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## Homework Statement

Greetings,

In Griffiths, Introduction to Electrodynamics, example 5.5 (page 216), calculating the B field a distance “S” away from a current carrying wire.

l' (dl’) is the horizontal current carry wire – will be segmented to dl’

[tex] tan(\theta) = \frac{l’}{s}[/tex]

In the next step, it is stated

[tex] dl’ = \frac{s}{cos^2(\theta)} d(\theta)[/tex]

I am stuck on this - I do not see how to get from the first equation to the second. Is there an approximation required?

## Homework Equations

## The Attempt at a Solution

[tex] tan(\theta) = \frac{l’}{s}[/tex]

and

[tex] r^2 = l^2 + s^2[/tex]

[tex] l' = s tan(\theta) [/tex]

I could square both sides

[tex]l^2 = s^2 tan^2(\theta)[/tex]

[tex]l^2 = s^2 \frac{sin^2(\theta)}{cos^2(\theta)}[/tex]

[tex]l^2 = (r^2-l^2) \frac{sin^2(\theta)}{cos^2(\theta)}[/tex]

[tex]\frac{l^2}{(r^2-l^2)} = \frac{sin^2(\theta)}{cos^2(\theta)}[/tex]

[tex]sin^2(\theta) - 1 = \frac{sin^2(\theta)}{cos^2(\theta)}[/tex]

I do not have this in terms of l' anymore.

bottom line how does Griffiths get to

[tex] dl’ = \frac{s}{cos^2(\theta)} d(\theta)[/tex]

from

[tex] tan(\theta) = \frac{l’}{s}[/tex]

Thanks for the help

-Sparky_