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 $\displaystyle x$ from its centre, is given by

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

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

How would you solve this question?

Thanks

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 $\displaystyle x$ from its centre, is given by

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

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

How would you solve this question?

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

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

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

Hence:

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

RonL

3. ## Re:

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

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

Thanks.

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

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

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

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

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

RonL

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

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

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

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

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

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

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

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

6. ## re:

Thanks for that Soroban!

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

Thanks

Thanks for that Soroban!

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

Thanks

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

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

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

RonL

8. ## re:

Originally Posted by Soroban
Combine the fractions: .$\displaystyle \frac{(x+L)^2 - (x-L)^2}{(x-L)^2(x+L)^2} \;=\;\frac{4xL}{(x^2-L^2)^2}$
Thanks for the post guys very useful!

9. Originally Posted by Soroban
Combine the fractions: .$\displaystyle \frac{(x+L)^2 - (x-L)^2}{(x-L)^2(x+L)^2} \;=\;\frac{4xL}{(x^2-L^2)^2}$
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: .$\displaystyle \frac{1}{(x-L)^2} - \frac{1}{(x+L)^2}$

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

. . $\displaystyle =\;\frac{(x+L)^2}{(x+L)^2(x-L)^2} - \frac{(x-L)^2}{(x+L)^2(x-L)^2} \;=\;\frac{(x+L)^2 - (x-L)^2}{(x+L)^2(x-L)^2}$

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

Got it?

11. Great

Thanks