Maxium Allowable Current to (+/-) 2% ?

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SUMMARY

The discussion focuses on calculating the maximum allowable current in an electric cable to maintain accuracy within ±2% for an experiment measuring the Earth's magnetic field. The relevant equation used is B = [I*μ]/[2*Pi*d], where μ represents the permeability of free space, and d is the distance from the cable. A specific calculation provided indicates that with a maximum current of 1750 A, the allowable deviation is calculated as 35 A, resulting in a maximum current of 1715 A to meet the accuracy requirement.

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  • Understanding of magnetic fields and their measurement
  • Familiarity with the equation B = [I*μ]/[2*Pi*d]
  • Basic knowledge of electrical current and its units
  • Concept of percentage error in experimental measurements
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  • Explore the effects of distance (d) on magnetic field strength
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Physics students, electrical engineers, and researchers conducting experiments involving magnetic fields and current measurements.

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Maxium Allowable Current to (+/-) 2% ?

Homework Statement



An experiment on the Earth's magnetic field is being carried out 7.00 from an electric cable.
What is the maximum allowable current in the cable if the experiment is to be accurate to 2.0 ? Assume that Earth’s magnetic field is .


Homework Equations



B = [I*μ]/[2*Pi*d]

The Attempt at a Solution



Check the picture to see how far I've gotten, but it's still wrong. I'm not sure how to attempt this problem.
 

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So any suggestions mates?

Melqarthos
 


you have to use 1750 * 2.0%

1750 * 0.02 = 35 A
 

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