Ch 33 Physics for Scientists and Engineers

Ch 33 Physics for Scientists and Engineers -...

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ELECTROMAGNETIC INDUCTION 33.1. Model: Assume the magnetic field is uniform. Visualize: Please refer to Figure Ex33.1. Since a motional emf was developed the field must be perpendicular to V . The positive charges experienced a magnetic force to the left. By the right-hand rule the field must be out of the page so that v x i is to the left. Solve: This is a straightforward use of Equation 33.3. We have = 0.10 T & 0.050 V vl B=-= (5.0 m / s)(O.lO m) Assess: This is reasonable. Laboratory fields are typically up to a few teslas in magnitude. 33.2. Visualize: Xi If To develop a motional emf the magnetic field needs to be perpendicular to both, so let’s say its direction is into the page. Solve: This is a straightforward use of Equation 33.3. We have = 2.0~10~ m/s E 1.0 v IB v=-= (1.0 m)(5.0 x 10” T) Assess: 33.3. visu*: x x x x x x x x x x-x This is an unreasonable speed for a car. It’s unlikely you’ll ever develop a volt. R xxxxx xxxxx xxxxxxxxxxx The wire is pulled with a constant force in a magnetic field. This results in a motional emf and produces a current in the circuit. From energy conservation, the mechanical power provided by the puller must appear as electrical power in the circuit. Solve: (a) Using Equation 33.6, 33-1
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33-2 Chapter 33 (b) Using Equation 33.6 again, p=- v212B2 R *B= /Y-\j(O.ZOR)(I.ON) -- (4.0 m/s)(O.lO m)* = 2.24 T Assess: This is reasonable field for the circumstances given. 33.4. Model: Visualize: Please refer to Figure Ex33.4. The directions of the fields are opposite, so some flux is positive and some negative. The total flux is the sum of the flux in the two regions. Solve: The field is constant in each region so we will use Equation 33.10. Take 2 to be into the page. Then, it is parallel to the field in the left region so the flux is positive, and it is opposite to the field in the right region so the flux is negative. The total flux is Assume the field changes abruptly at the boundary between the two sections. @ = A. B = ALBL COSO, + ARBR CosOR = (0.20 m)*(2.0 T)-(o.~o m)2(1.0 T) = 0.040 wb Assess: surface. The flux is positive because the areas are equal and the stronger field is parallel to the normal of the 33.5. Model: Consider the solenoid to be long so the field is constant inside and zero outside. Visualize: Please refer to Figure Ex33.5. The field of a solenoid is along the axis. The flux through the loop is only nonzero inside the solenoid. Since the loop completely surrounds the solenoid, the total flux through the loop will be the same in both the perpendicular and tilted cases. Solve: The field is constant inside the solenoid so we will use Equation 33.10. Take A to be in the same direction as the field. The magnetic flux is @ = AlmP . When the loop is tilted the component of through which the magnetic field lines cross is increased by the same factor. 33.6. Model: Visualize: Please refer to Figure Ex32.6. According to Lenz's law, the induced current creates an induced field that opposes the change in flux.
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This homework help was uploaded on 04/07/2008 for the course PHSX 211/212 taught by Professor Medvedev during the Spring '08 term at Kansas.

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Ch 33 Physics for Scientists and Engineers -...

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