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.

Respuesta :

Answer:

The current is  [tex]I = 1420 \ A[/tex]

Explanation:

From the question we are told that

   The  diameter of the wire is  [tex]d = 5.68 \ mm = 0.00568 \ m[/tex]

    The  magnetic field is  [tex]B = 0.100 \ T[/tex]

   

Generally the radius of the wire is mathematically evaluated as

       [tex]r = \frac{d}{2}[/tex]

substituting values

     [tex]r = \frac{ 0.00568}{2}[/tex]

     [tex]r = 0.00284 \ m[/tex]

Generally the magnetic field is mathematically represented as

       [tex]B = \frac{\mu_o * I}{ 2 \pi r }[/tex]

=>    [tex]I =\frac{ B * 2 \pi r }{\mu_o}[/tex]

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

substituting values

=>     [tex]I =\frac{ 0.100 * 2 * 3.142 * 0.00284 }{ 4 \pi * 10^{-7}}[/tex]

=>     [tex]I = 1420 \ A[/tex]