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

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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
 
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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
 
Yes I got the answer I forgot to note that lol sorry