chapter08_7th_solution

# chapter08_7th_solution - CHAPTER 8 8.1a. Find H in...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 Biot-Savart 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 Biot-Savart 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 innitely-long wire in cylindrical components, and transforming to cartesian components. The Biot-Savart 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 Biot-Savart 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 just-found 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....
View Full Document

## chapter08_7th_solution - CHAPTER 8 8.1a. Find H in...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online