Spring 11 Phys 121 - HW 10

Spring 11 Phys 121 - HW 10 - dzifanu (sd26397) hw 10...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: dzifanu (sd26397) hw 10 opyrchal (121102) 1 This print-out should have 11 questions. Multiple-choice 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 2 A, where the radius of the circular arc is 9 cm. The permeability of free space is 1 . 25664 10 6 T m / A . 2 A 9 c m O Determine the magnitude of the magnetic field at point O , the origin of the arc. Correct answer: 3 . 49066 T. Explanation: Let : I = 2 A and R = 9 cm = 0 . 09 m . For the straight sections d vectors r = 0. The quarter circle makes one-fourth 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) (2 A) 8 (0 . 09 m) = 3 . 49066 T , into the paper . 002 10.0 points A conductor consists of a circular loop of radius 0 . 0543 m and straight long sections, as shown. The wire lies in the plane of the paper and carries a current of 8 . 75 A. I R Find the magnitude of the magnetic field at the center of the loop. The permeability of free space is 1 . 25664 10 6 N / A 2 . Correct answer: 0 . 000133477 T. Explanation: Let : = 1 . 25664 10 6 N / A 2 , I = 8 . 75 A , and R = 0 . 0543 m . We can think of the total magnetic field as the superposition of the field due to the long straight wire having magnitude B straight = I 2 R and directed into the page and the field due to the circular loop having magnitude B loop = I 2 R , also directed into the page. Using appropriate right hand rules, both B fields are in the same direction, so the resultant magnetic field is B = B straight + B loop = parenleftbigg 1 + 1 parenrightbigg I 2 R = parenleftbigg 1 + 1 parenrightbigg (1 . 25664 10 6 N / A 2 ) (8 . 75 A) 2 (0 . 0543 m) = . 000133477 T . 003 (part 1 of 2) 10.0 points dzifanu (sd26397) hw 10 opyrchal (121102) 2 A conductor in the shape of a square, whose sides are of length 0 . 439 m, carries a clockwise 20 A current as shown in the figure below. . 439 m 20 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: 51 . 5431 T. Explanation: Let : = 1 . 25664 10 6 T m / A , = 0 . 439 m and I = 20 A . By the Biot-Savart law, dB = 4 I dvectors r r 2 . Consider a thin, straight wire carring a con- stant current I along the x-axis 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...
View Full Document

This note was uploaded on 09/16/2011 for the course PHYS PHYS 121 taught by Professor Prodan during the Spring '09 term at NJIT.

Page1 / 7

Spring 11 Phys 121 - HW 10 - dzifanu (sd26397) hw 10...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online