ch26-p050 - 10 5 J. 50. Assuming the current is along the...

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where k = 2.75 × 10 10 A/m 4 and R = 0.00300 m. The rate of thermal energy generation is found from Eq. 26-26: P = iV = 210 W. Assuming a steady rate, the thermal energy generated in 40 s is QP t =∆= (210 J/s)(3600 s) = 7.56
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Unformatted text preview: 10 5 J. 50. Assuming the current is along the wire (not radial) we find the current from Eq. 26-4: i = | J | dA = 2 2 R kr rdr = 1 2 k R 4 = 3.50 A...
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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