{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework_14

# Homework_14 - Nguyen Thanh Homework 14 Due 7:00 pm Inst...

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

Nguyen, Thanh – Homework 14 – Due: Mar 14 2008, 7:00 pm – Inst: Weathers 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A solenoid 6 . 1 cm in radius and 31 m in length has 12900 uniformly spaced turns and carries a current of 6 . 7 A. Consider a plane circular surface of radius 2 . 9 cm located at the center of the solenoid with its axis coincident with the axis of the solenoid. What is the magnetic flux Φ through this circular surface?(1 Wb = 1 T m 2 ) Correct answer: 9 . 25675 × 10 - 6 Wb. Explanation: Let : = 31 m , N = 12900 , I = 6 . 7 A , and r = 2 . 9 cm = 0 . 029 m . Basic Concepts: B = μ 0 N I l I = μ 0 n I Φ B = Z ~ B · d ~ A The magnetic field in the solenoid is given by B = μ 0 I N Call A the area of the plane circular surface. The the magnetic flux through this surface is Φ = B A = μ 0 I N π r 2 = μ 0 (6 . 7 A)(12900) 31 m π (0 . 029 m) 2 = 9 . 25675 × 10 - 6 Wb . The radius of the solenoid is not required since it is smaller that the plane circular surface. keywords: 002 (part 1 of 1) 10 points A rectangular loop located a distance from a long wire carrying a current is shown in the figure. The wire is parallel to the longest side of the loop. 4 . 8 cm 3 . 71 cm 20 . 3 cm 0 . 0289 A Find the total magnetic flux through the loop. Correct answer: 6 . 71885 × 10 - 10 Wb. Explanation: Let : c = 4 . 8 cm , a = 3 . 71 cm , b = 20 . 3 cm , and I = 0 . 0289 A . c a b r dr I From Amp` ere’s law, the strength of the magnetic field created by the current-carrying wire at a distance r from the wire is (see figure.) B = μ 0 I 2 π r , so the field varies over the loop and is directed perpendicular to the page. Since ~ B is parallel

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

View Full Document
Nguyen, Thanh – Homework 14 – Due: Mar 14 2008, 7:00 pm – Inst: Weathers 2 to d ~ A , the magnetic flux through an area element dA is Φ Z B dA = Z μ 0 I 2 π r dA . Note: ~ B is not uniform but rather depends on r , so it cannot be removed from the integral. In order to integrate, we express the area element shaded in the figure as dA = b dr . Since r is the only variable that now appears in the integral, we obtain for the magnetic flux Φ B = μ 0 I 2 π b Z a + c c d r r = μ 0 I b 2 π ln r fl fl fl a + c c = μ 0 I b 2 π ln a + c c = μ 0 (0 . 0289 A)(0 . 203 m) 2 π ln a + c c = μ 0 (0 . 0289 A)(0 . 203 m) 2 π (0 . 572626) = 6 . 71885 × 10 - 10 Wb . keywords: 003 (part 1 of 2) 10 points A 0 . 1304 A current is charging a capacitor that has circular plates, 8 . 84 cm in radius. The plate separation is 2 . 62 mm. The permitivity or free space is 8 . 85 × 10 - 12 and the permeability of free space is 1 . 25664 × 10 - 6 T · m / A. What is the time rate of increase of electric field between the plates? Correct answer: 6 . 00178 × 10 11 V / m s. Explanation: Let : I = 0 . 1304 A , r = 8 . 84 cm = 0 . 0884 m , A = π r 2 = 0 . 0245502 m 2 , and ² 0 = 8 . 85 × 10 - 12 . Let Φ E be the flux of the electric field, defined as Φ E = Z ~ E · d ~ A . Thus d Φ E dt = d dt ( E A ) = I ² 0 Time rate of increase of electric field between the plates is P = d E dt = I ² 0 A = 0 . 1304 A (8 . 85 × 10 - 12 )(0 . 0245502 m 2 ) = 6 . 00178 × 10 11 V / m s .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}