Question

Calculate the critical current for a wire of lead having a diameter of 1mm at 4.2 K. Critical temperature for lead is 7.18 K and H0 = 6.5 X 104 A/m.​

Answer 1

Step-by-step 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

A

A

Hope this answer is helpful.

Answer 2

Answer:

Critical current = 13.4A

Step-by-step explanation:

Given,

Diameter of the wire = 1mm = 1×$10^{-3}$

Radius of the wire =r =  $\frac{1}{2} X10^{-3}$

Temperature = T = 4.2K

Critical Temperature = $T_c$ = 7.18K

Critical field = $H_0$ = 6.5 X $10^4$ A/m.

To find

Critical current($I_c$)

Recall the formula

Critical current$I_c = 2\pi rH_c$

Critical Magnetic field = $H_c = H_0[1-(\frac{T}{T_c})^2]$

Solution:

$H_c = H_0[1-(\frac{T}{T_c})^2]$

$H_c = 6.5 X10^4[1-(\frac{4.2}{7.18})^2]$

$H_c = 6.5 X10^4[1-0.3421]$

$= 6.5 X10^4X0.6579$

$=4.2764 X10^4$ A/m.

$I_c = 2\pi rH_c$

=2 × 3.14 ×  $\frac{1}{2} X10^{-3}$ ×$4.2764 X10^4$

=13.4A

∴ Critical current = 13.4A

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