Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 1 / 16B / iii

Use the identity

\nabla \times B = \nabla (\nabla \cdot A) - \nabla^{2} A

. The Laplacian identity for the Coulomb kernel gives, component by component,

- \nabla^{2} A = - \frac{μ _{0}}{4 π} \int J (r^{'}) \nabla^{2} (s^{- 1}) d V^{'} = μ_{0} J (r) .

(1)

Also

\nabla (s^{- 1}) = - \nabla^{'} (s^{- 1})

, so integration by parts yields

\nabla \cdot A = - \frac{μ _{0}}{4 π} \int J (r^{'}) \cdot \nabla^{'} (s^{- 1}) d V^{'} = \frac{μ _{0}}{4 π} \int \frac{\nabla ^{'} \cdot J ( r ^{'} )}{s} d V^{'} .

(2)

Here the surface integral at infinity vanishes under the assumed localization of the current. As usual, this assumption means sufficient decay for these integrals and the boundary limit, rather than merely pointwise convergence to zero. Combining the two terms gives

\nabla \times B = μ_{0} J + \frac{μ _{0}}{4 π} \nabla \int \frac{\nabla ^{'} \cdot J ( r ^{'} )}{∣ r - r ^{'} ∣} d V^{'} .

(3)

 Ancestors (10)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version

Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 1 / 16B / iii

Solution

 Ancestors (10)

 Discussion (0)  Subscribe (1) New discussion

 Articles by others on the same topic (0)

 Discussion (0) New discussion