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309-2008-Solutions1

309-2008-Solutions1 - ECE 309 Electromagnetic Fields...

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ECE 309 — Electromagnetic Fields University of Virginia Fall 2008 Homework # 1 Solutions 1. There are a number of ways to approach this problem. We are given that ~ A is a known vector and ~ X is an unknown vector. Also, the following products are known, p = ~ A · ~ X and ~ B = ~ A × ~ X To find ~ X , we need to find a basis — that is, a set of vectors that can be used to express ~ X . Since p is the projection of ~ A onto ~ X , ~ A is an easy choice for one of the basis vectors. Also note that the vector ~ B × ~ A is perpendicular to ~ A (and also to ~ B ). Thus we can use this as the second basis vector and write ~ X as, ~ X = a ~ A + b ( ~ B × ~ A ) where a and b are constants (scalars) that we need to find. Consider the dot product given above, p = ~ A · ~ X = a ~ A · ~ A + b ~ A · ( ~ B × ~ A ) The last term above is zero (because ~ A is perpendicular to the vector ~ B × ~ A ), so noting that ~ A · ~ A = A 2 , a = p A 2 Now let’s consider the cross product, ~ B = ~ A × ~ X = a ~ A × ~ A + b ~ A × ( ~ B × ~ A ) Noting that ~ A × ~ A = 0 and using the vector identity, ~ A × ( ~ B × ~ C ) = ~ B ( ~ A · ~ C ) - ~ C ( ~ A · ~ B ) we have ~ B = ~ A × ~ X = b ~ B ( ~ A · ~ A ) - ~ A ( ~ A · ~ B ) Again the last term is zero because ~ A and ~ B are perpendicular. Thus,
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b = 1 A 2 and we have, ~ X = p ~ A + ~ B × ~ A A 2 Note that a quicker way to the answer is to expand ~ A × ~ B , and use the identity we exploited previously (with ~ X replacing ~ C ): ~ A × ~ B = ~ A × ( ~ A × ~ X ) = ~ A ( ~ A · ~ X ) - ~ X ( ~ A · ~ A ) = p ~ A - A 2 ~ X Now, we can solve for ~ X algebraically, ~ X = p ~ A + ~ B × ~ A A 2 , as before! 2. To take the gradients, let’s write the vector ~ r in Cartesian coordinates, ~ r = ( x - x 0 x + ( y - y 0 y + ( z - z 0 z Thus, ~ ( r 2 ) = ~ ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 = ˆ x ∂x ( x - x 0 ) 2 y ∂y ( y - y 0 ) 2 z ∂z ( z - z 0 ) 2 = 2 ( x - x 0 x +( y - y 0 y +( z - z 0 z and we have, ~ ( r 2 ) = 2 ~ r
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Now consider, ~ 1 r = ~ h ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 - 1 2 i = - ( x - x 0 ) r 3 ˆ x - ( y - y 0 ) r 3 ˆ y - ( z - z 0 ) r 3 ˆ z, or ~ 1 r = - ~ r r 3 = - ˆ r r 2 where ˆ r ~ r/r , which is a unit vector in the ~ r -direction.
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