# Ch29Word - Chapter 29 Page 1 CHAPTER 29 Electromagnetic...

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Chapter 29 Page 1 CHAPTER 29 – Electromagnetic Induction and Faraday’s Law 1. The average induced emf is å = – N ? Φ B /? t = – (2)[(+58 Wb) – (– 80 Wb)]/(0.72 s) = – 3.8 × 10 2 V . 2. Because the plane of the coil is perpendicular to the magnetic field, the initial flux through the loop is maximal. The magnitude of the average induced emf is å = – ? Φ B /? t = – A ? B /? t = – p(0.13 m) 2 (0 – 0.90 T)/(0.15 s) = 0.32 V . 3. As the coil is pushed into the field, the magnetic flux increases into the page. To oppose this increase, the flux produced by the induced current must be out of the page, so the induced current is counterclockwise . 4. As the magnet is pushed into the coil, the magnetic flux increases to the right. To oppose this increase, the flux produced by the induced current must be to the left, so the induced current in the resistor will be from right to left . 5. The magnitude of the average induced emf is å = – ? Φ B /? t = – A ? B /? t = – p(0.036 m) 2 (0 – 1.3 T)/(0.20 s) = 0.026 V . 6. We choose up as the positive direction. The average induced emf is å = – ? Φ B /? t = – A ? B /? t = – p(0.046 m) 2 [(– 0.25 T) – (+ 0.63 T)]/(0.15 s) = 3.9 × 10 –2 V = 39 mV . B N S R B

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Chapter 29 Page 2 7. ( a ) The increasing current in the wire will cause an increasing field into the page through the loop. To oppose this increase, the induced current in the loop will produce a flux out of the page, so the direction of the induced current will be counterclockwise . ( b ) The decreasing current in the wire will cause a decreasing field into the page through the loop. To oppose this decrease, the induced current in the loop will produce a flux into the page, so the direction of the induced current will be clockwise . ( c ) Because the current is constant, there will be no change in flux, so the induced current will be zero . ( d ) The increasing current in the wire will cause an increasing field into the page through the loop. To oppose this increase, the induced current in the loop will produce a flux out of the page, so the direction of the induced current will be counterclockwise . 8. ( a ) As the resistance is increased, the current in the outer loop will decrease. Thus the flux through the inner loop, which is out of the page, will decrease. To oppose this decrease, the induced current in the inner loop will produce a flux out of the page, so the direction of the induced current will be counterclockwise . ( b ) If the small loop is placed to the left, the flux through the small loop will be into the page and will decrease. To oppose this decrease, the induced current in the inner loop will produce a flux into the page, so the direction of the induced current will be clockwise . 9. As the solenoid is pulled away from the loop, the magnetic flux to the right through the loop decreases. To oppose this decrease, the flux produced by the induced current must be to the right, so the induced current is counterclockwise as viewed from the solenoid.
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## This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch29Word - Chapter 29 Page 1 CHAPTER 29 Electromagnetic...

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