Chapter 23 - Chapter 23 4. Picture the Problem: A house has...

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Chapter 23 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: Insight: The flux through the vertical walls of the house is determined by the horizontal component of the magnetic field instead of the vertical component. 11. Picture the Problem : The figure shows the flux through a single loop coil as a function of time. Strategy: Use equation 23-3 to calculate the emf at the times t = 0.05 s, 0.15 s, and 0.50 s. Use the graph to find the change in flux. Solution: 1. (a) Calculate the emf at t = 0.05 s: 2. (b Calculate the emf at t = 0.15 s: 3. (c) Calculate the emf at t = 0.50 s: Insight: When the slope of the flux is constant, the emf is constant. The emf is 0.04 kV from t = 0.2 s to t = 0.6 s. 13. Picture the Problem : The magnetic flux through a coil oscillates in time as indicated by the graph at the right. Strategy: Use equation 23-4 to estimate the magnitude of the emf from the graph. Solution: 1. (a) The magnitude of the induced emf is greater near t = 0.5 s, because is changing more rapidly there. (In fact, it is not changing at all at the instant t = 0.4 s.)
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This note was uploaded on 09/03/2008 for the course PHYS 104 taught by Professor Dr.g during the Spring '08 term at Gwinnett Technical College.

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Chapter 23 - Chapter 23 4. Picture the Problem: A house has...

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