Physics Group - II (3110018)

Here, ${\mathrm{T}}_{\mathrm{c}}$ = Critical Temperature $= 4 \mathrm{K}$

Here, $\mathrm{T}$ = Temperature $= 2 \mathrm{K}$

Here, $\mathrm{r}$ = Redius $= \frac{\mathrm{D}}{2} = \frac{2}{2} \mathrm{mm} = 1 \mathrm{mm} = 1 \times {10}^{-3} \mathrm{m}$

Here, ${\mathrm{H}}_0$ = Critical Field at 0 $= 6.5 \times {10}^4 \mathrm{T}$

Now, ${\mathrm{H}}_{\mathrm{c}} = {\mathrm{H}}_0 \left[ 1 - {\left(\frac{\mathrm{T} }{ {\mathrm{T}}_{\mathrm{c}} }\right)}^2 \right]$

     $= 6.5 \times {10}^4 \times \left[ 1 - {\left(\frac{2 }{ 4 }\right)}^2 \right]$

     $= 6.5 \times {10}^4 \times \left[ 1 - 0.25 \right]$

     $= 4.8 \times {10}^4 \mathrm{T}$

Now, Critical current 

${\mathrm{I}}_{\mathrm{c}} = 2 \mathrm{\pi } \mathrm{r} {\mathrm{H}}_{\mathrm{c}}$

$\Rightarrow {\mathrm{I}}_{\mathrm{c}} = 2 \times 3.14 \times 1 \times {10}^{-3} \times 4.8 \times {10}^4$

$\Rightarrow {\mathrm{I}}_{\mathrm{c}} = 301.44 \mathrm{A}$