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anonymous

  • one year ago

The current in a solenoid is changed from 1.1 A to 4.4 A. By what factor does the magnitude of the magnetic field inside the solenoid increase? A. 2 B. 4 C. 0.5 D. 0.25

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  1. Michele_Laino
    • one year ago
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    let's suppose that our solenoid has N turns and its length is L, then the field H inside it is given by the subsequent formula: \[\Large H = {\mu _0}\frac{{NI}}{L},\quad {\mu _0} = 12.56 \times {10^{ - 6}}Henry/meter\]

  2. anonymous
    • one year ago
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    1.256E-5 ?

  3. Michele_Laino
    • one year ago
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    yes! so we can write: \[\Large \begin{gathered} {H_1} = {\mu _0}\frac{{N{I_1}}}{L},\quad {H_2} = {\mu _0}\frac{{N{I_2}}}{L} \hfill \\ \hfill \\ \frac{{{H_2}}}{{{H_1}}} = \frac{{{I_2}}}{{{I_1}}} \hfill \\ \end{gathered} \]

  4. anonymous
    • one year ago
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    ok! what do we plug in?

  5. Michele_Laino
    • one year ago
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    and therefore: \[\Large \frac{{{H_2}}}{{{H_1}}} = \frac{{{I_2}}}{{{I_1}}} = \frac{{4.4}}{{1.1}} = ...\]

  6. anonymous
    • one year ago
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    4! so choice B (4) is our solution? :O

  7. Michele_Laino
    • one year ago
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    yes!

  8. anonymous
    • one year ago
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    yay!! thanks!!:D

  9. Michele_Laino
    • one year ago
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    :)

  10. Directrix
    • 9 months ago
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    @Daniellelovee

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