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Unformatted text preview: CHAPTER 8 8.1a. Find H in cartesian components at P (2 , 3 , 4) if there is a current lament 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 R 2 = Z Idz a z [2 a x + 3 a y + (4 z ) a z ] 4 ( z 2 8 z + 29) 3 / 2 = Z Idz [2 a y 3 a x ] 4 ( z 2 8 z + 29) 3 / 2 Using integral tables, this evaluates as H a = I 4 2(2 z 8)(2 a y 3 a x ) 52( z 2 8 z + 29) 1 / 2 = I 26 (2 a y 3 a x ) Then with I = 8 mA, we nally obtain H a = 294 a x + 196 a y A / m b. Repeat if the lament 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 ( z 2 8 z + 26) 3 / 2 = Z Idz [3 a y a x ] 4 ( z 2 8 z + 26) 3 / 2 Evaluating as before, we obtain with I = 8 mA: H b = I 4 2(2 z 8)(3 a y a x ) 40( z 2 8 z + 26) 1 / 2 = I 20 (3 a y a x ) = 127 a x + 382 a y A / m c. Find H if both laments 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 A / m This problem can also be done (somewhat more simply) by using the known result for H from an innitelylong wire in cylindrical components, and transforming to cartesian components. The BiotSavart method was used here for the sake of illustration. 8.2. A lamentary conductor is formed into an equilateral triangle with sides of length ` carrying current I . Find the magnetic eld intensity at the center of the triangle. I will work this one from scratch, using the BiotSavart law. Consider one side of the triangle, oriented along the z axis, with its end points at z = `/ 2. Then consider a point, x , on the x axis, which would correspond to the center of the triangle, and at which we want to nd H associated with the wire segment. We thus have Id L = Idz a z , R = p z 2 + x 2 , and a R = [ x a x z a z ] /R . The dierential magnetic eld at x is now d H = Id L a R 4 R 2 = Idz a z ( x a x z a z ) 4 ( x 2 + z 2 ) 3 / 2 = I dz x a y 4 ( x 2 + z 2 ) 3 / 2 where a y would be normal to the plane of the triangle. The magnetic eld at x is then H = Z `/ 2 `/ 2 I dz x a y 4 ( x 2 + z 2 ) 3 / 2 = I z a y 4 x p x 2 + z 2 `/ 2 `/ 2 = I` a y 2 x p ` 2 + 4 x 2 1 8.2. (continued). Now, x lies at the center of the equilateral triangle, and from the geometry of the triangle, we Fnd that x = ( `/ 2)tan(30 ) = `/ (2 3). Substituting this result into the justfound expression for H leads to H = 3 I/ (2 ` ) a y . The contributions from the other two sides of the triangle eectively multiply the above result by three. The Fnal answer is therefore H net = 9 I/ (2 ` ) a y A / m . It is also possible to work this problem (somewhat more easily) by using Eq. (9), applied to the triangle geometry. using Eq....
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 Spring '08
 CITRIN
 Electromagnet

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