- #1
Sparky_
- 227
- 5
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_