ε===ddtAdBdtLdBdtBΦloop24πwhere the rate of change of the field is dB/dt= 0.0100 T/s. Consequently, we obtain ()222 22233232286(/ 4 ) (/)(1.00 10 m) (0.500 m)0.0100 T/s/(/ 4)6464 (1.69 10m)3.68 10 W .BdtdLdBPRLddtεπρππρ−−−×⎛⎞==⎜⎟×Ω⋅⎝⎠=×29. Thermal energy is generated at the rate P= 2/R(see Eq. 27-23). Using Eq. 27-16, the resistance is given by R = L/A, where the resistivity is 1.69 ×10–8Ω·m (by Table 27-1) and A = πd2/4 is the cross-sectional area of the wire (
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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.