1. ## 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

2. Originally Posted by dadon
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

3. ## Re:

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}
$
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.

4. Originally Posted by dadon
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

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

6. ## 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

7. Originally Posted by dadon
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

8. ## re:

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!

9. 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

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?

11. Great

Thanks