EMLecture11 - Lecture 11 Notes, 95.657 Electromagnetic...

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Lecture 11 Notes, 95.657 Electromagnetic Theory I, Fall 2010 Dr. Christopher S. Baird, UMass Lowell 1. Magnetic Fields of a Localized Current Distribution - Similar to the electrostatic multipole expansion, if we have a localized current distribution and want to know the magnetic fields far away, we can make a magnetostatic multiplole expansion and only keep the first few terms. - Consider the Biot-Savart Law for the vector potential: A = 0 4 J x ' x x ' d x ' where B x =∇× A and this is a solution to ∇× B x = 0 J x - The primed coordinates refer to the localized currents, so that an expansion in powers of the primed variables will allow us to drop higher order terms: 1 x x ' = 1 x x x ' x 3 ... - Using this expansion: A = 0 4 1 x J x ' d x ' 0 4 x x 3 x ' J x ' d x ' ... - The first term corresponds to the magnetic field generated by a magnetic monopole moment. We already know that there are no monopoles, so that this term should be zero. Let us prove it. - Consider a functions f and g and make an expansion using integration by parts: V g J ⋅∇ ' f d x ' = S f g J ⋅ x ' da V f ' ⋅ g J d x ' - Because the current is localized, if the bounding surface is larger than the current distribution, the current density will be everywhere zero on the surface, so that: V g J ⋅∇ ' f d x ' =− V f ' ⋅ g J d x ' - Expand the right side using a vector identity V g J ⋅∇ ' f d x ' =− V f J ⋅∇ ' g d x ' V f g ' J d x ' - Collect terms: V g J ⋅∇ ' f f J ⋅∇ ' g f g ' J d x ' = 0 - This holds for any functions f and g and vector J . We now apply it to our problem. Make J the current density vector. The definition of magnetostatics is that there is no divergence: ' J = 0 V g J ⋅∇ ' f f J ⋅∇ ' g d x ' = 0
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- Set f = 1 and g = x i V J i d x ' = 0 - If the integral of every component is zero, than the integral of the whole vector is zero and we have proven that the first term in the expansion is zero. If only the second term in the expansion is kept and all higher order terms are dropped, the potential far from a localized current density becomes: A = 0 4 1 x 3 x x '  J x ' d x ' - Express this explicitly in terms of components: A = 0 4 1 x 3 j x j x j '  i x i J i d x ' - Break the terms in half and shuffle things around: A =− 1 2 0 4 1 x 3 d x ' i x i j x j [− x j ' J i − x j ' J i ] - If we set f = x i ' and g = x j ' in the identity on the bottom of the previous page, we get V − x j ' J i d x ' = V x i ' J j d x ' - Use this to evaluate the integral in the potential equation: A =− 1 2 0 4 1 x 3 d
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This note was uploaded on 02/13/2012 for the course PHYSICS 95.657 taught by Professor Staff during the Fall '11 term at UMass Lowell.

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EMLecture11 - Lecture 11 Notes, 95.657 Electromagnetic...

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