MATH 2B Lecture Notes - Lecture 10: Thermistor, Short Circuit, Chemical Energy
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MATH 2B Full Course Notes
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Evaluate: compared to typical charges of objects in electrostatics, this is a huge amount of charge. 2 (5. 8 10 )(1. 60 10 c)( (1. 3 10 m) ) A dv is smaller than in example 25. 1, because i is smaller in this problem. The charge of an electron has magnitude n q v d. Q it (5. 00 a)(1. 00 s) 5. 00 c. I a would decrease and dv would decrease. 25. 5. passing through the light bulb in 1. 00 s would not change. (a) identify: by definition, j = i/a and radius is one-half the diameter. Set up: solve for the current: i = ja = j (d/2)2. Execute: i = (1. 50 106 a/m2)( )[(0. 00102 m)/2]2 = 1. 23 a. Evaluate: this is a realistic current. (b) identify: the current density is j = nqvd. Set up: solve for the drift velocity: vd = j/nq. Execute: since most laboratory wire is copper, we use the value of n for copper, giving v .
