A large flat coil of 500 turns and radius R = 20 cm carries a steady current of 4.0 A…

Step 1. Field at the centre (a). $B_0 = \dfrac{\mu_0N_1I}{2R} = \dfrac{4\pi \times 10^{-7}\ \text{T m/A} \times 500 \times 4.0\ \text{A}}{2 \times 0.20\ \text{m}} = 6.28\ \text{mT}$.

Step 2. Steepest slope (b). $\dfrac{dB}{dz} = -\dfrac{3\mu_0N_1IR^2z}{2(R^2 + z^2)^{5/2}}$. Its size is greatest where $\dfrac{d}{dz}\left[\dfrac{z}{(R^2 + z^2)^{5/2}}\right] = 0$, that is $R^2 + z^2 = 5z^2$, so $z = R/2 = 10.0\ \text{cm}$.

Step 3. Greatest emf (c). There $\left|\dfrac{dB}{dz}\right| = \dfrac{3\mu_0N_1I}{4R^2}\left(\dfrac45\right)^{5/2} = 0.0270\ \text{T/m}$. With $A = \pi(0.01\ \text{m})^2 = 3.142 \times 10^{-4}\ \text{m}^2$:

Step 4. Charge (d). $Q = \int \dfrac{\mathcal{E}}{r}dt = \dfrac{N_2A}{r}\left[B(0) - B(\infty)\right] = \dfrac{N_2AB_0}{r} = 197\ \mu\text{C}$.

Step 5. Heat (e). With $dt = dz/v$:

Twice as fast. The greatest emf doubles, as $v$. The charge does not change: it depends only on the flux at the start and the end. The heat doubles too: the power goes as $v^2$ but the trip takes half the time. This is why a search coil joined to a charge meter (a ballistic galvanometer) measures a field whatever the hand's speed.