An electric bulb is rated as $$200$$ W. What will be the peak magnetic field at $$4$$ m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with $$3.5\%$$ efficiency.

We are given an electric bulb rated at $$200$$ W with $$3.5\%$$ efficiency, and we need to determine the peak magnetic field at a distance of $$4$$ m.

Since the bulb converts $$200 \times \frac{3.5}{100} = 200 \times 0.035 = 7$$ W into radiated power, we have:

$$P_{rad} = 200 \times \frac{3.5}{100} = 200 \times 0.035 = 7 \text{ W}$$

This radiated power spreads uniformly over a spherical surface of radius $$4$$ m, so the intensity is:

$$I = \frac{P_{rad}}{4\pi r^2} = \frac{7}{4\pi (4)^2} = \frac{7}{4\pi \times 16} = \frac{7}{64\pi} \text{ W/m}^2$$

From the relation between the intensity of an electromagnetic wave and its peak magnetic field $$B_0$$,

$$I = \frac{c B_0^2}{2\mu_0}$$

we solve for $$B_0$$:

$$B_0 = \sqrt{\frac{2\mu_0 I}{c}}$$

Substituting $$\mu_0 = 4\pi \times 10^{-7}$$ T m/A and $$c = 3 \times 10^8$$ m/s yields:

$$B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times \frac{7}{64\pi}}{3 \times 10^8}}$$

Simplifying the numerator gives:

$$2 \times 4\pi \times 10^{-7} \times \frac{7}{64\pi} = \frac{2 \times 4 \times 7}{64} \times 10^{-7} = \frac{56}{64} \times 10^{-7} = \frac{7}{8} \times 10^{-7}$$

Hence,

$$B_0 = \sqrt{\frac{0.875 \times 10^{-7}}{3 \times 10^8}} = \sqrt{\frac{0.875}{3} \times 10^{-15}}$$

$$B_0 = \sqrt{0.29167 \times 10^{-15}} = \sqrt{2.9167 \times 10^{-16}}$$

$$B_0 = 1.708 \times 10^{-8} \text{ T} \approx 1.71 \times 10^{-8} \text{ T}$$

Therefore, the correct option is Option B: $$1.71 \times 10^{-8}$$ T.