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Unformatted text preview: malsam (wgm329) – hw21 – turner – (56705) 1 This printout should have 16 questions. Multiplechoice 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
 Turner

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