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 3.08-mm-diameter niobium wire can carry and remain superconducting Need Help?Read It

Answer 1

Given data magnetic field exceeds (B)_max=0.100

Diameter=3.08 mm

Radius =D/2

             =3.08/2

              1.54

By ampere’s rule, the magnetic field at distance r from center of wire is given by the formula

B=μ 0 I max/2 π r

Here we have to find I _max

We know the I _max formula is

I _max=2π×1.54×10^-3×0.100/4π×10^-7

I _max=2π×1.54×10^-3+7×0.100/4π

I _max=2π×1.54×10⁴×0.100/4π

I max=770 A

(OR)

We can do in another method it is mentioned below

We know the formula for magnetic field on surface of conductor of radius R is

B=μ0 I/2πR

Here I=B×2πR/μ₀

I=B×2πR/4π×10^-7

I=BR/2×10^-7

I=0.1×1.54×10^-3/2×10^-7

I=0.1×1.54×10⁴/2

I=770 A

Maximum current is 770 A