A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of $$40\pi$$ rad s$$^{-1}$$ about its axis, perpendicular to its plane. If the magnetic field at its centre is $$3.8 \times 10^{-9}$$ T, then the charge carried by the ring is close to $$\mu_0 = 4\pi \times 10^{-7}$$ N/A$$^2$$.

First we write every quantity in SI units. The radius of the ring is $$R = 10\ \text{cm} = 0.10\ \text{m}$$ and the angular speed is $$\omega = 40\pi\ \text{rad s}^{-1}$$. The magnetic permeability of free space is given as $$\mu_0 = 4\pi \times 10^{-7}\ \text{N A}^{-2}$$, and the magnetic field at the centre is $$B = 3.8 \times 10^{-9}\ \text{T}$$.

When a ring of total charge $$Q$$ rotates with angular speed $$\omega$$, every complete rotation carries the charge once around the axis. If the time for one rotation (the time period) is $$T$$, the current is defined by $$I = \dfrac{Q}{T}$$. Because angular speed and time period are related by the formula $$\omega = \dfrac{2\pi}{T}$$, we can rewrite the current as

$$I = \dfrac{Q}{T} = Q \left(\dfrac{\omega}{2\pi}\right).$$

The magnetic field at the centre of a circular loop of radius $$R$$ carrying a current $$I$$ is given by the formula

$$B = \dfrac{\mu_0 I}{2R}.$$

We now substitute $$I = Q\omega / 2\pi$$ in this expression:

$$B = \dfrac{\mu_0}{2R}\left(\dfrac{Q\omega}{2\pi}\right) = \dfrac{\mu_0 Q \omega}{4\pi R}.$$

To isolate the charge $$Q$$ we rearrange:

$$Q = \dfrac{B \,(4\pi R)}{\mu_0 \,\omega}.$$

Now we substitute the numerical values step by step. First calculate the factor $$4\pi R$$:

$$4\pi R = 4 \times \pi \times 0.10\ \text{m} = 0.40\pi\ \text{m}.$$

Next write the denominator $$\mu_0 \omega$$:

$$\mu_0 \omega = (4\pi \times 10^{-7}) \times (40\pi) = 4 \times 40 \times \pi^2 \times 10^{-7} = 160\pi^2 \times 10^{-7}.$$

Putting these into the expression for $$Q$$ gives

$$Q = \dfrac{(3.8 \times 10^{-9}) \times (0.40\pi)}{160\pi^2 \times 10^{-7}}.$$

Notice that one factor of $$\pi$$ cancels immediately:

$$Q = \dfrac{(3.8 \times 10^{-9}) \times 0.40}{160\pi \times 10^{-7}}.$$

Multiply the numbers in the numerator:

$$3.8 \times 0.40 = 1.52,$$

so

$$Q = \dfrac{1.52 \times 10^{-9}}{160\pi \times 10^{-7}}.$$

Write the powers of ten together:

$$Q = \dfrac{1.52}{160\pi} \times 10^{-9+7} = \dfrac{1.52}{160\pi} \times 10^{-2}.$$

Compute the numerical ratio $$\dfrac{1.52}{160} = 0.0095.$$ Using $$\pi \approx 3.14$$ gives

$$\dfrac{0.0095}{\pi} \approx \dfrac{0.0095}{3.14} \approx 0.0030.$$

Finally, restoring the power of ten:

$$Q \approx 0.0030 \times 10^{-2}\ \text{C} = 3.0 \times 10^{-5}\ \text{C}.$$

Thus the charge on the ring is approximately $$3 \times 10^{-5}\ \text{coulomb}$$.

Hence, the correct answer is Option B.