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Unformatted text preview: SOLUTIONS, ASSIGNMENT 14 25.1. I DENTIFY : / I Q t = . S ET U P : 1.0 h 3600 s = E XECUTE : 4 (3.6 A)(3.0)(3600 s) 3.89 10 C. Q It = = = E VALUATE : Compared to typical charges of objects in electrostatics, this is a huge amount of charge. 25.4. (a) I DENTIFY : By definition, J = I/A and radius is one-half the diameter. S ET U P : Solve for the current: I = JA = J ( D /2) 2 E XECUTE : I = (1.50 10 6 A/m 2 )( )[(0.00102 m)/2] 2 = 1.23 A E VALUATE : This is a realistic current. (b) I DENTIFY : The current density is J = nqv d S ET U P : Solve for the drift velocity: v d = J/nq E XECUTE : Since most laboratory wire is copper, we use the value of n for copper, giving 6 2 d (1.50 10 A/m ) v = /[(8.5 10 28 el/m 3 )(1.60 19 10- C) = 1.1 4 10- m/s = 0.11 mm/s 25.7. I DENTIFY and S ET U P : Apply Eq. (25.1) to find the charge dQ in time dt . Integrate to find the total charge in the whole time interval....
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