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Unformatted text preview: malsam (wgm329) – hw21 – turner – (56705) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points To monitor the breathing of a hospital pa- tient, a thin belt is girded around the patient’s chest. The belt is a 242 turns coil. When the patient inhales, the area encircled by the coil increases by 56 . 1 cm 2 . Find the average induced emf in the coil during this time. Assume the magnitude of the Earth’s magnetic field is 69 . 5 μ T and makes an angle of 20 . 8 ◦ with the plane of the coil. A patient takes 1 . 43 s to inhale. Correct answer: 23 . 4307 μ V. Explanation: Let : N = 242 turns , Δ A = 56 . 1 cm 2 , Δ t = 1 . 43 s , B = 69 . 5 μ T , and θ = 20 . 8 ◦ . The magnetic flux through one turn of the coil is Φ B = B A cos θ , where θ = 90 ◦ − 20 . 8 ◦ = 69 . 2 ◦ is the angle between the loop vector and magnetic field, so by Faraday’s Law E = − N d Φ B dt = − N d ( B A cos θ ) dt = − N Δ A Δ t B cos θ = − (242 turns) (0 . 00561 m 2 ) 1 . 43 s × (69 . 5 μ T) cos 69 . 2 ◦ = 23 . 4307 μ V . 002 10.0 points An electron under the influence of some central force moves at speed v i in a counter- clockwise circular orbit of radius R . A uni- form magnetic field B perpendicular to the plane of the orbit is turned on (see figure). B increasing v electron circular path of electron radius R Suppose that the magnitude of the field changes at a given rate d B dt . What is the magnitude of the electric field induced at the radius of the electron orbit? 1. E = R 2 d B dt 2. E = 2 π R d B dt 3. E = R d B dt 4. E = π R 2 d B dt 5. E = π R d B dt 6. E = π 2 R d B dt 7. E = π R 2 d B dt 8. E = π 2 R 2 d B dt 9. E = 2 π 2 R d B dt 10. E = R 2 d B dt correct Explanation: Basic concepts: Faraday’s law states E = − d Φ dt . Magnetic flux, Orbital magnetic moment of an electron, and Newton’s second law may be germane. malsam (wgm329) – hw21 – turner – (56705) 2 Solution: To determine the induced elec- tric field apply Faraday’s law contintegraldisplay vector E · dvectors = − d Φ B dt . The line integral around the circular electric path reduces to E integraldisplay ds = E 2 π R , and the magnetic flux through the surface enclosed by the electron’s path is Φ B = π R 2 B . Since we are interested in the magnitude of E , we will drop the “ − ” sign in Faraday’s law. Thus E 2 π R = d ( π R 2 B ) dt , E = R 2 d B dt . The induced electric field at the electron gen- erates an induced magnetic field which op- poses the increase of the magnetic flux. Thus the induced magnetic field points downward, correspondingly the induced electronic field pointing clockwise as viewed from above....
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This note was uploaded on 11/22/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
- Spring '08