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
Unformatted text preview: aljabr (faa335) Hw13 Ross (89251) 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points The segment of wire in the figure carries a current of 6 A, where the radius of the circular arc is 5 cm. The permeability of free space is 1 . 25664 10 6 T m / A . 6 A 5 c m O Determine the magnitude of the magnetic field at point O , the origin of the arc. Correct answer: 18 . 8496 T. Explanation: Let : I = 6 A and R = 5 cm = 0 . 05 m . For the straight sections d vectors r = 0. The quarter circle makes onefourth the field of a full loop B = I 8 R into the paper. Or, you can use the equation B = o I 4 , where = 2 . Thus the magnetic field is B = I 8 R , = (1 . 25664 10 6 T m / A) (6 A) 8 (0 . 05 m) = 18 . 8496 T , into the paper . 002 (part 1 of 2) 10.0 points A conductor in the shape of a square, whose sides are of length 0 . 825 m, carries a counter clockwise 21 . 5 A current as shown in the figure below. . 825 m 21 . 5 A P What is the magnitude of the magnetic field at point P (at the center of the square loop) due to the current in the wire? Correct answer: 29 . 4842 T. Explanation: Let : = 1 . 25664 10 6 T m / A , = 0 . 825 m and I = 21 . 5 A . By the BiotSavart law, dB = 4 I dvectors r r 2 . Consider a thin, straight wire carring a con stant current I along the xaxis with the y axis pointing towards the center of the square, as in the following figure. y x O r P x I r ds a Let us calculate the total magnetic field at the point P located at a distance a from the wire. An element dvectors is at a distance r from P . The direction of the field at P due to this element is out of the paper, since dvectors r is out of the paper. In fact, all elements give a contribution directly out of the paper at P . Therefore, we have only to determine the aljabr (faa335) Hw13 Ross (89251) 2 magnitude of the field at P . In fact, taking the origin at O and letting P be along the positive y axis, with k being a unit vector pointing out of the paper, we see that dvectors r = k  dvectors r  = k ( dx sin ) ....
View Full
Document
 Spring '10
 Gavrine
 Current

Click to edit the document details