\\[10pt]\textbf{Where:} \\\bullet \vec{B} = \text{Magnetic field} \\\bullet m = \text{Mass of the charge carrier (usually electron)} \\\bullet q = \text{Charge of the carrier} \\\bullet n = \text{Number density of superconducting carriers} \\\bullet c = \text{Speed of light in vacuum}​​

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