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Math Help - proof2

  1. #1
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    Post proof2

    Hey MathForum,

    Im not sure if this is for this section.

    The field strength of a magnet (H) at a point on the axis, distance x from its centre, is given by

     H = \frac{M}{2l} \left\{ \frac{1}{(x-l)^2} - \frac{1}{(x+l)^2}\right\}

    where 2l = length of magnet and M = moment. Show that, if l is very small compared with x, then  H\approx\frac{2M}{x^3}

    How would you solve this question?

    Thanks
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  2. #2
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    Quote Originally Posted by dadon View Post
    Hey MathForum,

    Im not sure if this is for this section.

    The field strength of a magnet (H) at a point on the axis, distance x from its centre, is given by

     H = \frac{M}{2l} \left\{ \frac{1}{(x-l)^2} - \frac{1}{(x+l)^2}\right\}

    where 2l = length of magnet and M = moment. Show that, if l is very small compared with x, then  H\approx\frac{2M}{x^3}

    How would you solve this question?

    Thanks
    If l<<x we expand the two terms in the curly brackets:

    <br />
\frac{1}{(x-l)^2} - \frac{1}{(x+l)^2}=<br />
\frac{1}{x^2(1-l/x)^2}-\frac{1}{x^2(1+l/x)^2}<br />

    ... =\frac{1}{x^2}(1+(-2)(-l/x)+O((l/x)^2)-\frac{1}{x^2}(1+(-2)(l/x)+O((l/x)^2) =\frac{4 \,l}{x^3}+O(l^2/x^4)<br />

    Hence:

    H= \frac{2M}{x^3}+O(l/x^4) \approx \frac{2M}{x^3}

    RonL
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    Post Re:

    Quote Originally Posted by CaptainBlack View Post
    If l<<x we expand the two terms in the curly brackets:
    <br />
\frac{1}{(x-l)^2} - \frac{1}{(x+l)^2}=<br />
\frac{1}{x^2(1-l/x)^2}-\frac{1}{x^2(1+l/x)^2}<br />
    Thank you for your post.

    What do you do on the first line? I don't know how you get x^2(1-l/x)^2


    Thanks.
    Last edited by dadon; January 28th 2007 at 08:15 AM.
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    Quote Originally Posted by dadon View Post
    Thank you for your post.

    Would do you do on the first line? I don't know how you get x^2(1-l/x)^2


    Thanks.
    Its a standard trick that is used in these problems:

    [x-l]^2=[x(1-l/x)]^2=x^2(1-l/x)^2

    to see that this is the case try it the other way:

    x^2(1-l/x)^2=[x(1-l/x)]^2=[x-l]^2

    RonL
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    Hello, dadon!

    The field strength of a magnet H at a point on the axis,

    distance x from its centre, is given by: .  H \;= \;\frac{M}{2L} \left\{ \frac{1}{(x-L)^2} - \frac{1}{(x+L)^2}\right\}
    where 2L = length of magnet and M = moment.

    Show that, if L is very small compared with x, then: .  H \:\approx\:\frac{2M}{x^3}

    Combine the fractions: . \frac{(x+L)^2 - (x-L)^2}{(x-L)^2(x+L)^2} \;=\;\frac{4xL}{(x^2-L^2)^2}

    The equation becomes: . H \;=\;\frac{M}{2L}\,\frac{4xL}{(x^2-L^2)^2} \;=\;\frac{2Mx}{(x^2-L^2)^2}


    If L is very small compared to x, .then L^2 is even smaller.

    If we consider L^2 to be negligible (virtually zero),

    . . the formula becomes: . H \;\approx\;\frac{2Mx}{(x^2)^2} \;=\;\frac{2M}{x^3}

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    Post re:

    Thanks for that Soroban!

    Good method you use. But how did you expand (x-L)^2(x+L)^2 to equal (x^2-L^2)^2. Im doing it the long way and it taking ages lol. what method did you use?

    Thanks

    Dadon.
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    Quote Originally Posted by dadon View Post
    Thanks for that Soroban!

    Good method you use. But how did you expand (x-L)^2(x+L)^2 to equal (x^2-L^2)^2. Im doing it the long way and it taking ages lol. what method did you use?

    Thanks

    Dadon.
    (x-L)^2(x+L)^2=[(x-L)(x+L)]^2

    Now what is inside the square brackets simplifies to the difference of x^2 and l^2

    (x-L)^2(x+L)^2=[(x-L)(x+L)]^2=[x^2-L^2]^2

    RonL
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    re:

    Quote Originally Posted by Soroban View Post
    Combine the fractions: . \frac{(x+L)^2 - (x-L)^2}{(x-L)^2(x+L)^2} \;=\;\frac{4xL}{(x^2-L^2)^2}
    Thanks for the post guys very useful!
    Last edited by dadon; January 31st 2007 at 11:57 AM.
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    Quote Originally Posted by Soroban View Post
    Combine the fractions: . \frac{(x+L)^2 - (x-L)^2}{(x-L)^2(x+L)^2} \;=\;\frac{4xL}{(x^2-L^2)^2}
    what do you mean when you say combine fractions? do you mulityply the denominators ?


    kind regards

    dadon
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    Hello, dadon!

    What do you mean when you say combine fractions?
    It's some old-fashioned algebra ... the most annoying kind:
    . . adding or subtracting fractions.

    We have: . \frac{1}{(x-L)^2} - \frac{1}{(x+L)^2}


    Get a common denominator: . \frac{(x+L)^2}{(x+L)^2}\cdot\frac{1}{(x-L)^2} - \frac{x-L)^2}{(x-L)^2}\cdot\frac{1}{(x+L)^2}


    . . =\;\frac{(x+L)^2}{(x+L)^2(x-L)^2} - \frac{(x-L)^2}{(x+L)^2(x-L)^2} \;=\;\frac{(x+L)^2 - (x-L)^2}{(x+L)^2(x-L)^2}<br />


    \. . = \;\frac{x^2 + 2xL + L^2 - x^2 + 2xL - L^2}{\left[(x+L)(x-L)\right]^2} \;=\;\frac{4xL}{(x^2-L^2)^2}


    Got it?

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