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Math Help - Magnetic flow

  1. #1
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    Magnetic flow

    Hiya need a little help with this please

    The Magnetic flow; M, through a rektangular coil with the area A kan be discribled with M=B*Acos(100pi*T) where B is the magnetic flows density. when the coil rotates - it has 50 rotations per second - inducing a AC current that gives dm/dt. find the amplitude of the AC if its magnetic flows density B=150 mT and the area A=2.5 dm^2

    I think i need to derv M=B*Acos(100pi*T) and then put in the numbers but am unsure how do derv this. maybe somthing like M=B*-Asin(100pi*T)+100 ?

    if anyone could give me a hand i would be thankful.
    Last edited by JBswe; February 4th 2008 at 02:41 PM.
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  2. #2
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    Quote Originally Posted by JBswe View Post
    Hiya need a little help with this please

    The Magnetic flow; M, through a rektangular coil with the area A kan be discribled with M=B*Acos(100pi*T) where B is the magnetic flows density. when the coil rotates - it has 50 rotations per second - inducing a AC current that gives dm/dt. find the amplitude of the AC if its magnetic flows density B=150 mT and the area A=2.5 dm^2

    I think i need to derv M=B*Acos(100pi*T) Mr F says: Don't think, know. You have obviously studied induced E.M.F. and Faraday's Law ......

    and then put in the numbers but am unsure how do derv this. maybe somthing like M=B*-Asin(100pi*T)+100 ?

    if anyone could give me a hand i would be thankful.
    If y = A \cos (kt) then \frac{dy}{dt} = - k A \sin (kt) .....
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  3. #3
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    Ok so M=B*Acos(100pi*T)

    dy/dt = -100*Asin(100T)

    A is = 2.5
    B is =150

    I swap in the values

    = -100*150*2.5sin(100T)

    so the amplitude of the AC is 37500? That seems wrong
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  4. #4
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    Quote Originally Posted by JBswe View Post
    Hiya need a little help with this please

    The Magnetic flow; M, through a rektangular coil with the area A kan be discribled with M=B*Acos(100pi*T) where B is the magnetic flows density. when the coil rotates - it has 50 rotations per second - inducing a AC current that gives dm/dt. find the amplitude of the AC if its magnetic flows density B=150 mT and the area A=2.5 dm^2

    I think i need to derv M=B*Acos(100pi*T) and then put in the numbers but am unsure how do derv this. maybe somthing like M=B*-Asin(100pi*T)+100 ?

    if anyone could give me a hand i would be thankful.
    \frac{dM}{dt} = -100 \pi B A \sin(100 \pi T) so the amplitude is 100 \pi B A.

    But you've gotta get the units right, sport:

    B = 150 mT = 150 \times 10^{-3} = 0.150 Tesla.

    A = 2.5 dm^2 = 2.5 \times 10^{-2} = 0.025 m^2.

    Note: 1 dm = 1/10 m therefore 1 dm^2 = 1/100 m^2.

    So the amplitude is (100)(\pi)(0.150)(0.025) = (15)(\pi)(0.025) = ......
    And the unit will be ....?

    Wrong units mean that answers can be wrong by many orders of magnitude. Make it your business to get them right.
    Last edited by mr fantastic; February 5th 2008 at 12:10 AM.
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  5. #5
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    By now you know how to get the correct answer. Just so you understand your mistakes:

    Quote Originally Posted by JBswe View Post
    Ok so M=B*Acos(100pi*T)

    dy/dt = -100*Asin(100T) Mr F says: There's a \pi missing from the amplitude ... The 'k' in M = B A \cos( 100 \pi T) is 100 \pi.

    A is = 2.5 Mr F says: What are the units!!

    B is =150 Mr F says: What are the units!!

    The unit problem I pointed out earlier means that your answer will be out by a factor of 10^{-5}. But then things have to get multiplied by the missing \pi.


    I swap in the values

    = -100*150*2.5sin(100T)

    so the amplitude of the AC is 37500? That seems wrong Mr F says: At least you realised .... many students would not have.
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  6. #6
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    ok units have always been a problem for me but i think the answer should be in mT
    <br />
\frac{dM}{dt} = -100 \pi B A \sin(100 \pi T)<br />

    <br />
\frac{dM}{dt} = -100 \pi 150*10^{-3}*2.5*10^{-2} \sin(100 \pi T)<br />

    so the amplitude is 1.18 mT? but that seems very small. maybe 1.18 T
    Last edited by JBswe; February 5th 2008 at 02:44 AM.
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  7. #7
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    Quote Originally Posted by JBswe View Post
    ok units have always been a problem for me but i think the answer should be in mT
    <br />
\frac{dM}{dt} = -100 \pi B A \sin(100 \pi T)<br />

    <br />
\frac{dM}{dt} = -100 \pi 150*10^{-3}*2.5*10^{-2} \sin(100 \pi T)<br />

    so the amplitude is 1.18 mT? but that seems very small. maybe 1.18 T
    Correct value but wrong unit.

    All data has been put into the equation in SI units (m^2, T). So the answer will be in SI units. And the SI unit of E.M.F (induced or otherwise) is ..... drum roll please ....... VOLT.

    So the amplitude is 1.18 Volt.
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