Key physical fact

For a superconductor, the resistance behaves as:

• $R \neq 0$ for $T > T_c(B)$
• $R = 0$ for $T < T_c(B)$

Also, the critical temperature decreases when magnetic field increases: $B_2 > B_1 \implies T_c(B_2) < T_c(B_1).$

Meaning for the $R$ vs $T$ graph

Since $B_2$ is larger than $B_1$, superconductivity is destroyed more easily in field $B_2$. Therefore the transition to zero resistance must occur at a lower temperature for $B_2$ than for $B_1$.

So in the $R$–$T$ plot:

• the curve for $B_1$ should drop to zero at a higher temperature,
• the curve for $B_2$ should drop to zero at a lower temperature.

Relative positions of the two curves

For temperatures between the two critical temperatures, $T_c(B_2) < T < T_c(B_1),$ we have:

• at $B_1$: material is superconducting, so $R=0$,
• at $B_2$: material is still normal, so $R>0$.

Hence the $B_2$ curve must lie to the left/lower-temperature side of the $B_1$ transition.

Correct option

Therefore the correct graph is the one in which the resistance drop for $B_2$ occurs at a lower temperature than for $B_1$.

That corresponds to Option A.

Comparison with stored answer

Stored correct answer = A.

My derived answer = A.

So they agree.