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Math Help - Ridiculously hard question

  1. #1
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    Angry Ridiculously hard question

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

    H = M/2l [ [(1/(x-l)^2] - [1/(x+l)^2] ], where 2l = length of magnet and M = moment. Show that, if l is very small compared with x, then H = 2M/x^3.

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  2. #2
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    Quote Originally Posted by nmanik90 View Post
    The field strength of a magnet (H) at a point on an axis, distance x from its centre, is given by the following equation:

    H = M/2l [ [(1/(x-l)^2] - [1/(x+l)^2] ], where 2l = length of magnet and M = moment. Show that, if l is very small compared with x, then H = 2M/x^3.

    Simplest approach is to use the linear approaximation f(x + h) \approx f(x) + h f'(x).

    Use it with f(x) = \frac{1}{x^2} and

    1. h = -l
    2. h = l

    to get first order approximations for \frac{1}{(x-l)^2} and \frac{1}{(x+l)^2}.

    The difference between the two approaximations works out to be \frac{4l}{x^3} ......
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  3. #3
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    Well I don't think it's as hard as you think actually.

    Ok so here is what you have now:

    H = \frac{M}{2l} \Big[ \frac{1}{(x-l)^2} - \frac{1}{(x+l)^2} \Big] = \frac{M}{2l} \Big[ \frac{(x+l)^2 - (x - l)^2}{(x-l)^2(x+l)^2} \Big]

    Work out the top:

    = \frac{M}{2l} \Big[ \frac{(x^2 + 2lx + l^2) - (x^2 - 2lx + l^2)}{(x-l)^2(x+l)^2} \Big]

    Now reduce what you can on top:

    = \frac{M}{2l} \Big[ \frac{4lx}{(x-l)^2(x+l)^2} \Big] = \frac{M2x}{(x-l)^2(x+l)^2}

    You were told that l is considerably smaller than x, so you can do this:

     = \frac{M2x}{\underbrace{(x-l)^2}_{acts \ like \ x^2} \underbrace{(x+l)^2}_{acts \ like \ x^2}} = \frac{M2x}{x^4} = \boxed{\frac{2M}{x^3}}

    Please don't just copy that understand. You should understand why I did each step. Most of these kinds of problems can be solved quite easily by simplifying them first.
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  4. #4
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    Hello, nmanik90!

    TrevorP beat me to it. My solution is longer/messier
    because I wanted to include the final steps.


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

    . . 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 \;= \;\frac{2M}{x^3}
    If L is very small compared with x, then the fraction \frac{L}{x} is very small.

    Simplify the equation: . H \;=\;\frac{M}{2L}\left[\frac{(x+L)^2 - (x-L)^2}{(x-L)^2(x+L)^2}\right] \;=\;\frac{M}{2L}\cdot\frac{4Lx}{(x-L)^2(x+L)^2}

    . . and we have: . H \;=\;\frac{2Mx}{(x-L)^2(x+L)^2}


    In denominator, multiply by \frac{x^2}{x^2} twice: . H \;=\;\frac{2Mx}{\frac{x^2}{x^2}(x-L)^2\cdot\frac{x^2}{x^2}(x+L)^2}

    . . = \;\frac{2Mx}{x^4\cdot\frac{(x-L)^2}{x^2}\cdot\frac{(x+L)^2}{x^2} } \;=\;\frac{2M}{x^3\cdot(\frac{x-L}{x})^2\cdot(\frac{x+L}{x})^2} \;=\;\frac{2M}{x^3\cdot(1 - \frac{L}{x})^2\cdot(1 + \frac{L}{x})^2}


    Now consider what happens when \frac{L}{x} is very small . . .

    . . \lim_{\frac{L}{x}\to0}\left[\frac{2M}{x^3\cdot(1 - \frac{L}{x})^2\cdot(1 + \frac{L}{x})^2}\right] \;=\;\frac{2M}{x^3\cdot(1 - 0)^2\cdot(1-0)^2} \;=\;{\color{blue}\frac{2M}{x^3}} . . . . ta-DAA!



    I need a nap . . .
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  5. #5
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    Quote Originally Posted by nmanik90 View Post
    The field strength of a magnet (H) at a point on an axis, distance x from its centre, is given by the following equation:

    H = M/2l [ [(1/(x-l)^2] - [1/(x+l)^2] ], where 2l = length of magnet and M = moment. Show that, if l is very small compared with x, then H = 2M/x^3.

    For the record, if you are in the Physical Sciences or Engineering or something like that, remember this method. You will likely be seeing a lot of it in the future.

    -Dan
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