What is the magnetic flux through the desk?

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SUMMARY

The magnetic flux through a horizontal desk with an area of 900 cm², given the magnetic field intensity of 57569 nT and an inclination of +75 degrees, can be calculated using the formula Φ = BAcos(θ). The correct angle to use is 15 degrees (90 degrees - 75 degrees). The calculated magnetic flux is 5.0050 x 10^-6 Tm², which must be expressed in scientific notation and accurate to five significant figures for submission.

PREREQUISITES
  • Understanding of magnetic flux and its calculation
  • Familiarity with the formula Φ = BAcos(θ)
  • Knowledge of SI units, specifically Tesla (T) and square meters (m²)
  • Basic trigonometry for angle conversions
NEXT STEPS
  • Review the concept of magnetic flux and its applications in physics
  • Learn about the Earth's magnetic field and its variations
  • Study the significance of significant figures in scientific calculations
  • Explore the use of scientific notation in reporting scientific data
USEFUL FOR

Students studying physics, particularly those focused on electromagnetism, as well as educators looking for examples of magnetic flux calculations.

monnapomona
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Homework Statement


The total magnetic field intensity at the location of Edmonton was 57652 nT on March 20, 2012, when I first wrote this question. Now, on Feb. 19, 2013, it is 57569 nT. The magnetic inclination is +75 degrees (i.e., it points downward at an angle of 75 degrees with respect to the horizontal). This site will tell you the current value of the Earth's magnetic field parameters at any location you choose.

Consider the desk you sit at in our class. Let us estimate that it has an area of about 900 cm^2 and is horizontal. Given the values of the magnetic field I listed for Feb. 19, 2013, what was the magnetic flux through the desk? Use SI units.

Note: Calculate and enter your answer with at least 5 significant digits. I had to make the accepted range of values in this answer quite small.


Homework Equations



\Phi = BAcos\vartheta


The Attempt at a Solution



I used A = 0.09 m^2 and B = 57569 x 10^-9. For the angle, I used 90 deg - 75 deg = 15 deg.

\Phi = (57569 x 10^-9 T)(0.09 m^2)cos(15)
= 0.0000050050 Tm^2

This was wrong in my online assignment. Not sure where I'm going wrong with this... Am I using the wrong angle?
 
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monnapomona said:
\Phi = (57569 x 10^-9 T)(0.09 m^2)cos(15)
= 0.0000050050 Tm^2

This was wrong in my online assignment. Not sure where I'm going wrong with this... Am I using the wrong angle?

Everything looks good, except the answer you give is not quite accurate to 5 significant figures. Also, scientific notation will make it look nicer.:approve:
 
TSny said:
Everything looks good, except the answer you give is not quite accurate to 5 significant figures. Also, scientific notation will make it look nicer.:approve:

Oh I see now! Thank you for your help and clarification! :smile:
 

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