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# ass14 - SOLUTIONS ASSIGNMENT 14 25.1 IDENTIFY I = Q t SET...

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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 π( D /2) 2 E XECUTE : I = (1.50 × 10 6 A/m 2 )( π )[(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. E XECUTE : (a) dQ I dt = ( 29 ( 29 ( 29 ( 29 8.0 s 8.0 s 2 2 2 3 0 0 55 A 0.65 A/s 55 A 0.217 A/s Q t dt t t = - = - ( 29 ( 29 ( 29 ( 29 3 2 55 A 8.0 s 0.217 A/s 8.0 s 330 C Q = - = (b) 330 C = 41 A 8.0 s Q I t = = E VALUATE : The current decreases from 55 A to 13.4 A during the interval. The decrease is not linear

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ass14 - SOLUTIONS ASSIGNMENT 14 25.1 IDENTIFY I = Q t SET...

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