A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s$$^{-1}$$ in a uniform horizontal magnetic field of $$3.0 \times 10^{-2}$$ T. The maximum emf induced in the coil will be _________ $$\times 10^{-2}$$ volt (rounded off to the nearest integer).

We have a circular coil with $$N = 20$$ turns and radius $$R = 8.0\text{ cm}$$. First, we convert the radius into SI units because the magnetic field $$B$$ is given in tesla.

$$R = 8.0\text{ cm} = 8.0 \times 10^{-2}\text{ m} = 0.08\text{ m}$$

The angular speed of rotation is $$\omega = 50\ \text{rad s}^{-1}$$ and the magnetic field is uniform with magnitude $$B = 3.0 \times 10^{-2}\ \text{T}$$.

For a coil rotating in a uniform magnetic field, the maximum (peak) emf induced is given by the well-known generator formula

$$\mathcal{E}_{\max} = N\,B\,A\,\omega,$$

where $$A$$ is the area of one turn of the coil.

Now, we calculate the area. The coil is circular, so

$$A = \pi R^{2}.$$

Substituting $$R = 0.08\ \text{m},$$ we get

$$A = \pi \,(0.08)^{2} = \pi \times 0.0064 = 0.0064\pi\ \text{m}^{2}.$$

Using $$\pi \approx 3.1416,$$

$$A \approx 0.0064 \times 3.1416 = 0.02010624\ \text{m}^{2}.$$

Now we substitute all known values into the emf formula:

$$\mathcal{E}_{\max} = N\,B\,A\,\omega$$

$$\mathcal{E}_{\max} = (20)\,(3.0 \times 10^{-2})\,(0.02010624)\,(50).$$

First, multiply the number of turns with the magnetic field:

$$20 \times 3.0 \times 10^{-2} = 0.6.$$

Next, multiply this result by the area:

$$0.6 \times 0.02010624 = 0.012063744.$$

Finally, multiply by the angular speed:

$$0.012063744 \times 50 = 0.6031872\ \text{V}.$$

The question expresses the answer in units of $$\times 10^{-2}$$ volt. Dividing by $$10^{-2}$$ (that is, multiplying by 100) converts our result:

$$0.6031872\ \text{V} = \frac{0.6031872}{0.01} = 60.31872 \times 10^{-2}\ \text{V}.$$

Rounding this to the nearest integer gives $$60.$$

So, the answer is $$60$$.