A 10.5 cm long solenoid contains 891 turns and carries a current of 5.95 A . What is the strength of the magnetic field at the center of thi

Question

A 10.5 cm long solenoid contains 891 turns and carries a current of 5.95 A . What is the strength of the magnetic field at the center of this solenoid?

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Amity 5 years 2021-08-29T09:32:11+00:00 1 Answers 11 views 0

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  1. ngocdiep
    0
    2021-08-29T09:33:38+00:00
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    Answer:

    B=0.06345T

    Explanation:

    The magnetic field at the center of the solenoid can be calculated with the formula:

    B=\frac{\mu_0NI}{L}

    where \mu_0=4\pi\times10^{-7}N/A^2 is the vacuum permeability, and for our values is:

    B=\frac{(4\pi\times10^{-7}N/A^2)(891)(5.95A)}{(0.105m)}=0.06345T

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