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Chapter 9: Problem 80

What is the critical magnetic field for lead at \(T=2.8 \mathrm{K} ?\)

Short Answer

Expert verified
The critical magnetic field for lead at \(T = 2.8 \mathrm{K}\) is approximately \(5.51 \times 10^4 \mathrm{A/m}\).

Step by step solution

01

Set up the formula

Write out the formula for the critical magnetic field in terms of temperature: \[B_c(T) = B_{c0}\Biggl[1-\Biggl(\frac{T}{T_c}\Biggr)^2\Biggr]\]
02

Plug in the values

Plug the given values for \(B_{c0}\), \(T_c\), and \(T\) into the formula: \[B_c(2.8) = 6.5 \times 10^4 \mathrm{A/m} \Biggl[1-\Biggl(\frac{2.8}{7.2}\Biggr)^2\Biggr]\]
03

Compute the fraction

Compute the fraction inside the brackets: \[\frac{2.8}{7.2} \approx 0.3889\]
04

Square the fraction

Square the calculated fraction: \[(0.3889)^2 \approx 0.1513\]
05

Calculate the expression inside the brackets

Subtract the squared fraction from 1: \[1 - 0.1513 \approx 0.8487\]
06

Multiply the result by \(B_{c0}\)

Multiply the calculated expression by the value of \(B_{c0}\): \[B_c(2.8) = 6.5 \times 10^4 \mathrm{A/m} \times 0.8487 \approx 5.51 \times 10^4 \mathrm{A/m}\] The critical magnetic field for lead at \(T = 2.8 \mathrm{K}\) is approximately \(5.51 \times 10^4 \mathrm{A/m}\).

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