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(B) superconductor exhibits the Meissner effect,which is the expulsion of magnetic field lines from its interior when it is cooled below its critical

Vedclass · Accepted Answer

$B_s = 0$

Vedclass · Answer

(B) superconductor exhibits the Meissner effect,which is the expulsion of magnetic field lines from its interior when it is cooled below its critical temperature in an external magnetic field.This phenomenon makes a superconductor a perfect diamagnet.For a perfect diamagnet,the magnetic susceptibility $\chi = -1$.The magnetic field inside the material is given by $B_s = \mu_0(H + M)$. Since $M = \chi H = -H$,we get $B_s = \mu_0(H - H) = 0$.Therefore,the magnetic field inside the superconductor is $B_s = 0$.

An example of a perfect diamagnet is a superconductor. This implies that when a superconductor is placed in an external magnetic field of intensity $B$,the magnetic field $B_s$ inside the superconductor will be such that

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