866_PartUniversity Physics Solution

866_PartUniversity Physics Solution - 29-16 Chapter 29...

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29-16 Chapter 29 29.52. IDENTIFY: Apply the results of Example 29.4, generalized to N loops: max . N BA ω = E vr = . SET UP: In the expression for max , E must be in rad/s. 30 rpm 3.14 rad/s = EXECUTE: (a) Solving for A we obtain 2 0 5 9.0 V 18 m . (3.14 rad/s)(2000 turns)(8.0 10 T) A NB == = × E (b) Assuming a point on the coil at maximum distance from the axis of rotation we have 2 18 m (3.14 rad/s) 7.5 m s. A ω ππ = = EVALUATE: The device is not very feasible. The coil would need a rigid frame and the effects of air resistance would be appreciable. 29.53. IDENTIFY: Apply Faraday&s law in the form av B N t ΔΦ =− Δ E to calculate the average emf. Apply Lenz&s law to calculate the direction of the induced current. SET UP: B BA Φ= . The flux changes because the area of the loop changes. EXECUTE: (a) 22 av (0.0650/2 m) (0.950 T) 0.0126 V. 0.250 s B Ar BB tt t ΔΦ Δ ==== = ΔΔ Δ E (b) Since the magnetic field is directed into the page and the magnitude of the flux through the loop is decreasing, the induced current must produce a field that goes into the page. Therefore the current flows from point a through the resistor to point b . EVALUATE: Faraday&s law can be used to find the direction of the induced current. Let A ! be into the page. Then B Φ is positive and decreasing in magnitude, so / 0. B dd t Φ < Therefore 0 > E and the induced current is clockwise around the loop. 29.54. IDENTIFY: By Lenz&s law, the induced current flows to oppose the flux change that caused it. SET UP: When the switch is suddenly closed with an uncharged capacitor, the current in the outer circuit immediately increases from zero to its maximum value. As the capacitor gets charged, the current in the outer circuit gradually decreases to zero. EXECUTE: (a) (i) The current in the outer circuit is suddenly increasing and is in a counterclockwise direction. The magnetic field through the inner circuit is out of the paper and increasing. The magnetic flux through the inner circuit is increasing, so the induced current in the inner circuit is clockwise ( a to b ) to oppose the flux increase. (ii) The current in the outer circuit is still counterclockwise but is now decreasing, so the magnetic field through the inner circuit is out of the page but decreasing. The flux through the inner circuit is now decreasing, so the induced current is counterclockwise ( b to a ) to oppose the flux decrease. (b) The graph is sketched in Figure 29.54. EVALUATE: Even though the current in the outer circuit does not change direction, the current in the inner circuit does as the flux through it changes from increasing to decreasing. Figure 29.54 29.55. IDENTIFY: Use Faraday&s law to calculate the induced emf and Ohm&s law to find the induced current. Use Eq.(27.19) to calculate the magnetic force I F on the induced current. Use the net force I FF in Newton&s 2nd law to calculate the acceleration of the rod and use that to describe its motion.
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Electromagnetic Induction 29-17 (a) SET UP: The forces in the rod are shown in Figure 29.55a.
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866_PartUniversity Physics Solution - 29-16 Chapter 29...

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