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Abhisar

  • one year ago

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.

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  1. Abhisar
    • one year ago
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    @Michele_Laino

  2. IrishBoy123
    • one year ago
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    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

  3. Michele_Laino
    • one year ago
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    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

  4. Abhisar
    • one year ago
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    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:

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