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Unformatted text preview: CHAPTER 22 ELECTROMAGNETIC INDUCTION ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. 3.5 m/s 2. (e) The work done by the hand equals the energy dissipated in the bulb. The energy dissipated in the bulb equals the power used by the bulb times the time. Since the time is the same in each case, more work is done when the power used is greater. The power, however, is the voltage squared divided by the resistance of the bulb, according to Equation 20.6c, so that a smaller resistance corresponds to a greater power. Thus, more work is done when the resistance of the bulb is smaller. 3. (c) The magnetic flux that passes through a surface is Ф cos BA φ Φ = (Equation 22.2), where B is the magnitude of the magnetic field, A is the area of the surface, and φ is the angle between the field and the normal to the surface. Knowing and Ф A , we can calculate cos / B A φ = Φ , which is the component of the field parallel to the normal or perpendicular to the surface. 4. (b) The magnetic flux that passes through a surface is Ф cos BA φ Φ = (Equation 22.2), where B is the magnitude of the magnetic field, A is the area of the surface, and φ is the angle between the field and the normal to the surface. It has the greatest value when the field strikes the surface perpendicularly ( 29 φ = ° and a value of zero when the field is parallel to a surface ( 29 90 φ = ° . The field is more nearly perpendicular to face 1 ( 29 20 φ = ° than to face 3 ( 29 70 φ = ° and is parallel to face 2. 5. (d) Faraday’s law of electromagnetic induction states that the average emf ξ induced in a coil of N loops is ( 29 / N t ξ =  ∆Φ ∆ (Equation 22.3), where is the change in magnetic ∆Ф flux through one loop and ∆ t is the time interval during which the change occurs. Reducing the time interval ∆ t during which the field magnitude increases means that the rate of change of the flux will increase, which will increase (not reduce) the induced emf. 6. 3.2 V 7. (c) According to Faraday’s law, the magnitude of the induced emf is the magnitude of the change in magnetic flux divided by the time interval over which the change occurs (see Equation 22.3). In each case the field is perpendicular to the coil, and the initial flux is zero since the coil is outside the field region. Therefore, the changes in flux are as follows: 2 2 A B C , 2 BL B L ∆Φ = ∆Φ = ∆Φ = (see Equation 22.2). The corresponding time intervals are 166 ELECTROMAGNETIC INDUCTION A B C , 2 t t t t t ∆ = ∆ = ∆ ∆ = ∆ . Dividing gives the following results for the magnitudes of the emfs: 2 2 2 2 A B C 2 2 , , 2 BL B L B L BL t t t t ξ ξ ξ = = = = ∆ ∆ ∆ ∆ . 8. (a) An induced current appears only when there is an induced emf to drive it around the loop....
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This note was uploaded on 03/30/2012 for the course PHYSICS 201 taught by Professor Rollino during the Fall '11 term at Rutgers.
 Fall '11
 rollino
 Physics, Energy, Power, Work

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