PHY
ohw20-solutions

# ohw20-solutions - Reilly(mar3978 ohw20 turner(56725 This...

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Reilly (mar3978) – ohw20 – turner – (56725) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A toroid having a rectangular cross section ( a = 1 . 27 cm by b = 2 . 76 cm) and inner radius 2 . 38 cm consists of N = 280 turns of wire that carries a current I = I 0 sin ω t , with I 0 = 31 . 1 A and a frequency f = 82 . 7 Hz. A loop that consists of N = 22 turns of wire links the toroid, as in the figure. b a N R N l Determine the maximum E induced in the loop by the changing current I . Correct answer: 0 . 194681 V. Explanation: Basic Concept: Faraday’s Law E = - d Φ B dt . Magnetic field in a toroid B = μ 0 N I 2 π r . Solution: In a toroid, all the flux is confined to the inside of the toroid B = μ 0 N I 2 π r . So, the flux through the loop of wire is Φ B 1 = integraldisplay B dA = μ 0 N I 0 2 π sin( ω t ) integraldisplay b + R R a dr r = μ 0 N I 0 2 π a sin( ω t ) ln parenleftbigg b + R R parenrightbigg . Applying Faraday’s law, the induced emf can be calculated as follows E = - N d Φ B 1 dt = - N μ 0 N I 0 2 π ω a ln parenleftbigg b + R R parenrightbigg cos( ω t ) = -E 0 cos( ω t ) where ω = 2 πf was used. The maximum magnitude of the induced emf , E 0 , is the coefficient in front of cos( ω t ). E 0 = - N d Φ B 1 dt = - N μ 0 N I 0 ω 2 π a ln bracketleftbigg b + R R bracketrightbigg = - (22 turns) μ 0 (280 turns) × (31 . 1 A) (82 . 7 Hz) (1 . 27 cm) × ln bracketleftbigg (2 . 76 cm) + (2 . 38 cm) (2 . 38 cm) bracketrightbigg = - 0 . 194681 V |E| = 0 . 194681 V . 002 10.0 points The two-loop wire circuit is 111 . 591 cm wide and 74 . 3942 cm high. The wire circuit in the figure is located in a magnetic field whose magnitude varies with time according to the expression B = (0 . 001 T / s) t and its direction is out of the page. Assume The resistance per length of the wire is 0 . 105 Ω / m. B B P Q 37 . 1971 cm 74 . 3942 cm 74 . 3942 cm

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Reilly (mar3978) – ohw20 – turner – (56725) 2 When the magnetic field is 0 . 5 T, find the magnitude of the current through middle leg PQ of the circuit. Correct answer: 590 . 43 μ A. Explanation: Let : = 0 . 743942 m , A l = 1 2 2 = 0 . 55345 m 2 / s , A r = 2 = 1 . 1069 m 2 / s , δ = 0 . 105 Ω / m , and d B dt = d dt α t = α = 0 . 001 T / s . Basic Concept: Faraday’s Law is E = - d Φ B dt , where Φ B = A B . Ohm’s Law is V = I R . Solution: The instantaneous value of the magnetic field ( B = 0 . 5 T) is not germane to this problem. B B P Q 2 I r I l I PQ The resistance for the wire is proportional to the length of the wire. For a length of 74 . 3942 cm, the resistance is R = δ ℓ = (0 . 105 Ω / m) (0 . 743942 m) = 0 . 0781139 Ω . Using Ohm’s law for the right perimeter and left perimeter of the left loops in the circuit, we have E r = 4 ℓ δ I r = 4 R I r , and (1) E l = 3 ℓ δ I l = 3 R I l . (2) Note: When the magnetic field changes with time, there is an induced emf in both the right-hand side and the left-hand side. From Faraday’s law, the magnitude of the induced emf in the loops equals E r = A r d B dt = 2 d B dt , and (3) E l = A l d B dt = 1 2 2 d B dt . (4) Equating Eqs. 1 and 3, and Eqs. 2 and 4, we have E r = 4 ℓ δ I r = 2 d B dt , so I r = 1 4 δ α and (5) E l = 3 ℓ δ I l = 1 2 2 d B dt , so I l = 1 6 δ α .
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• Fall '10
• TURNER,J
• Magnetic Field, Reilly, bar magnet

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