Question

Calculate the critical current of a wire of lead having the diameter 5mm.The critical temperature of lead is 7.18K and H=6.5×10^4 A m^-1

Answer 1

the critical current of a wire of lead is 1020.5 A

Explanation:

Given,

The diameter of the wire d = 5 mm

Therefore, the radius of the wire r = 2.5 mm

Critical Magnetic Field Hc = 6.5 × 10⁴ A/m

We know that critical Current flowing through a wire is given by

$I_c=2\pi rH_c$

$\implies I_c=2\pi\times 2.5\times 10^{-3}\times 6.5\times 10^4$

$\implies I_c=2\times 3.14\times 25\times 6.5$ A

$\implies I_c=1020.5$ A

Hope this answer is helpful.

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