spring_04_exam2_sol_phy2054

spring_04_exam2_sol_phy2054 - Solution to Exam 2 Paul...

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Unformatted text preview: Solution to Exam 2 Paul Avery, Charles Thorn PHY2049, Spring 2004 March 5, 2004 1. A tiny current of 10- 9 A exists in a copper wire whose diameter is 2 mm. Assuming the current is uniform over the wire cross section, calculate the electron drift speed in m/s. Take the free electron density to be 7 10 28 electrons/m 3 . Solution: The current can be written as i = nev d A , where n is the free electron density, e is the electron charge and v d is the drift velocity. Substituting yields v d = 2 . 8 10- 14 m/s. 2. In the circuit shown E = 12 V , R 1 = 16 , and R is variable. If the value of R is chosen so that the ideal battery will transfer energy to the resistors at the maximum possible rate, what is that rate (in watts)? R R 1 R 1 E Solution: The total resistance is R 1 + RR 1 / ( R + R 1 ). Clearly the value of R must be zero to minimize the resistance and therefore maximize the power. The total resistance is 16 so the total power delivered is P = E 2 / 16 = 9 watts. 3. An electron is accelerated from rest by a potential difference of 2.0 kV. It then enters a uniform magnetic field of magnitude 1.0 T with its velocity perpendicular to the direction of the field. Calculate the radius in meters of its path in the magnetic field....
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spring_04_exam2_sol_phy2054 - Solution to Exam 2 Paul...

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