Problem 22.41

22.41.
Consider a small strip of the metal rod of width dr. Then, the is an amount of current dI flowing through the rod where we realize that
$dI = I(\frac{dr}{w})$
We know that the magnetic field a distance r away from a wire carrying current dI is given by
$dB = \frac{\mu_{0} dI}{2\pi r}$
So, to find the total magnetic field, we have to integrate.
$B = \int dB$
$B = \int_{b}^{b+w} \frac{\mu_{0}I dr}{2 \pi w r}$
$B = \frac{\mu_{0}I}{2 \pi w} \int_{b}^{b+w} \frac{dr}{r}$
$B = \frac{\mu_{0}I}{2 \pi w} \ln (1 + \frac{w}{b})$
or $\vec{B} = \frac{\mu_{0}I}{2 \pi w} \ln (1 + \frac{w}{b}) \hat{k}$