Answer to Question #251555 in Physics for Jashwanth

Question

A flux quantum (fluxoid) is approximately equal to 2×10^-7 gauss−cm2.A type II superconductor is placed in a small magnetic field, which is then slowly increased till the field starts penetrating the superconductor. The strength of the field at this point is 2/π×10^5 gauss

The penetration depth of this superconductor is

 100 Å

 200 Å

 1000 Å

 314 Å

Answer 1

The relation between critical field and penetration depth is given by

$$H_c=\frac{\Phi_0}{\pi\lambda^2}$$

Hence, the penetration depth

$$\lambda=\sqrt{\frac{\Phi_0}{\pi H_c}}$$

$$=\sqrt{\frac{2×10^{-7}\:\rm gauss*cm^2}{\pi *2/π×10^5\:\rm gauss}}=10^{-6}\:\rm cm=100\:\mathring{A}$$

Answer: $100\rm\:\mathring{A}$