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

AI Thread Summary
The discussion focuses on calculating the magnetic flux through a disk-shaped area positioned in a solenoid. The initial calculation used an incorrect area formula, incorporating the length of the cylinder, which led to a discrepancy in the expected flux value. The correct approach involves using the cross-sectional area of the solenoid, specifically πr², where r is the radius of the solenoid. The participants clarify that the magnetic field outside the solenoid is zero, emphasizing the importance of using the correct radius for calculations. Ultimately, the correct flux value can be achieved by applying the appropriate area formula with the solenoid's dimensions.
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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
 
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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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