Griffiths - example 5.5 Electrodynamics - Calculate B Field

[Sparky_]

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’

$$tan(\theta) = \frac{l’}{s}$$

In the next step, it is stated

$$dl’ = \frac{s}{cos^2(\theta)} d(\theta)$$

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

$$tan(\theta) = \frac{l’}{s}$$

and
$$r^2 = l^2 + s^2$$

$$l' = s tan(\theta)$$

I could square both sides

$$l^2 = s^2 tan^2(\theta)$$

$$l^2 = s^2 \frac{sin^2(\theta)}{cos^2(\theta)}$$

$$l^2 = (r^2-l^2) \frac{sin^2(\theta)}{cos^2(\theta)}$$

$$\frac{l^2}{(r^2-l^2)} = \frac{sin^2(\theta)}{cos^2(\theta)}$$

$$sin^2(\theta) - 1 = \frac{sin^2(\theta)}{cos^2(\theta)}$$

I do not have this in terms of l' anymore.bottom line how does Griffiths get to

$$dl’ = \frac{s}{cos^2(\theta)} d(\theta)$$

from

$$tan(\theta) = \frac{l’}{s}$$

Thanks for the help

-Sparky_

 

[TSny]

$$tan(\theta) = \frac{l’}{s}$$

In the next step, it is stated

$$dl’ = \frac{s}{cos^2(\theta)} d(\theta)$$

It's easy if you recall the formula for the derivative of the tangent function.

 

[Sparky_]

ah crap - chain rule

I was thinking straight algebra type identity or small angle approximation or some such

did not see the forest for the trees

THANKS!
Sparky_