Answer
a. $0.015H$
b. 75 Turns
Work Step by Step
a. Example 21-13 (page 609) derives the expression for the inductance of an air-filled solenoid.
$$L=\frac{\mu_oN^2A}{\mathcal{l}}$$
$$L=\frac{(4 \pi \times10^{-7}(T \cdot m)/A)(2600)^2\pi(0.0125m)^2}{0.282m}=0.015H$$
b. Apply the same equation. This time, solve for the number of turns N and use the permeability of iron.
$$L=\frac{\mu N^2A}{\mathcal{l}}$$
$$N=\sqrt{\frac{L\mathcal{l}}{\mu A }}$$
$$=\sqrt{\frac{(0.01479H)(0.282m)}{(1200)(4 \pi \times10^{-7}(T \cdot m)/A) \pi(0.0125m)^2}}$$
$$=75 turns$$
