Homework 10-solutions

Homework 10-solutions - soto(rrs946 – Homework 10 –...

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Unformatted text preview: soto (rrs946) – Homework 10 – Hoffmann – (56605) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A long, straight wire carries a current and lies in the plane of a rectangular loops of wire, as shown in the figure. 6 . 6cm 26 cm 14cm (33A)sin bracketleftBig (159rad / s) t + δ bracketrightBig → 130 loops (turns) Determine the maximum emf , |E| , induced in the loop by the magnetic field created by the current in the straight wire. Correct answer: 30 . 5059 mV. Explanation: Let : N = 130 , ω = 159 rad / s , ℓ = 14 cm , a = 6 . 6 cm , b = 26 cm , and I = 33 A . a b ℓ r dr I = I sin( ωt + δ ) Magnetic field near a long wire is B = μ I 2 π r . Faraday’s Law is E =- d Φ B dt . The magnitude of the magnetic field is B = μ I 2 π r . Thus the flux linkage is Φ B = N Φ B 1 = μ N I ℓ 2 π integraldisplay a + b a dr r = μ N I ℓ 2 π ln bracketleftbigg a + b a bracketrightbigg sin( ω t + δ ) . Finally, the induced emf E is E =- d Φ B dt =- μ N I ℓ ω 2 π ln bracketleftbigg a + b a bracketrightbigg cos( ω t + δ ) , which is a maximum when the cosine function yields 1. E max = μ N 2 π I ℓ ω ln bracketleftbigg a + b a bracketrightbigg = (1 . 25664 × 10 − 6 N / A 2 )(130) 2(3 . 14159) × (33 A)(14 cm)(159 rad / s) × ln bracketleftbigg (6 . 6 cm) + (26 cm) (6 . 6 cm) bracketrightbigg × parenleftBig 10 − 2 m cm parenrightBig parenleftbigg 10 3 mV V parenrightbigg = 30 . 5059 mV . soto (rrs946) – Homework 10 – Hoffmann – (56605) 2 002 (part 2 of 2) 10.0 points If at t = 0, δ =0, and a positive I denotes an upward moving current in the figure, then which statement below is correct at t = 0. 1. The sense of current is not determined with the information given. 2. The current in the loop is counter- clockwise. correct 3. The current in the loop is clockwise. 4. There is zero current in the loop. Explanation: Note: d I dt ( t =0) = ω I . At t = 0, the magnetic field from the long straight wire is increasing into the loop from above which leads to a finite time rate of change of magnetic flux, although B ( t =0) = 0 . The induced current in the loop then at- tempts to produce magnetic field through the loop that penetrates the loop from below ( i.e. , from Lenz’s law it is attempting to resist the change of flux). By the right hand rule, this requires counter-clockwise induced current in the loop. 003 10.0 points A toroid having a rectangular cross section ( a = 1 . 58 cm by b = 2 . 71 cm) and inner radius 2 . 74 cm consists of N = 410 turns of wire that carries a current I = I sin ω t , with I = 35 A and a frequency f = 44 . 2 Hz. A loop that consists of N ℓ = 18 turns of wire links the toroid, as in the figure....
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This note was uploaded on 11/10/2011 for the course PHYSICS 303L taught by Professor Hoffman during the Fall '11 term at University of Texas.

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Homework 10-solutions - soto(rrs946 – Homework 10 –...

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