Find the magnitude and direction of a magnetic field(Biort-Savart Law)

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SUMMARY

The discussion centers on calculating the magnetic field at point P due to a current-carrying arc using the Biot-Savart Law. The user correctly applies the formula B = (μ₀Iθ) / (4πa) with μ₀ = 4π × 10⁻⁷ T·m/A, I = 4 A, θ = π/3, and a = 0.4 m, resulting in a magnetic field of 1.0472E-6 Tesla. The user mistakenly converts this value to nanoTesla as 0.0010472 nT instead of the correct 1.0472 nT, as 1 T equals 10⁹ nT. The discrepancy arises from confusion over unit conversion.

PREREQUISITES
  • Understanding of the Biot-Savart Law
  • Familiarity with magnetic field calculations
  • Knowledge of unit conversions, specifically between Tesla and nanoTesla
  • Basic trigonometry for angle conversions
NEXT STEPS
  • Study the Biot-Savart Law in detail, focusing on its applications in different geometries
  • Learn about unit conversions in physics, particularly between different scales of measurement
  • Explore examples of magnetic field calculations for various current configurations
  • Review common mistakes in unit conversions and how to avoid them
USEFUL FOR

Students in physics, particularly those studying electromagnetism, educators teaching magnetic field concepts, and anyone involved in solving problems related to magnetic fields and current-carrying conductors.

Raziel2701
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Homework Statement



A current path shaped as shown in the figure(I included the picture at the end of post) produces a magnetic field at P, the center of the arc. If the arc subtends an angle of θ = 60.0° and the radius of the arc is 0.400 m, what are the magnitude and direction of the field produced at P if the current is 4.00 A?

Homework Equations



[tex]B=\frac{\mu_{0}I \theta}{4\pi\,a}[/tex]

It's straight from the book. And I know it's relevance due to the shape I'm dealing with.

The Attempt at a Solution



Doing that calculation I get an answer of 1.0472E-6 Tesla. The book wants the answer in nT, which I assume to be nano Tesla. I figure I need to move the decimal point three spaces. Giving me a result of .0010472 nano Tesla. The book, well, webassign is telling me this is wrong. I'll try to get the picture in here if I can get it to work.

What I substituted into the equation shown above is the following:

[tex]\mu_0=4\pi * 10^{-7}[/tex]
I= 4A
[tex]\theta=\frac{\pi}{3}[/tex]
a=0.4m

I noticed a discrepancy between the angle shown in the drawing and the one given in the problem, but I think it's a matter of webassign directly taking the drawing from the book and placing it here. The problem in the book does have an angle of 30 degrees, but webassign randomizes the given numbers. Meaning that I don't think the drawing's angle reflects my given numbers.

I don't see what I am doing wrong. The image of the problem is number two by the way.

ME7rb.png


What am I missing?
 
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"Doing that calculation I get an answer of 1.0472E-6 Tesla.
The book wants the answer in nT, which I assume to be nano Tesla.
I figure I need to move the decimal point three spaces. Giving me a
result of .0010472 nano Tesla. "

If your high is 1.7 m and somebody wants it in cm-s, you would say that you are 0.017 cm high, as you needed to move the decimal point by two?

Your first result 1.0472E-6 T is correct. And 1 T=109 nT. Do you get a smaller or higher number if you give the result in smaller units?

ehild
 
How could I be so dense? It hit me like a train.

Thank you good madame/sir!
 

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