Thin and Thick Wires - Revision history

Rkeefe3: /* References */

2015-12-06T11:22:45Z

References

← Older revision		Revision as of 07:22, 6 December 2015
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	==References==		==References==
	~~Studd~~
			Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4th Edition

			http://farside.ph.utexas.edu/teaching/302l/lectures/node26.html

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← Older revision		Revision as of 07:14, 6 December 2015
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	'''Solution:''' Using the equation <math>i=nAuE</math>, one can solve this problem. Rearrange the equation and plug in the numbers.		'''Solution:''' Using the equation <math>i=nAuE</math>, one can solve this problem. Rearrange the equation and plug in the numbers.

	<math>E_(D)=\frac{i}{nAU}=\frac{510^{18}}{410^{28}\pi(0.210^{-3})^{2}0.00006}=1.66 V/m		<math>E_(D)=\frac{i}{nAU}=\frac{510^{18}}{410^{28}\pi(0.210^{-3})^{2}0.00006}=1.66 V/m</math>

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Example 2

← Older revision		Revision as of 06:58, 6 December 2015
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	'''Solution:''' Using the equation <math>i=nAuE</math>, one can solve this problem. Rearrange the equation and plug in the numbers.		'''Solution:''' Using the equation <math>i=nAuE</math>, one can solve this problem. Rearrange the equation and plug in the numbers.

	<math>E_(D)=\frac{i}{nAU}=\frac{510^{18}}{410^{28}\pi(0.210^{-3}){2}0.00006}=1.66 V/m		<math>E_(D)=\frac{i}{nAU}=\frac{510^{18}}{410^{28}\pi(0.210^{-3})^{2}0.00006}=1.66 V/m

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Example 2

← Older revision		Revision as of 06:58, 6 December 2015
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	'''Solution:''' Using the equation <math>i=nAuE</math>, one can solve this problem. Rearrange the equation and plug in the numbers.		'''Solution:''' Using the equation <math>i=nAuE</math>, one can solve this problem. Rearrange the equation and plug in the numbers.

	<math>E_(D)=\frac{i}{nAU}=\frac{510^{18}}{410^(28)\pi(0.210^(-3))^20.00006}=1.66 V/m		<math>E_(D)=\frac{i}{nAU}=\frac{510^{18}}{410^{28}\pi(0.210^{-3}){2}0.00006}=1.66 V/m

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Example 2

← Older revision		Revision as of 06:57, 6 December 2015
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	'''Solution:''' Using the equation <math>i=nAuE</math>, one can solve this problem. Rearrange the equation and plug in the numbers.		'''Solution:''' Using the equation <math>i=nAuE</math>, one can solve this problem. Rearrange the equation and plug in the numbers.

	<math>E_(D)=\frac{i}{nAU}=\frac{510^(18)}{410^(28)\pi(0.210^(-3))^20.00006}=1.66 V/m		<math>E_(D)=\frac{i}{nAU}=\frac{510^{18}}{410^(28)\pi(0.210^(-3))^20.00006}=1.66 V/m

2015-12-06T10:56:54Z

Example 2

← Older revision		Revision as of 06:56, 6 December 2015
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	'''Solution:''' Using the equation <math>i=nAuE</math>, one can solve this problem. Rearrange the equation and plug in the numbers.		'''Solution:''' Using the equation <math>i=nAuE</math>, one can solve this problem. Rearrange the equation and plug in the numbers.

	<math>E_(D)=\frac{i}{nAU}=\frac{510^18}{410^28\pi(0.210^-3)^20.00006}=1.66 V/m		<math>E_(D)=\frac{i}{nAU}=\frac{510^(18)}{410^(28)\pi(0.210^(-3))^20.00006}=1.66 V/m

2015-12-06T10:56:14Z

Example 2

← Older revision		Revision as of 06:56, 6 December 2015
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	===Example 2===		===Example 2===
	Using the picture from Example 1, let's say that the electron current is 5E18 <math>s^(-1)</~~math>~~		Using the picture from Example 1, let's say that the electron current is 5E18 1/s, there are 4E28 mobile electrons per cubic meter, the electron mobility is 0.00006 (m/s)(V/m), and that the radius of the thing wire is 0.2mm. What is the electric field strength around point D?

			'''Solution:''' Using the equation <math>i=nAuE</math>, one can solve this problem. Rearrange the equation and plug in the numbers.

			<math>E_(D)=\frac{i}{nAU}=\frac{510^18}{410^28\pi(0.210^-3)^20.00006}=1.66 V/m