Question

Niobium metal becomes a superconductor when cooled below 9 K. Its superconductivity is destroyed when the surface magnetic field exceeds 0.100 T. In the absence of any external magnetic field, determine the maximum current a 5.68-mm-diameter niobium wire can carry and remain superconducting.

Answer 1

Answer:

The current is  $I = 1420 \ A$

Explanation:

From the question we are told that

   The  diameter of the wire is  $d = 5.68 \ mm = 0.00568 \ m$

    The  magnetic field is  $B = 0.100 \ T$

   

Generally the radius of the wire is mathematically evaluated as

       $r = \frac{d}{2}$

substituting values

     $r = \frac{ 0.00568}{2}$

     $r = 0.00284 \ m$

Generally the magnetic field is mathematically represented as

       $B = \frac{\mu_o * I}{ 2 \pi r }$

=>    $I =\frac{ B * 2 \pi r }{\mu_o}$

Here $\mu_o$ is the permeability of free space  with value $\mu_o = 4 \pi *10^{-7} N/A^2$

substituting values

=>     $I =\frac{ 0.100 * 2 * 3.142 * 0.00284 }{ 4 \pi * 10^{-7}}$

=>     $I = 1420 \ A$