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anonymous

  • one year ago

A current of 3.20 A in a solenoid produces a magnetic field of 4.0 × 10^-3 T. The current is changed to 6.40 A. What is the magnitude of the magnetic field after the change? A. 1.0 * 10^-3 T B. 2.0 * 10^-3 T C. 8.0 * 10^-3 T D. 16 * 10^-3 T image drawn inside!!

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  1. anonymous
    • one year ago
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    |dw:1434435467105:dw|

  2. Michele_Laino
    • one year ago
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    the magnitude of the magnetic field inside a solenoid , is given by the subsequent formula: \[\Large B = {\mu _0}\frac{{NI}}{L}\] where I is the current which flows along the solenoid, L is the xial length of the solenoid, N are the turns of our solenoid , and as usual: \[\Large {\mu _0} = 4\pi {10^{ - 6}}Henry/meter\]

  3. anonymous
    • one year ago
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    ok! what does that translate into for this problem? :/

  4. Michele_Laino
    • one year ago
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    so we can write: \[\Large {B_1} = {\mu _0}\frac{{N{I_1}}}{L},\quad {B_2} = {\mu _0}\frac{{N{I_2}}}{L}\]

  5. anonymous
    • one year ago
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    ok! what do we substitute in? :O

  6. Michele_Laino
    • one year ago
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    dividing side by side, we get: \[\Large \frac{{{B_1}}}{{{B_2}}} = \frac{{{I_1}}}{{{I_2}}}\]

  7. Michele_Laino
    • one year ago
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    now we have: I_1=3.20 A B_1=4.0 × 10^-3 T I_2=6.40 A so we can solve that equation for B_2, and we get: \[\Large {B_2} = \frac{{{I_2}}}{{{I_1}}}{B_1} = \frac{{6.4}}{{3.2}} \times 4 \times {10^{ - 3}} = ...\]

  8. anonymous
    • one year ago
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    and we get 0.008?

  9. Michele_Laino
    • one year ago
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    that's right!

  10. anonymous
    • one year ago
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    so would that mean choice A is the solution?

  11. Michele_Laino
    • one year ago
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    hint: 0.008 = 8*10^-3

  12. anonymous
    • one year ago
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    ohh okay! so choice C! thank you:)

  13. Michele_Laino
    • one year ago
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    :)

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