ass14 - SOLUTIONS, ASSIGNMENT 14 25.1. I DENTIFY : / I Q t...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

Page1 / 2

ass14 - SOLUTIONS, ASSIGNMENT 14 25.1. I DENTIFY : / I Q t...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online