fall_06_exam2_sol_phy2049

# fall_06_exam2_sol_phy2049 - PHY2049 Fall 2006 Prof. Yasu...

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PHY2049 Fall 2006 1 Prof. Yasu Takano P r o f . P a u l A v e r y Oct. 18, 2006 Exam 2 Solutions 1. ( WebAssign 26.6 ) A current of 30 A flows through a copper wire whose radius is 1.25 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 . (1) 5.5 × 10 4 (2) 7.1 × 10 3 (3) 1.4 × 10 5 (4) 2.5 × 10 2 (5) 1.7 × 10 7 The total current I = JA is given by ed I JA en v A = = , where J is the current density, e is the electronic charge, n e is the free electron density, v d is the drift velocity and A = π R 2 is the area. With the numbers provided, we obtain 4 5.5 10 d v m/s. 2. ( Homework 29.37 ) Use the information in the previous problem. What is the magnetic field (in T) at a radius r = 1 mm? (1) 3.8 × 10 3 (2) 7.1 × 10 3 (3) 6.0 × 10 3 (4) 2.5 × 10 4 (5) 1.6 × 10 2 The field B inside the wire can be calculated from Ampere’s law ( ) 0enc 2 B ri πμ = , where i enc is the current enclosed at r = 1 mm. The enclosed current is proportional to the enclosed area, or () 2 enc 30 1/1.25 19.2 i == A. Solving for B yields 3 3.8 10 B T.

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PHY2049 Fall 2006 2 3. ( WebAssign 26.20 ) A battery is attached to a resistive rod as shown. The rod consists of two sections of the same material of equal lengths but different radii. (The relative radii may be dif- ferent than shown in the figure.) The resulting electric field in the top section, whose radius is 3 mm, is 2000 V/m. The electric field in the bottom section is 5000 V/m. What is the radius of the bottom section in millimeters?
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## This test prep was uploaded on 04/17/2008 for the course PHY 2054 taught by Professor Avery during the Spring '08 term at University of Florida.

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fall_06_exam2_sol_phy2049 - PHY2049 Fall 2006 Prof. Yasu...

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