Textbook 29

Textbook 29 - Assignment 7 Solutions Chpt 29 15. Each of...

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Assignment 7 Solutions Chpt 29 15. Each of the semi-infinite straight wires contributes R i π μ 4 / 0 emiinfinite B (Eq. 29-7) to the field at the center of the circle (both contributions pointing “out of the page”). The current in the arc contributes a term given by Eq. 29-9 pointing into the page, and this is able to produce zero total field at that location if arc s 2.00 B = , or 00 2.00 ii R R μφ ππ ⎛⎞ = ⎜⎟ 44 ⎝⎠ which yields φ = 2.00 rad. 36. Using Eq. 29-13, the force on, say, wire 1 (the wire at the upper left of the figure) is along the diagonal (pointing towards wire 3 which is at the lower right). Only the forces (or their components) along the diagonal direction contribute. With θ = 45°, we find the force per unit meter on wire 1 to be () 22 11 2 1 3 1 4 1 2 1 3 2 2 3 | | 2 cos 2 cos 45 2 4 T m A 15.0A 3 1.12 N/m. 8.50 10 m FFFF F F aa a μμ =+ += + = ° + = ×⋅ == × × rrr −7 −3 π π π1 0 10 2 0 i The direction of is along 1 F r ˆˆ ˆ (i j)/ 2 r =− . In unit-vector notation, we have 1 (1.12 N/m) ˆ ˆˆˆ (i j) (7.94 N/m)i ( 7.94 N/m)j 2 F × = × + × r -3 -4 10 47. For , ra 2 0en c 0 0 0 0 0 2 3 rr iJ r B r J r rdr J a a = = ∫∫ pp p .
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Textbook 29 - Assignment 7 Solutions Chpt 29 15. Each of...

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