Useful Constants:k = 9.00 × 10ºNm²C28.85 x 10-12 C²Nm2%3De = 1.6 x 10-19Cme = 9.11 x 10mp = 1.67 × 10-27kg%3D-27mn= 1.68 x 10%3D The room temperature resistivity of gold is p = 2.44 x 108.m.A gold wire that is 0.75 mm in diameter and 21 cm longcarries a current of 800 mA. What is the magnitude of the electric field in the wire?%3D0.020 V/m0.044 V/mO 1.5 V/mO 0.076 V/m0.37 V/m

Answered: Image /qna-images/answer/a207093b-6e16-48c3-80b6-c8e474cfb7f0.jpg

The room temperature resistivity of gold isp= 2.44 x 10-8 N. m.Agold wire that is 0.75 mm in diameter and 21 cm long carries a current of 800 mA. What is the magnitude of the electric field in the wire? O 0.020 V/m O 0.044 V/m

The room temperature resistivity of gold isp= 2.44 x 10-8 N. m.Agold wire that is 0.75 mm in diameter and 21 cm long carries a current of 800 mA. What is the magnitude of the electric field in the wire? O 0.020 V/m O 0.044 V/m

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter30: Atomic Physics

Section: Chapter Questions

Problem 29CQ: Which of the following are not allowed? State which rule is violated for any that are not allowed....

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Concept explainers

Microscopic view of current

In physics, the term current is expressed as the rate of flow of electric charges across the area per unit time (I=q/t). It propagates along in the direction of the motion of positive charges. When the current is caused by the motion of electrons, the direction of the current is opposite to the flow.

Question

Useful Constants:
k = 9.00 × 10º
Nm²
C2
8.85 x 10-12 C²
Nm2
%3D
e = 1.6 x 10-19
C
me = 9.11 x 10
mp = 1.67 × 10-27kg
%3D
-27
mn
= 1.68 x 10
%3D

The room temperature resistivity of gold is p = 2.44 x 108.m.A gold wire that is 0.75 mm in diameter and 21 cm long
carries a current of 800 mA. What is the magnitude of the electric field in the wire?
%3D
0.020 V/m
0.044 V/m
O 1.5 V/m
O 0.076 V/m
0.37 V/m

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