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Phys121-11sol

# Phys121-11sol - Anderson Matthew Homework 11 Due 9:00 am...

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Anderson, Matthew – Homework 11 – Due: Apr 15 2008, 9:00 am – Inst: Vitaly 1 This print-out should have 11 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) 15 points A plane loop of wire of area A is placed in a region where the magnetic field is perpendicu- lar to the plane. The magnitude of B varies in time according to the expression B = B 0 e - at . That is, at t = 0 the field is B 0 , and for t > 0, the field decreases exponentially in time. Find the induced emf, E , in the loop as a function of time. 1. E = a B 0 t 2. E = a A B 0 e - 2 at 3. E = a A B 0 e - at correct 4. E = A B 0 e - at 5. E = a B 0 e - at 6. E = a A B 0 Explanation: Basic Concepts: Faraday’s Law: E ≡ I E · ds = - d Φ B dt Solution: Since B is perpendicular to the plane of the loop, the magnetic flux through the loop at time t > 0 is Φ B = B A = A B 0 e - a t Also, since the coefficient AB 0 and the pa- rameter a are constants, and Faraday’s Law says E = - d Φ B dt the induced emf can be calculated the from Equation above: E = - d Φ B dt = - A B 0 d dt e - a t = a A B 0 e - a t That is, the induced emf decays exponentially in time. Note: The maximum emf occurs at t = 0 , where E = a A B 0 . B = B 0 e - at B 0 0 0 ~ t The plot of E versus t is similar to the B versus t curve shown in the figure above. keywords: 002 (part 1 of 1) 15 points The plane of a rectangular coil, 7 . 2 cm by 11 . 4 cm, is perpendicular to the direction of a uniform magnetic field B . If the coil has 38 turns and a total resistance of 11 . 1 Ω, at what rate must the magnitude of B change to induce a current of 0 . 14 A in the windings of the coil? Correct answer: 4 . 9823 T / s. Explanation: Given : x = 7 . 2 cm = 0 . 072 m , y = 11 . 4 cm = 0 . 114 m , N = 38 turns , r = 11 . 1 Ω , and I = 0 . 14 A . The induced emf is E = I R = N d Φ dt = N d ( B A ) dt = N A d B dt , so d B dt = I R N ( x y ) = (0 . 14 A) (11 . 1 Ω) 38 (0 . 072 m) (0 . 114 m) = 4 . 9823 T / s .

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Anderson, Matthew – Homework 11 – Due: Apr 15 2008, 9:00 am – Inst: Vitaly 2 keywords: 003 (part 1 of 1) 15 points A conducting bar moves as shown near a long wire carrying a constant I = 30 A current. I a v L A B If a = 2 . 4 mm, L = 140 cm, and v = 26 m / s, what is the potential difference, Δ V V A - V B ? Correct answer: 91 mV. Explanation: Given; E = - d Φ B dt = - d dt ( B ‘ x ) = - B ‘ dx dt = - B ‘ v . From Ampere’s law, the strength of the magnetic field created by the long current- carrying wire at a distance a from the wire is B = μ 0 I 2 π a .
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