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Department
Physics
Course
PHY 303L
Professor
Storr
Semester
Summer

Description
rashid (sar3553) – HW9 electromagneticInduction – storr – (37537) 1 This print-out should have 11 questions. Multiple-choice questions may continue on I the next column or page – find all choices before answering. R 001 10.0 points A coil is wrapped with 125 turns of wire on the perimeter of a circular frame (of radius What is the magnitude of the average emf 25 cm). Each turn has the same area, equal generated in the loop? to that of the frame. A uniform magnetic field is directed perpendicular to the plane ofCorrect answer: 51941.7 ▯V. the coil. This field changes at a constant rate Explanation: from 28 mT to 49 mT in 22 ms. The initial angle between the normal to the What is the magnitude of the induced E in plane of the circular loop and the magnetic the coil at the instant the magnetic field has field is a magnitude of 43 mT? ◦ ◦ ◦ θi= 51 − 90 = −39 Correct answer: 23.4281 V. θ = −39 + 180 = 141 . ◦ f Explanation: dΦ B From Faraday’s Law E = −N for a Basic Concepts: dt solenoid the average emf is dΦ B E = −N dt Eavg= −N ∆Φ B = −N ∆B · A Z ∆t ∆t B π r2 ΦB ≡ B · dA = B · A = −N (cosθf− cosθ i ∆t (74 ▯T)π (0.66 m)2 Solution: = −(29.7 turn) 0.09 s dΦ B × (cos141 − cos−39 ) E = −N dt = 51941.7 ▯V = −N A ∆B ∆t |Eavg| = 51941.7 ▯V . (B − B ) = −N π r 2 2 1 ∆t 003 10.0 points = −(125)π (25 cm)2 A magnetic field of 0.0913 T exists in the re- (49 mT) − (28 mT) gion enclosed by a solenoid that has 510 turns × 22 ms and a diameter of 9.41 cm. Within what period of time must the field = −23.4281 V |E| = 23.4281 V. be reduced to zero if the average magnitude of the induced emf within the coil during this time interval is to be 15.1 kV? 002 10.0 points Correct answer: 2.14453 × 10 s. A 29.7 turn circular loop of wire has a diam- eter of 1.32 m. The plane of the circular loopExplanation: makes an angle of 51 to the Earth’s magnetic Basic Concept: field. In 0.09 s the coil is flipped at a locationraday’s Law: where the magnitude of the Earth’s magnetic dΦ B field is 74 ▯T. E = −N dt rashid (sar3553) – HW9 electromagneticInduction – storr – (37537) 2 From Faraday’s Law, we get a = 7.6 cm, b = 19 cm, and |E| = −N dΦ B I = 54 A. dt 0 dB = N A dr dt ∆B = N A ) ∆t δ B + = N A t ω t r ℓ So, the time needed equals sin( I N AB t = I E (510 turns)(0.00695455 m )(0.0913 T) a b = (15.1 kV) = 2.14453 × 10−5 s. Magnetic field near a long wire is ▯0I 004 (part 1 of 2) 10.0 points B = 2π r . A long, straight wire carries a current and liFaraday’s Law is in the plane of a rectangular loops of wire, as shown in the figure. dΦ E = − B . dt → i δ 120 loops (turns) The magnitude of the magnetic field is + t ▯0I s) B = 2π r . / Thus the flux linkage is 29 cm h (205 rad ΦB = N Φ B1 ▯ N I ℓ Z a+b dr = 0 19 cm 2π a r . cm ▯ N I ℓ a + b (54 A)sin = 0 0 ln sin(ω t + δ). 2π a Determine the maximum emf, |E|, induced Finally, the induced emf E is in the loop by the magnetic field created by the current in the straight wire. dΦ B E = − dt Correct answer: 96.5219 mV. ▯0N I 0ω a + b = − 2π ln a cos(ω t + δ), Explanation: which is a maximum when the cosine function Let : N = 120, yields 1. ω = 205 rad/s, ▯0N a + b E max= I0ℓω ln ℓ = 29 cm, 2π a rashid (sar3553) – HW9 electromagneticInduction – storr – (37537) 3 (1.25664 × 106 N/A )(120) = 2(3.14159) × (54 A)(29 cm)(205 rad/s) R l Fapp × ln (7.6 cm) + (19 cm) (7.6 cm) m mV × 10 −2 103 cm V At what speed should the bar be moved to produce a current of 0.44 A in the resistor? = 96.5219 mV . Correct answer: 0.40357 m/s. Explanation: Let : R = 5.96 Ω, 005 (part 2 of 2) 10.0 points If at t = 0, δ=0, and a positive I denotes ℓ = 1.9 m, B = 3.42 T, and an upward moving current in the figure, then which statement below is correct at t = 0. I = 0.44 A. The emf inside the loop can be calculated 1. There is zero current in the loop. using Ohm’s Law E = I R 2. The current in th
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