A 100 W lightbulb actually emits around only 10 W of light. What is the intensity of the light 1 cm away if the light is emitted perfectly spherically? What is the magnitude of the electric field emitted by the lightbulb? What about the magnetic field?

• A. Intensity = 318 W/m²; Electric field intensity = 489 N/C; Magnetic field intensity = 1.63μT
• B. Intensity = 7958 W/m²; Electric field intensity = 2448 N/C; Magnetic field intensity = 8.16μT
• C. Intensity = 31830 W/m²; Electric field intensity = 4897 N/C; Magnetic field intensity = 16.3μT
• D. Intensity = 7958 W/m²; Electric field intensity = 2448 N/C; Magnetic field intensity = 2.17 μT

Answer: (B)

Verified step by step guidance

1. First, calculate the intensity of the light at a distance of 1 cm from the lightbulb. Intensity (I) is defined as the power (P) per unit area (A). Since the light is emitted spherically, the area is the surface area of a sphere with radius r. Use the formula: I = P / A, where A = 4πr².
2. Convert the distance from centimeters to meters for consistency in units. Since 1 cm = 0.01 m, the radius r = 0.01 m.
3. Substitute the given power of the light (10 W) and the calculated area into the intensity formula: I = 10 W / (4π(0.01 m)²).
4. Next, calculate the magnitude of the electric field (E) using the relationship between intensity and electric field: I = (1/2)ε₀cE², where ε₀ is the permittivity of free space (8.85 x 10⁻¹² C²/N·m²) and c is the speed of light (3 x 10⁸ m/s). Rearrange the formula to solve for E: E = sqrt(2I / (ε₀c)).
5. Finally, calculate the magnitude of the magnetic field (B) using the relationship between the electric field and magnetic field in an electromagnetic wave: B = E / c. Substitute the calculated electric field value to find B.