What Is the Flux Through a Disk-Shaped Area in a Solenoid?

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SUMMARY

The discussion focuses on calculating the magnetic flux through a disk-shaped area positioned in a solenoid. The magnetic field inside the solenoid is given by the formula B = (μ₀ I N) / l, where μ₀ is the permeability of free space, I is the current, N is the number of turns, and l is the length of the solenoid. The correct area for flux calculation is the cross-sectional area of the solenoid, πr², where r is the radius of the solenoid, which is 1.25 cm. The final flux calculation yields a value of approximately 7.40 µWb, aligning with the expected answer from the textbook.

PREREQUISITES
  • Understanding of magnetic flux and its calculation.
  • Familiarity with solenoid properties and formulas.
  • Knowledge of the permeability of free space (μ₀).
  • Basic geometry involving circles and areas.
NEXT STEPS
  • Study the derivation of the magnetic field inside a solenoid using Ampère's Law.
  • Learn about the concept of magnetic flux and its applications in electromagnetism.
  • Explore the effects of varying current and number of turns on the magnetic field strength in solenoids.
  • Investigate the relationship between magnetic flux and induced electromotive force (EMF) as described by Faraday's Law.
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Students studying electromagnetism, physics educators, and anyone involved in electrical engineering or related fields seeking to understand magnetic fields and flux calculations in solenoids.

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Homework Statement




Calculate the flux through the surface of a disk-shaped area of radius R = 5.00 cm that is
positioned perpendicular to and centered on the axis of the solenoid as shown below


The Attempt at a Solution



\oint \vec{B} \cdot \vec{dA} = \Phi_{B}

\vec{B} = \frac{\mu_0 I N}{l}

\vec{B} \cdot (\pi R^2 l) = \Phi_{B}

2\mu_0 \pi INR^2 = 7.106 \times 10^{-5}Wb

My book has 7.40uWb.

For my "surface area", I had \pi r^2 for the circle and I times it by l because that's how long the cylinder is. I know that is volume but I already tried other possible "areas" like just πr2 and 2πrl
 
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I think you forgot to attach the diagram. Based on your description alone, I think your solution is correct, except that area should be pi*r^2 instead of pi*r^2*l. (I know you already tried that, but unless the diagram shows something I haven't considered, it should definitely be correct.)
 
ideasrule said:
I think you forgot to attach the diagram. Based on your description alone, I think your solution is correct, except that area should be pi*r^2 instead of pi*r^2*l. (I know you already tried that, but unless the diagram shows something I haven't considered, it should definitely be correct.)

My idiocy is truly unparalleled

http://img857.imageshack.us/i/78891006.png/

Uploaded with ImageShack.us
 
Last edited by a moderator:
The figure does indeed show something I haven't considered. B is 0 outside of the solenoid, so the effective area you should use in calculating the flux should be the cross-sectional area of the solenoid. Are you given that value?
 
Yeah just let me wash my hands first because I was eating and writing at the same time and I just spilled my drink...
 
Yup, the radius of the solenoid is r = 1.25cm. The length of the of the solenoid is 30.0cm, the current through it is 12A and there are 300 turns
 
So have you tried using pi*r^2 as your area, with r=1.25 cm?
 
Yes I got the answer I forgot to note that lol sorry
 

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