Ampere-Maxwell Law - Revision history

Rohitnaras: /* Connectedness */

2026-04-27T03:36:53Z

Connectedness

← Older revision		Revision as of 23:36, 26 April 2026
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	==Connectedness==		==Connectedness==
			'''How is this topic connected to something that you are interested in?'''

	As a computer engineering student, the Ampere-Maxwell Law is foundational to almost every piece of hardware I work with. Every PCB trace carrying a high-frequency signal radiates because of changing electric fields between adjacent conductors, this is exactly the displacement current term in action. Signal integrity engineering, antenna design, and even the GHz-range coupling that causes crosstalk in dense chip layouts are all governed by this equation. Without Maxwell's correction, we would have no theoretical basis for wireless communication, RF circuits, or the high-speed serial links inside modern processors.		As a computer engineering student, the Ampere-Maxwell Law is foundational to almost every piece of hardware I work with. Every PCB trace carrying a high-frequency signal radiates because of changing electric fields between adjacent conductors, this is exactly the displacement current term in action. Signal integrity engineering, antenna design, and even the GHz-range coupling that causes crosstalk in dense chip layouts are all governed by this equation. Without Maxwell's correction, we would have no theoretical basis for wireless communication, RF circuits, or the high-speed serial links inside modern processors.

Rohitnaras: /* Simple */

2026-04-27T03:36:07Z

Simple

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	===Simple===		===Simple===

	'''Question:''' A 30 A current is charging a parallel-plate capacitor with circular plates of radius ''R'' = 2.5 mm. The plate separation is 4 mm. What is the magnitude of the magnetic field between the plates at a radial distance ''r'' = 1 mm from the central axis?		'''Question:''' A 30 A current is charging a parallel-plate capacitor with circular plates of radius R = 2 mm. The plate separation is 5 mm. What is the magnitude of the magnetic field between the plates at a radial distance r = 1 mm from the central axis?

	[[File:Example1.png]]		[[File:Example1.png]]
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	'''Solution:'''		'''Solution:'''

	Apply the Ampere-Maxwell Law to a circular Amperian loop of radius ''r'' centered on the axis between the plates. There is no conduction current between the plates, so:		Apply the Ampere-Maxwell Law to a circular Amperian loop of radius ''r'' = 1 mm centered on the axis between the plates. There is no conduction current piercing this loop (we are between the plates, not in the wire), so only the displacement current term contributes:

	:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>		:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}</math>

	The electric field between the plates of a capacitor is ~~<math>E = \frac{Q}{\epsilon_0 A_{~~plate~~}}</math>, where <math>A_{plate} = \pi R^2</math>. The flux through our Amperian loop (~~area ~~<math>\pi r^2</math>) is~~:		The electric field between the plates of a parallel-plate capacitor is uniform across the plate area:

	:<math>\Phi_E = E \cdot \pi r^2 = \frac{Q}{\epsilon_0 \pi R^2}\cdot \pi r^2 = \frac{Q r^2}{\epsilon_0 R^2}</math>		:<math>E = \frac{Q}{\epsilon_0 A_{plate}} = \frac{Q}{\epsilon_0 \pi R^2}</math>

			The electric flux through our small Amperian loop of area <math>\pi r^2</math> is:

			:<math>\Phi_E = E \cdot \pi r^2 = \frac{Q}{\epsilon_0 \pi R^2} \cdot \pi r^2 = \frac{Q r^2}{\epsilon_0 R^2}</math>

	Taking the time derivative and recognizing <math>dQ/dt = I</math>:		Taking the time derivative and recognizing <math>dQ/dt = I</math>:
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	:<math>\frac{d\Phi_E}{dt} = \frac{I r^2}{\epsilon_0 R^2}</math>		:<math>\frac{d\Phi_E}{dt} = \frac{I r^2}{\epsilon_0 R^2}</math>

	~~By symmetry, B is constant in magnitude around~~ the ~~loop, so~~ <math>\oint \vec{B}\cdot d\vec{l} = B(2\pi r)</math>. Solving:		Substituting back into the Ampere-Maxwell Law:

			:<math>\oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \cdot \frac{I r^2}{\epsilon_0 R^2} = \frac{\mu_0 I r^2}{R^2}</math>

			By the symmetry of the problem, '''B''' is constant in magnitude along the Amperian circle and is everywhere tangent to it, so the line integral simplifies to <math>B(2\pi r)</math>:

			:<math>B(2\pi r) = \frac{\mu_0 I r^2}{R^2}</math>

			Solving for ''B'':

	:<math>B = \frac{\mu_0 I r}{2\pi R^2}</math>		:<math>B = \frac{\mu_0 I r}{2\pi R^2}</math>

	Plugging in ''I'' = 30 A, ''r'' = 0.001 m, ''R'' = 0.~~0025~~ m:		Plugging in ''I'' = 30 A, ''r'' = 0.001 m, ''R'' = 0.002 m:

			:<math>B = \frac{(4\pi \times 10^{-7}\ \text{T·m/A})(30\ \text{A})(0.001\ \text{m})}{2\pi (0.002\ \text{m})^2}</math>

			:<math>B = \frac{(4\pi \times 10^{-7})(0.03)}{2\pi (4 \times 10^{-6})} = 1.5 \times 10^{-3}\ \text{T}</math>

			:<math>\boxed{B = 1.5\ \text{mT}}</math>

	~~:<math>B = \frac{(4\pi \times 10^{-7})(30)(0.001)}{2\pi (0.0025)^2} \approx 9.6 \times 10^{-4}\ \text{T} \approx 0.96\ \text{mT}</math>~~
	===Middling===		===Middling===

Rohitnaras: /* Computational Model */

2026-04-27T03:32:38Z

Computational Model

← Older revision		Revision as of 23:32, 26 April 2026
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	The full GlowScript code is also provided below for transparency and reproducibility.		The full GlowScript code is also provided below for transparency and reproducibility.

	~~<iframe src="~~https://trinket.io~~/embed~~/glowscript/faf9ad543e9a~~?start=result" width="100%" height="356" frameborder="0" marginwidth="0" marginheight="0" allowfullscreen></iframe>~~		Link: https://trinket.io/glowscript/faf9ad543e9a

	==Examples==		==Examples==

Rohitnaras: /* Computational Model */

2026-04-27T03:31:58Z

Computational Model

← Older revision		Revision as of 23:31, 26 April 2026
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	The full GlowScript code is also provided below for transparency and reproducibility.		The full GlowScript code is also provided below for transparency and reproducibility.

			<iframe src="https://trinket.io/embed/glowscript/faf9ad543e9a?start=result" width="100%" height="356" frameborder="0" marginwidth="0" marginheight="0" allowfullscreen></iframe>

	==Examples==		==Examples==

Rohitnaras at 03:29, 27 April 2026

2026-04-27T03:29:27Z

Show changes

Rohitnaras at 03:00, 27 April 2026

2026-04-27T03:00:42Z

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	~~Claimed by Maria Rivero~~
	~~Claimed by Yeon Jae Cho (FALL 2016)~~

	~~'''Claimed by Stella Chen (Fall 2017)'''~~
	'''Claimed by Rohit Naras (Spring 2026)'''		'''Claimed by Rohit Naras (Spring 2026)'''

Rohitnaras at 02:59, 27 April 2026

2026-04-27T02:59:54Z

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	'''Claimed by Stella Chen (Fall 2017)'''		'''Claimed by Stella Chen (Fall 2017)'''
	'''Claimed by Rohit Naras (Spring 2026)'''/n		'''Claimed by Rohit Naras (Spring 2026)'''

Rohitnaras at 02:59, 27 April 2026

2026-04-27T02:59:42Z

← Older revision		Revision as of 22:59, 26 April 2026
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	'''Claimed by Stella Chen (Fall 2017)'''		'''Claimed by Stella Chen (Fall 2017)'''
	'''Claimed by Rohit Naras (Spring 2026)'''		'''Claimed by Rohit Naras (Spring 2026)'''/n

Rohitnaras at 02:59, 27 April 2026

2026-04-27T02:59:24Z

← Older revision		Revision as of 22:59, 26 April 2026
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	'''Claimed by Stella Chen (Fall 2017)'''		'''Claimed by Stella Chen (Fall 2017)'''
			'''Claimed by Rohit Naras (Spring 2026)'''


	==Ampere-Maxwell Law==		==Ampere-Maxwell Law==

Schen498: /* Difficult */

2017-11-30T04:55:04Z

Difficult

← Older revision		Revision as of 00:55, 30 November 2017
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	the expression for the normal current is:		the expression for the normal current is:

	:<math>I = \frac{dq(t)}{t}</math>		:<math>I = \frac{dq(t)}{dt}</math>

	the current is directly related to charge on left capacitor plate		the current is directly related to charge on left capacitor plate