W11HW7 - 1 Problem 12.7 1.1 Part a We know that the ~ E eld...

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Unformatted text preview: 1 Problem 12.7 1.1 Part a We know that the ~ E eld from the particle is: ~ E = q ~ r ~ r j ~ r ~ r j 3 If we let ~ r = ~ r ~ r , this becomes: ~ E = q ~ r r 3 = q x ^ x + y ^ y + z ^ z ( x 2 + y 2 + z 2 ) 3 = 2 Plugging this and ~ B = B ^ z into Jackson's equation 12.106: P eld = 1 4 c Z ~ E ¢ ~ B d 3 x = 1 4 c Z a x x = x Z 1 y = 1 Z 1 z = 1 q ( x 2 + y 2 + z 2 ) 3 = 2 ^ x ^ y ^ z x y z B d 3 x = qB 4 c Z a x x =0 x Z 1 y = 1 Z 1 z = 1 y ^ x x ^ y ( x 2 + y 2 + z 2 ) 3 = 2 d 3 x By antisymmetry in y , we see that the x-component must be zero. Expresssing the y- component in polar coordinats in the y- z plane{ that is, = p y 2 + z 2 : P eld = qB 4 c ^ y Z a x x = x Z 1 =0 Z 2 =0 x dx d d ( x 2 + 2 ) 3 = 2 Evaluating this integral in Maple for the three regions yields: P eld = qB c ^ y 8 > < > : a 2 x < a 2 x < x < a a 2 x > a The conjugate momentum of the particle comes from Jackson's equation 12.14: ~ P particle = ~m~v + q c ~ A Because r ¢ ~ A = ~ B and ~ B = B ^ z , ~ A = xB ^ y . At x = x , ~ P particle becomes: ~ P particle = ~m~v qx B c ^ y (1) 1 Summing ~ P eld and ~ P particle gives ~ G : ~ G = ~m~v qx B c ^ y + qB c ^ y 8 > < > : a 2 x < a 2 x < x < a a 2 x > a Note that our answer is not gauge invariant. The ~ A we chose is not the only choice in gauge which yields ~ B = B ^ z . Because our answer depends on this choice in gauge, it is not gauge invariant....
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W11HW7 - 1 Problem 12.7 1.1 Part a We know that the ~ E eld...

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