# proof2

• January 27th 2007, 01:11 PM
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
• January 27th 2007, 01:31 PM
CaptainBlack
Quote:

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< we expand the two terms in the curly brackets:

$
\frac{1}{(x-l)^2} - \frac{1}{(x+l)^2}=
\frac{1}{x^2(1-l/x)^2}-\frac{1}{x^2(1+l/x)^2}
$

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

Hence:

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

RonL
• January 28th 2007, 02:07 AM
Re:
Quote:

Originally Posted by CaptainBlack
If $l< we expand the two terms in the curly brackets:
$
\frac{1}{(x-l)^2} - \frac{1}{(x+l)^2}=
\frac{1}{x^2(1-l/x)^2}-\frac{1}{x^2(1+l/x)^2}
$

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

Thanks.
• January 28th 2007, 03:29 AM
CaptainBlack
Quote:

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
• January 28th 2007, 05:56 AM
Soroban

Quote:

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

• January 28th 2007, 08:14 AM
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

• January 28th 2007, 08:53 AM
CaptainBlack
Quote:

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

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

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
• January 30th 2007, 12:15 PM
re:
Quote:

Originally Posted by Soroban
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!
• January 31st 2007, 11:58 AM
Quote:

Originally Posted by Soroban
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

• February 3rd 2007, 09:08 AM
Soroban

Quote:

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}
$

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

Got it?

• February 3rd 2007, 01:26 PM