What is the average magnitude of the Poynting vector 5.00 miles from a radio transmitter broadcasting isotropically (equally in all directions) with an average power of 250 kW?

Question

Asked on June 13, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-06-13T07:11:12.895Z

Here are the solutions to your questions: 1. What is the average magnitude of the Poynting vector 5.00 miles from a radio transmitter broadcasting isotopically (equally in all directions) with an average power of 250 kW? The average magnitude of the Poynting vector is the intensity (I), which is power per unit area. For isotropic radiation, the area is that of a sphere. Step 1: Convert given values to SI units. r = 5.00 miles × 1609.34 mmile = 8046.7 m P = 250 kW = 250 × 10^3 W Step 2: Calculate the surface area of the sphere. A = 4 r^2 = 4 (8046.7 m)^2 = 8.138 × 10^8 m^2 Step 3: Calculate the average magnitude of the Poynting vector (intensity). I = (P)/(A) = 250 × 10^3 W8.138 × 10^8 m^2 = 3.07 × 10^-4 Wm^2 The average magnitude of the Poynting vector is 3.07 × 10^-4 (W)/(m^2). 2. A parallel plate capacitor has circular plates with radius of 7.0cm. If the electric field between the plates is changing at a rate of 2 × 10^12 V/m/s. Find the displacement current. The displacement current (I_d) is given by the formula I_d = _0 A (dE)/(dt). Step 1: Convert the radius to meters and calculate the area of the plates. r = 7.0 cm = 0.07 m A = r^2 = (0.07 m)^2 = 0.01539 m^2 Step 2: Substitute the values into the displacement current formula. (dE)/(dt) = 2 × 10^12 Vm · s _0 = 8.854 × 10^-12 Fm I_d = (8.854 × 10^-12 Fm) (0.01539 m^2) (2 × 10^12 Vm · s) Step 3: Calculate the displacement current. I_d = 0.272 A The displacement current is 0.272 A. 3. The electric field amplitude of a laser beam is 6.0 × 10^8 V/m. Calculate its average and maximum intensity. The maximum intensity (I_max) and average intensity (I_avg) are related to the electric field amplitude (E_0). Step 1: Calculate the maximum intensity. I_max = c _0 E_0^2 Where c = 3.00 × 10^8 m/s and _0 = 8.854 × 10^-12 F/m. I_max = (3.00 × 10^8 ms) (8.854 × 10^-12 Fm) (6.0 × 10^8 Vm)^2 I_max = (3.00 × 10^8) (8.854 × 10^-12) (3.6 × 10^17) Wm^2 I_max = 9.56 × 10^14 Wm^2 Step 2: Calculate the average intensity. For a sinusoidal electromagnetic wave, the average intensity is half of the maximum intensity. I_avg = (1)/(2) I_max I_avg = (1)/(2) (9.56 × 10^14 Wm^2) = 4.78 × 10^14 Wm^2 The maximum intensity is 9.56 × 10^14 (W)/(m^2) and the average intensity is 4.78 × 10^14 (W)/(m^2). 4. A microwave oven has an average intensity of 8.33 × 10^3 W/m^2. Find the peak electric field strength (E_0). The average intensity (I_avg) is related to the peak electric field strength (E_0) by the formula I_avg = (1)/(2) c _0 E_0^2. Step 1: Rearrange the formula to solve for E_0. E_0^2 = 2 I_avgc _0 E_0 = sqrt(2 I_avg)c _0 Step 2: Substitute the given values and constants. I_avg = 8.33 × 10^3 Wm^2 c = 3.00 × 10^8 ms _0 = 8.854 × 10^-12 Fm E_0 = sqrt(2 × (8.33 × 10^3 W)m^2)(3.00 × 10^8 ms) (8.854 × 10^-12 Fm) E_0 = sqrt((1.666 × 10^4)/(2.6562 × 10^-3)) Vm E_0 = sqrt(6.272 × 10^6) Vm Step 3: Calculate E_0. E_0 = 2504.4 Vm The peak electric field strength is 2504.4 (V)/(m). 5. A 100W light bulb radiates energy uniformly in all directions. What is the intensity 2.0m away? The intensity (I) is the power (P) distributed over the surface area (A) of a sphere at a given distance (r). Step 1: Calculate the surface area of the sphere at r = 2.0 m. A = 4 r^2 = 4 (2.0 m)^2 = 4 (4.0 m^2) = 16 m^2 ≈ 50.265 m^2 Step 2: Calculate the intensity. I = (P)/(A) = 100 W50.265 m^2 = 1.99 Wm^2 The intensity 2.0m away is 1.99 (W)/(m^2). Drop the next question! 📸

Accepted Answer · 2026-06-13T07:11:12.895Z

Here are the solutions to your questions: 1. What is the average magnitude of the Poynting vector 5.00 miles from a radio transmitter broadcasting isotopically (equally in all directions) with an average power of 250 kW? The average magnitude of the Poynting vector is the intensity (I), which is power per unit area. For isotropic radiation, the area is that of a sphere. Step 1: Convert given values to SI units. r = 5.00 miles × 1609.34 mmile = 8046.7 m P = 250 kW = 250 × 10^3 W Step 2: Calculate the surface area of the sphere. A = 4 r^2 = 4 (8046.7 m)^2 = 8.138 × 10^8 m^2 Step 3: Calculate the average magnitude of the Poynting vector (intensity). I = (P)/(A) = 250 × 10^3 W8.138 × 10^8 m^2 = 3.07 × 10^-4 Wm^2 The average magnitude of the Poynting vector is 3.07 × 10^-4 (W)/(m^2). 2. A parallel plate capacitor has circular plates with radius of 7.0cm. If the electric field between the plates is changing at a rate of 2 × 10^12 V/m/s. Find the displacement current. The displacement current (I_d) is given by the formula I_d = _0 A (dE)/(dt). Step 1: Convert the radius to meters and calculate the area of the plates. r = 7.0 cm = 0.07 m A = r^2 = (0.07 m)^2 = 0.01539 m^2 Step 2: Substitute the values into the displacement current formula. (dE)/(dt) = 2 × 10^12 Vm · s _0 = 8.854 × 10^-12 Fm I_d = (8.854 × 10^-12 Fm) (0.01539 m^2) (2 × 10^12 Vm · s) Step 3: Calculate the displacement current. I_d = 0.272 A The displacement current is 0.272 A. 3. The electric field amplitude of a laser beam is 6.0 × 10^8 V/m. Calculate its average and maximum intensity. The maximum intensity (I_max) and average intensity (I_avg) are related to the electric field amplitude (E_0). Step 1: Calculate the maximum intensity. I_max = c _0 E_0^2 Where c = 3.00 × 10^8 m/s and _0 = 8.854 × 10^-12 F/m. I_max = (3.00 × 10^8 ms) (8.854 × 10^-12 Fm) (6.0 × 10^8 Vm)^2 I_max = (3.00 × 10^8) (8.854 × 10^-12) (3.6 × 10^17) Wm^2 I_max = 9.56 × 10^14 Wm^2 Step 2: Calculate the average intensity. For a sinusoidal electromagnetic wave, the average intensity is half of the maximum intensity. I_avg = (1)/(2) I_max I_avg = (1)/(2) (9.56 × 10^14 Wm^2) = 4.78 × 10^14 Wm^2 The maximum intensity is 9.56 × 10^14 (W)/(m^2) and the average intensity is 4.78 × 10^14 (W)/(m^2). 4. A microwave oven has an average intensity of 8.33 × 10^3 W/m^2. Find the peak electric field strength (E_0). The average intensity (I_avg) is related to the peak electric field strength (E_0) by the formula I_avg = (1)/(2) c _0 E_0^2. Step 1: Rearrange the formula to solve for E_0. E_0^2 = 2 I_avgc _0 E_0 = sqrt(2 I_avg)c _0 Step 2: Substitute the given values and constants. I_avg = 8.33 × 10^3 Wm^2 c = 3.00 × 10^8 ms _0 = 8.854 × 10^-12 Fm E_0 = sqrt(2 × (8.33 × 10^3 W)m^2)(3.00 × 10^8 ms) (8.854 × 10^-12 Fm) E_0 = sqrt((1.666 × 10^4)/(2.6562 × 10^-3)) Vm E_0 = sqrt(6.272 × 10^6) Vm Step 3: Calculate E_0. E_0 = 2504.4 Vm The peak electric field strength is 2504.4 (V)/(m). 5. A 100W light bulb radiates energy uniformly in all directions. What is the intensity 2.0m away? The intensity (I) is the power (P) distributed over the surface area (A) of a sphere at a given distance (r). Step 1: Calculate the surface area of the sphere at r = 2.0 m. A = 4 r^2 = 4 (2.0 m)^2 = 4 (4.0 m^2) = 16 m^2 ≈ 50.265 m^2 Step 2: Calculate the intensity. I = (P)/(A) = 100 W50.265 m^2 = 1.99 Wm^2 The intensity 2.0m away is 1.99 (W)/(m^2). Drop the next question! 📸

What is the average magnitude of the Poynting vector 5.00 miles from a radio transmitter broadcasting isotropically (equally in all directions) with an average power of 250 kW?

Need help with your own homework?

More Physics Questions

What is the average magnitude of the Poynting vector 5.00 miles from a radio transmitter broadcasting isotropically (equally in all directions) with an average power of 250 kW?

Need help with your own homework?

More Physics Questions