How Does Ampere's Law Calculate Current in a Uniform Magnetic Field?

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Homework Help Overview

The discussion revolves around applying Ampere's Law to calculate the current enclosed by a circular path in a uniform magnetic field. The problem specifies a magnetic field strength and radius, prompting participants to explore the relationship between these quantities and the current.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Ampere's Law, with one individual attempting to calculate the current and arriving at a specific value. Others question the consistency of the magnetic field value used and clarify the distinction between vacuum permittivity and permeability constants.

Discussion Status

The discussion is ongoing, with participants providing insights and questioning assumptions about the constants used in the calculations. There is an acknowledgment of differing results, and some guidance has been offered regarding the correct constants to apply.

Contextual Notes

Participants are navigating potential discrepancies in the values of the magnetic field and constants, which may affect the calculations. The original poster's calculations are under scrutiny, and there is a need for clarification on the parameters involved.

ineedhelp5454
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Homework Statement


Along a circular path the magnetic field is measured and is found to be uniform in magnitude and always tangent to the circular path.
If the radius of the path is 0.03 m and B along the path is 1.4e-06 T, use Ampere's law to calculate the magnitude of the conventional current enclosed by the path.

Homework Equations


ampere's law -- integral(B*dl) = mu I

The Attempt at a Solution


i keep getting 3.19e4 A
 
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Hello ineedhelp5454,

Welcome to Physics Forums!
ineedhelp5454 said:
i keep getting 3.19e4 A
My answer is different by 5 or so orders of magnitude.

Show us your steps and maybe we can help figure out what's going on.
 
1.5e-6 * 2 * pi * .03 = 2.87e-7 (left half of ampere's equation)
2.87e-7 = 8.85e-12(mu) * I (right half)
I = 3.19e4
 
ineedhelp5454 said:
1.5e-6 * 2 * pi * .03 = 2.87e-7 (left half of ampere's equation)
Is the magnitude of B 1.5 x 10-6 or 1.4 x 10-6 T?
2.87e-7 = 8.85e-12(mu) * I (right half)
I = 3.19e4
You're using the vacuum permittivity constant ε0 = 8.854 x 10-12 F/m, but you need to use the vacuum permeability constant μ0= 4π x 10-7 H/m.

They're not the same thing, but they are related. The speed of light in a vacuum [itex]c = 1/\sqrt{\epsilon_0\mu_0}[/itex]
 
thanks!
 

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