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

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Homework Help Overview

The discussion revolves around calculating the magnetic flux through a disk-shaped area positioned within a solenoid. The original poster attempts to determine the flux using the magnetic field expression and area calculations, while comparing their result to a value from a textbook.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the appropriate area to use for calculating flux, questioning the inclusion of length in the area calculation. There is also a focus on the effective area for flux calculation based on the solenoid's properties.

Discussion Status

Some participants have provided guidance on the area to use for the flux calculation, suggesting that the cross-sectional area of the solenoid should be considered. There is acknowledgment of the need for a diagram to clarify the setup, and multiple interpretations of the area calculation are being explored.

Contextual Notes

Participants note the radius and length of the solenoid, as well as the current and number of turns, which are relevant to the problem but not fully resolved in the discussion.

flyingpig
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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



[tex]\oint \vec{B} \cdot \vec{dA} = \Phi_{B}[/tex]

[tex]\vec{B} = \frac{\mu_0 I N}{l}[/tex]

[tex]\vec{B} \cdot (\pi R^2 l) = \Phi_{B}[/tex]

[tex]2\mu_0 \pi INR^2 = 7.106 \times 10^{-5}Wb[/tex]

My book has 7.40uWb.

For my "surface area", I had [tex]\pi r^2[/tex] 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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