Answer:
The associated magnetic field, at a distance of 150E6 km from the beacon is
B = (2.177 × 10⁻¹⁴) T
Step-by-step explanation:
The intensity of waves, with power P, at a point r, from source is given as
I = (P/A)
The associated electric field for the electromagnetic wave at that point is given as
E = √(2Iμ₀c)
And the associated Magnetic field for the electromagnetic wave is related to the associated electric field through
B = (E/c)
Note that
P = power of the wave = 16.0 GW
= (16 × 10⁹) W
A = Surface area of an imaginary sphere at the point where the intensity of the wave is required = 4πr²
r = distance from the source of the wave to the point where the intensity is required
r = (150 × 10⁶) km = (150 × 10⁹) m
A = 4πr² = 4π × (150 × 10⁹)²
A = (2.827 × 10²³) m²
I = (P/A)
I = (16 × 10⁹) ÷ (2.827 × 10²³)
I = (5.659 × 10⁻¹⁴) W/m²
E = √(2Iμ₀c)
I = intensity of the wave at that point = (5.659 × 10⁻¹⁴) W/m²
μ₀ = magnetic constant or magnetic permeability of vacuum = (4π × 10⁻⁷) H/m
c = speed of light = (3.0 × 10⁸) m/s
E = √(2 × 5.659×10⁻¹⁴ × 4π×10⁻⁷ × 3.0×10⁸)
E = √(4.2668 × 10⁻¹¹)
E = 0.0000065321 N/C
= (6.5321 × 10⁻⁶) N/C
Associated Magnetic field for the electromagnetic wave is related to the associated electric field through
B = (E/c)
E = Electric field = (6.5321 × 10⁻⁶) N/C
c = speed of light = (3.0 × 10⁸) m/s
B = (6.5321 × 10⁻⁶) ÷ (3.0 × 10⁸)
B = (2.177 × 10⁻¹⁴) T
Hope this Helps!!!