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exam3_sol

# exam3_sol - PHY2049 Fall 2009 Exam 3 Solution Profs A...

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PHY2049 Fall 2009 Profs. A. Petkova, A. Rinzler, S. Hershfield Exam 3 Solution 1. In the figure R represents a variable resistor and the battery supplies an emf that drives current clockwise around the loop. The following is done. The resistance is steadily increased and then steadily decreased. For each stage of this sequence (in order), what is the direction of the induced emf by the inductor? CW denotes clockwise, CCW denotes counterclockwise, and NC means no change. Answer: CW, CCW Solution: When the resistance is increased, the current in the clockwise direction is decreased. To oppose that the induced emf will be in the clockwise direction. When the resistance is decreased, the current in the clockwise direction will increase. To oppose that the induced emf will be in the counterclockwise direction. 2. A solenoid that is 85.0 cm long has a cross-sectional area of 17.0 cm 2 . There are 950 turns of wire carrying current of 10.0 A. What is the total energy stored in the magnetic field inside the solenoid (neglect end effects)? Answer: 0.113 J Solution: The magnetic field in a solenoid is B = μ o nI , where n is the number of terms per unit length. The energy density is u = B 2 / (2 μ o ), and the net energy is U = u Vol., where the volume inside the solenoid is its length times its cross-sectional area. 3. A square wire loop with 2.00 m sides is perpendicular to a uniform magnetic field, with half the area of the loop in the field as shown. The loop contains an ideal battery with emf E = 15 . 0V. If the magnitude of the field varies with time according to B = 0 . 0620 - 0 . 870 t , with B in Tesla and t in seconds, what is the net emf in the circuit, and its direction? Answer: 16.7 V, CCW Solution: The area in the magnetic field is 0 . 5(2 m ) 2 = 2 m 2 . The rate of change of the magnetic flux is (2 m 2 ) dB/dt = 2 × 0 . 87 = 1 . 74 V , which is also the induced emf. Because the flux is out of the page and decreasing in magnitude, via Lenz’s law the induced current is in the CCW direction. Thus, the induced emf and the emf of the battery are in the same direction. The net emf is 15 . 0 V + 1 . 7 V . 4. The magnetic field of a cylindrical magnet that has a pole-face diameter of 3.3 cm can be varied sinusoidally between 28.0 and 30.0 T at a frequency of 15 Hz. At a radial distance of 1.4 cm, what is the amplitude of the electric field induced by the variation? (Note that the magnetic field is directed along the axis of the cylinder.) Answer: 0.66 V/m Solution: The amplitude of the magnetic field has the form B ( t ) = 29 T + 1 T sin(2 πft ), so the rate of change of flux is d Φ B /dt = 2 πf (1 T ) A cos(2 πt ). For a path of integration of radius r = 1 . 4 cm, Faraday’s law implies that contintegraldisplay vector E · dvectors = 2 πrE = - d Φ B dt == - 2 πf (1 T ) πr 2 cos(2 πt ) .

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exam3_sol - PHY2049 Fall 2009 Exam 3 Solution Profs A...

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