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

The current is
`I = 1420 \ A`

Step-by-step 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 = (d)/(2)`

substituting values

`r = ( 0.00568)/(2)`

`r = 0.00284 \ m`

Generally the magnetic field is mathematically represented as

`B = (\mu_o * I)/( 2 \pi r )`

=>
`I =( 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 =( 0.100 * 2 * 3.142 * 0.00284 )/( 4 \pi * 10^(-7))`

=>
`I = 1420 \ A`