Chapter 23: Magnetic Flux and Faraday’s Law of InductionAnswers to Even-Numbered Conceptual Questions2. The magnetic field will have little apparent effect, because the break in the ring will prevent a current fromflowing around its circumference. What the magnetic field will do, however, is produce a nonzero emfbetween the two sides of the break.4. The metal plate moving between the poles of a magnet experiences eddy currents that retard its motion. Thishelps to damp out oscillations in the balance, resulting in more accurate readings.6. Nothing. In this case, the break prevents a current from circulating around the ring. This, in turn, preventsthe ring from experiencing a magnetic force that would propel it into the air.8. As the penny begins to tip over, there is a large change in the magnetic flux through its surface, due to thegreat intensity of the MRI magnetic field. This change in magnetic flux generates an induced current in thepenny that opposes its motion. As a result, the penny falls over slowly, as if it were immersed in molasses.10. When the angular speed of the coil in an electric generator is increased, the rate at which the magnetic fluxchanges increases as well. As a result, the magnitude of the induced emf produced by the generatorincreases. Of course, the frequency of the induced emf increases as well.12. When the switch is opened in a circuit with an inductor, the inductor tries to maintain the original current.(In general, an inductor acts to resist any change in the current—whether the current is increasing ordecreasing.) Therefore, the continuing current may cause a spark to jump the gap.Solutions to Problems and Conceptual Exercises1. Picture the Problem: The image shows a ring of radius 3.1 cm oriented at an angle of θ = 16º from a B = 0.055 T magnetic field.Strategy: Solve equation 23-1 for the flux.Solution: Calculate the flux: Φ =BAcosθ= 0.055 T( )π 0.031 m( )2cos16°Φ =1.6×10−4 WbInsight: The maximum flux through this coil, 1.66×10− 4 Wb, occurs when the angle θ is zero.2. Picture the Problem: The image shows a box immersed in a vertical magnetic field.Strategy: Use equation 23-1 to calculate the flux through each side.Solution: 1. The sides of the box are parallel to the field, so the magnetic flux through the sides is zero . 2. Calculate the flux through the bottom: Φ =BAcosθ = 0.0250 T( )0.325 m( )0.120 m( )cos0°= 9.75×10−4 Wb .Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.23 – 1Chapter 23: Magnetic Flux and Faraday’s Law of Induction James S. Walker, Physics, 4th EditionInsight: The height of the box is not important in this problem.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.23 – 2Chapter 23: Magnetic Flux and Faraday’s Law of Induction James S. Walker, Physics, 4th Edition3. Picture the Problem: The image shows a rectangular loop oriented 42 degrees from a magnetic field.Strategy: Solve equation 23-1 for the magnetic field.Solution: Calculate the magnetic field: B =ΦAcosθ=4.8×10−5 Tm20.051 m( )0.068 m( )cos47°= 0.020 TInsight: The minimum magnetic field that would produce this flux would occur when the rectangle is parallel to the magnetic field.4. Picture the Problem: A house has a floor of dimensions 22 m by 18 m. The local magnetic field due to Earth has a horizontal component 2.6×10-5 T and a downward vertical component 4.2×10-5 T.Strategy: The horizontal component of the magnetic field is parallel to the floor, so it does not contribute to the flux. Use equation 23-1 to calculate the flux using the vertical component. Solution: Calculate the magnetic flux: Φ =BAcosθ =B⊥A = 4.2×10−5T( )22 m( )18 m( )= 1.7×10−2 WbInsight: The flux through the vertical walls of the house is determined by the horizontal component of the magnetic field instead of the vertical component.5. Picture the Problem: A solenoid of diameter 1.2 m produces a magnetic field of 1.7 T. Strategy: Multiply the magnetic field by the cross-sectional area of the solenoid to calculate the magnetic flux. Solution: Calculate the magnetic flux: Φ =BAcosθ = 1.7 T( )π1.2 m2⎛⎝⎜⎞⎠⎟2cos0°= 1.9 WbInsight: Note that the length of the solenoid does not affect the flux through the solenoid.6. Picture the Problem: A magnetic field of magnitude 5.9×10−5 T is directed 72 below the horizontal and passes through a horizontal region 130 cm by 82 cm.Strategy: Use equation 23-1 to calculate the flux, where the angle from the vertical is θ = 90 − 72.Solution: Calculate the flux: Φ =BAcosθ = 5.9×10−5T( )1.30 m( )0.82 m( )cos 90°−72°( )= 6.0×10−5 WbInsight: Increasing the angle from the horizontal increases the flux through the desk top. For example, if the angle were increased to 80 from the horizontal the total flux would increase to 6.2×10-5 Wb.7. Picture the Problem: When a current flows through a solenoid of diameter 15.0 cm and 375 turns per meter, a flux of 1.28×10 − 4 Wb is generated in the core of the solenoid.Strategy: Divide the flux by the cross-sectional area of the solenoid to calculate the magnetic field. Then use equation 22-12 to calculate the current in the solenoid.Solution: 1. (a) Calculate the magnetic field inside the solenoid: B =ΦA=1.28×10−4T ⋅m2π 0.0850 m( )2=5.64×10−3 T2. Calculate the current: B =μ0nII =Bμ0n=5.64×10−3 T4π ×10−7T ⋅m/A( )385 m−1( )=11.7 A3. (b) Because the current is inversely proportional to the square of the diameter of the solenoid, doubling the diameter would multiply the answer to part (a) by a factor of one-fourth. Therefore, the current would be cut to a fourth.Insight: If the diameter were 34.0 cm, the current would become 2.91 A, which is one-fourth that found in part (a).Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from
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