solution10_pdf

# solution10_pdf - hussain(wh4892 – hw 10 – Opyrchal...

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: hussain (wh4892) – hw 10 – Opyrchal – (121104) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A conductor consists of a circular loop of radius 0 . 157 m and straight long sections, as shown. The wire lies in the plane of the paper and carries a current of 6 . 08 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: 3 . 20776 × 10 − 5 T. Explanation: Let : μ = 1 . 25664 × 10 − 6 N / A 2 , I = 6 . 08 A , and R = 0 . 157 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 = μ I 2 π R and directed into the page and the field due to the circular loop having magnitude B loop = μ 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 π parenrightbigg μ I 2 R = parenleftbigg 1 + 1 π parenrightbigg (1 . 25664 × 10 − 6 N / A 2 ) (6 . 08 A) 2 (0 . 157 m) = 3 . 20776 × 10 − 5 T . 002 (part 1 of 2) 10.0 points The set up is shown in the figure, where 36 A is flowing in the wire segments, AB = CD = 49 m, and the wire segment arc has a radius 25 m subtending an angle of 90 ◦ . The permeability of free space is 1 . 25664 × 10 − 6 T · m / A . 49m 36A 3 6 A 49 m 36 A 2 5 m A B C D O Find the magnitude of the magnetic field at O due to the current segment ABCD . Correct answer: 2 . 26195 × 10 − 7 T. Explanation: Let : I = 36 A , μ = 1 . 25664 × 10 − 6 T · m / A , and a = 25 m . Along CD dvectors and ˆ r are antiparallel, so again dvectors × ˆ r = 0 . Therefore that segment of the current also creates no magnetic field at O . Along BC dvectors is perpendicular to ˆ r so | dvectors × ˆ r | = ds = a dθ . Also dvectors × ˆ r is in the same direction for all dvectors along BC , while r = a , so the magnitude of the magnetic field at O due to ABCD is B = μ 4 π I integraldisplay π/ 2 a dθ a 2 = μ 4 π a I θ vextendsingle vextendsingle vextendsingle π/ 2 = μ I 8 a = (1 . 25664 × 10 − 6 T · m / A) (36 A) 8 (25 m) = 2 . 26195 × 10 − 7 T . hussain (wh4892) – hw 10 – Opyrchal – (121104) 2 003 (part 2 of 2) 10.0 points Find the magnitude of the magnetic field at O due to the current segment AB alone. Correct answer: 0 T. Explanation: The Biot-Savart law is d vector B = μ 4 π I dvectors × ˆ r r 2 for the magnetic field due to a small current segment, I dvectors, at a point a distance r from the segment. ˆ r is the unit vector pointing from the segment to the point O. Along AB dvectors is parallel to ˆ r , so dvectors × ˆ r = 0 . Therefore d vector B =0 and vector B = integraldisplay d vector B = 0 T ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

solution10_pdf - hussain(wh4892 – hw 10 – Opyrchal...

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

View Full Document
Ask a homework question - tutors are online