The element niobium, which is a metal, is a superconductor (i.e., no electrical resistance) at temperatures below 9 K. However, the superconductivity is destroyed if the magnetic field at the surface of the metal reaches or exceeds 0.10 T.What is the maximum current in a straight, 3.2-mm-diameter superconducting niobium wire?

Magnetic field represented by B=0.1T, and is defined as"A field generated when an electric charge via electrons move through the space or in an electrical conductor."

I represent the current (variable os interest)

r represent the radius= 3.2/2 = 1.6mm = 0.0016m

2) Formula to use

For this case we can use the magnetic field due to a wire with current formula, given by:

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

[tex]\mu=4\pi x10^{-7} \frac{TA}{m}[/tex] is a constant

Solving for I from the last formula we got:

[tex]I=\frac{B(2\pi r)}{\mu}[/tex]

And replacing the values given we have:

[tex]I=\frac{0.1T(2\pix0.0016m)}{4\pi x10^{-7} \frac{TA}{m}}=800A[/tex]

So the current passing thorugh the wire would be 800 A.