47 from problem 46 we know that for a uniformly

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Unformatted text preview: eld due to the piece dy’ at the position -y. So the net field will be horizontal. The net field is: l/2 E = Ex = −l/2 1 E= xλ 2π 0 1 λdy cos θ = 2 + y2 4π 0 x l/2 2 2 −3/2 (x +y ) 0 l/2 −l/2 1 λdy 2 + y2 4π 0 x x2 x2 + y 2 l/2 y λx dy = 2π 0 x2 x + y2 = 0 l/2 =2 λx 2π 0 x2 0 1 λdy 2 + y2 4π 0 x l/2 x2 + l2 /4 = 2π 0 21.47 From problem 46, we know that for a uniformly charged straight wire with length l and point 0 at the midpoint, the field at a point P perpendicular distance x from 0 is: λ l 2 + 4x2 )1/2 2π 0 x(l 5 x2 + y 2 λl + 4x2 )1/2 x(l2 Appendix B of the textbook has a formula to evaluate the integral. The direction of the field is the +x direction. E= x For our problem, let’s consider one wire. Then the point P at a distance z above the center of the square loop is at th...
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