{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solutions 14

solutions 14 - Patel Kinal – Homework 14 – Due 7:00 pm...

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: Patel, Kinal – 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 7 . 6 cm in radius and 69 m in length has 6100 uniformly spaced turns and carries a current of 1 . 1 A. Consider a plane circular surface of radius 2 . 4 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: 2 . 21134 × 10- 7 Wb. Explanation: Let : ‘ = 69 m , N = 6100 , I = 1 . 1 A , and r = 2 . 4 cm = 0 . 024 m . Basic Concepts: B = μ N I l I = μ nI Φ B = Z ~ B · d ~ A The magnetic field in the solenoid is given by B = μ I N ‘ Call A the area of the plane circular surface. The the magnetic flux through this surface is Φ = B A = μ I N ‘ π r 2 = μ (1 . 1 A)(6100) 69 m π (0 . 024 m) 2 = 2 . 21134 × 10- 7 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. 7 . 41cm 5 . 12 cm 30 . 3cm . 0383A Find the total magnetic flux through the loop. Correct answer: 1 . 2192 × 10- 9 Wb. Explanation: Let : c = 7 . 41 cm , a = 5 . 12 cm , b = 30 . 3 cm , and I = 0 . 0383 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 I 2 π r , so the field varies over the loop and is directed perpendicular to the page. Since ~ B is parallel Patel, Kinal – 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 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 = bdr . Since r is the only variable that now appears in the integral, we obtain for the magnetic flux Φ B = μ I 2 π b Z a + c c dr r = μ I b 2 π ln r fl fl fl a + c c = μ I b 2 π ln µ a + c c ¶ = μ (0 . 0383 A)(0 . 303 m) 2 π ln µ a + c c ¶ = μ (0 . 0383 A)(0 . 303 m) 2 π (0 . 525295) = 1 . 2192 × 10- 9 Wb . keywords: 003 (part 1 of 2) 10 points A 0 . 1553 A current is charging a capacitor that has circular plates, 11 . 19 cm in radius. The plate separation is 4 . 53 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....
View Full Document

{[ snackBarMessage ]}

Page1 / 10

solutions 14 - Patel Kinal – Homework 14 – Due 7:00 pm...

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

View Full Document
Ask a homework question - tutors are online