Abhisar
  • Abhisar
A solenoid 4cm in diameter and 20cm in length has 250 turns and carries a current of 15A. Calculate the flux through the surface of a disc of 10cm radius that is positioned perpendicular to and centered on the axis of the solenoid.
Physics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

Abhisar
  • Abhisar
IrishBoy123
  • IrishBoy123
for the magnetic field inside the solenoid, flux density: \[B = \mu_o . \frac{N}{L}.I\] for the flux: \(\Phi = BA\) where A is for the solenoid as it has smaller rad than disc
Michele_Laino
  • Michele_Laino
Using the idea of @IrishBoy123 we can write the concatenated flux of magnetic field with the same solenoid, as below: \[\Large \Phi = NBA = {\mu _0}\frac{{{N^2}}}{L}IA\] where \(\Large A\) is the area of each turn of our solenoid. Such flux is exactly the concatenated flux with our disc, since the disc is larger than the solenoid, and out of the solenoid the magnetic field can neglected, because it ia close to zero. Of course, \(\Large I\) is the current of our solenoid, and \(\Large L \) is its axial length, furthermore \(\Large A\) is the geometrical area of each turn

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Abhisar
  • Abhisar
Oh, Now I get it. I was multiplying the magnetic field with the area of disc but of course any area outside the cross sectional area of solenoid will not receive any magnetic field !! Thanks a bunch both of you !!! c:

Looking for something else?

Not the answer you are looking for? Search for more explanations.