Chapter 34

# Chapter 34 - 34.1 Visualize To develop a motional emf the...

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34.1. Visualize: 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 34.3. We have () 4 5 1.0 V 2.0 10 m/s 1.0 m 5.0 10 T v lB == × E Assess: This is an unreasonable speed for a car. It’s unlikely you’ll ever develop a volt.

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34.2. Model: Assume the magnetic field is uniform. Visualize: Please refer to Figure Ex34.2. Since a motional emf was developed the field must be perpendicular to v G . 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 vB × G G is to the left. Solve: This is a straightforward use of Equation 34.3. We have () 0.050 V 0.10 T 5.0 m/s 0.10 m B vl == = E Assess: This is reasonable. Laboratory fields are typically up to a few teslas in magnitude.
34.3. Visualize: 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 34.6, pull pull 4.0 W 4.0 m/s 1.0 N P PFv v F =⇒ == = (b) Using Equation 34.6 again, () ( ) 22 2 pull 2 2 0.20 1.0 N 2.2 T 4.0 m/s 0.10 m RF vlB PB Rv l Ω = = = Assess: This is reasonable field for the circumstances given.

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34.4. Model: Assume the field changes abruptly at the boundary between the two sections. Visualize: Please refer to Figure Ex34.4. 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 34.10. Take A G 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 () ( ) ( ) LL L RR R 22 cos cos 0.20 m 2.0 T 0.20 m 1.0 T 0.040 Wb AB AB AB θθ Φ= ⋅ = + =−= G G Assess: The flux is positive because the areas are equal and the stronger field is parallel to the normal of the surface.
34.5. Model: Consider the solenoid to be long so the field is constant inside and zero outside. Visualize: Please refer to Figure Ex34.5. 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 34.10. Take A G to be in the same direction as the field. The magnetic flux is () ( ) 2 25 loop loop sol sol sol sol cos 0.010 m 0.20 T 6.3 10 Wb AB A B r B πθ π Φ= = = = = × GG When the loop is tilted the component of B G in the direction of A G is less, but the effective area of the loop surface through which the magnetic field lines cross is increased by the same factor.

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34.6.
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## This note was uploaded on 01/30/2011 for the course PHYS 131 taught by Professor E.salik during the Winter '10 term at Cal Poly Pomona.

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Chapter 34 - 34.1 Visualize To develop a motional emf the...

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