- #1
calvinjhfeng
- 32
- 0
(In SI units)
Start with London's 2nd equation in Superconductivity, curl J = 1/(μ*λ²), and Ampere's curl B = μ*j.
Then we curl both side curl curl B = μ* curl J and we do the substitution.
So curl curl B = 0 - del²B which is the laplacian operator.
My question is...how to integrate it?
the equation becomes -del²B = -B/λ²
And I am not sure how to integrate it to solve for B(x). Am I supposed to?
I don't understand how the books jump to B(x) = B*exp(-x / λ)
Please help and thank you very much. Any input is much appreciated too.
Start with London's 2nd equation in Superconductivity, curl J = 1/(μ*λ²), and Ampere's curl B = μ*j.
Then we curl both side curl curl B = μ* curl J and we do the substitution.
So curl curl B = 0 - del²B which is the laplacian operator.
My question is...how to integrate it?
the equation becomes -del²B = -B/λ²
And I am not sure how to integrate it to solve for B(x). Am I supposed to?
I don't understand how the books jump to B(x) = B*exp(-x / λ)
Please help and thank you very much. Any input is much appreciated too.