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Unformatted text preview: Chapter 8 – OddNumbered 8.1a. Find H in cartesian components at P (2 , 3 , 4) if there is a current filament on the z axis carrying 8 mA in the a z direction: Applying the BiotSavart Law, we obtain H a = Z ∞ −∞ Id L × a R 4 πR 2 = Z ∞ −∞ Idz a z × [2 a x + 3 a y + (4 − z ) a z ] 4 π ( z 2 − 8 z + 29) 3 / 2 = Z ∞ −∞ Idz [2 a y − 3 a x ] 4 π ( z 2 − 8 z + 29) 3 / 2 Using integral tables, this evaluates as H a = I 4 π · 2(2 z − 8)(2 a y − 3 a x ) 52( z 2 − 8 z + 29) 1 / 2 ¸ ∞ −∞ = I 26 π (2 a y − 3 a x ) Then with I = 8 mA, we finally obtain H a = − 294 a x + 196 a y µ A / m b. Repeat if the filament is located at x = − 1, y = 2: In this case the BiotSavart integral becomes H b = Z ∞ −∞ Idz a z × [(2 + 1) a x + (3 − 2) a y + (4 − z ) a z ] 4 π ( z 2 − 8 z + 26) 3 / 2 = Z ∞ −∞ Idz [3 a y − a x ] 4 π ( z 2 − 8 z + 26) 3 / 2 Evaluating as before, we obtain with I = 8 mA: H b = I 4 π · 2(2 z − 8)(3 a y − a x ) 40( z 2 − 8 z + 26) 1 / 2 ¸ ∞ −∞ = I 20 π (3 a y − a x ) = − 127 a x + 382 a y µ A / m c. Find H if both filaments are present: This will be just the sum of the results of parts a and b , or H T = H a + H b = − 421 a x + 578 a y µ A / m This problem can also be done (somewhat more simply) by using the known result for H from an infinitelylong wire in cylindrical components, and transforming to cartesian components. The BiotSavart method was used here for the sake of illustration. 8.3. Two semiinfinite filaments on the z axis lie in the regions −∞ < z < − a (note typographical error in problem statement) and a < z < ∞ . Each carries a current I in the a z direction. a) Calculate H as a function of ρ and φ at z = 0: One way to do this is to use the field from an infinite line and subtract from it that portion of the field that would arise from the current segment at − a < z < a , found from the BiotSavart law. Thus, H = I 2 πρ a φ − Z a − a I dz a z × [ ρ a ρ − z a z ] 4 π [ ρ 2 + z 2 ] 3 / 2 The integral part simplifies and is evaluated: Z a − a I dz ρ a φ 4 π [ ρ 2 + z 2 ] 3 / 2 = Iρ 4 π a φ z ρ 2 p ρ 2 + z 2 ¯ ¯ ¯ a − a = Ia 2 πρ p ρ 2 + a 2 a φ Finally, H = I 2 πρ " 1 − a p ρ 2 + a 2 # a φ A / m 82 8.3. (continued) b) What value of a will cause the magnitude of H at ρ = 1, z = 0, to be onehalf the value obtained for an infinite filament? We require " 1 − a p ρ 2 + a 2 # ρ =1 = 1 2 ⇒ a √ 1 + a 2 = 1 2 ⇒ a = 1 / √ 3 8.5. The parallel filamentary conductors shown in Fig. 8.21 lie in free space. Plot  H  versus y , − 4 < y < 4, along the line x = 0, z = 2: We need an expression for H in cartesian coordinates. We can start with the known H in cylindrical for an infinite filament along the z axis: H = I/ (2 πρ ) a φ , which we transform to cartesian to obtain: H = − Iy 2 π ( x 2 + y 2 ) a x + Ix 2 π ( x 2 + y 2 ) a y If we now rotate the filament so that it lies along the...
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This note was uploaded on 10/02/2011 for the course EE 1 taught by Professor Joshi during the Fall '08 term at UCLA.
 Fall '08
 Joshi
 Electrical Engineering

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