1. ## 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.

2. Originally Posted by nmanik90
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 $\frac{4l}{x^3}$ ......

3. 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.

4. 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 . . .

5. Originally Posted by nmanik90
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