A radio-activated garage-door opener responds to signals with average intensity

Question

A radio-activated garage-door opener responds to signals with average intensity as weak as 20 µW/m2. If the transmitter unit produces a 240-mW signal, broadcast in all directions, what is the maximum distance at which the transmitter will activate the door opener? What is the minimum value for the peak electric field to which the unit responds?

Explanation / Answer

Given that I = 20 µW/m2 P = 240-mW We know that I = P / 4 * pi * r^2 ==> r = [ P / 4 * pi * I] ^1/2 = [ 240 x 10^-3 / 4 * 3.14 * 20 x 10^-6] ^1/2 = 30.90 m Electric field E = [ I * c * uo]^1/2 = [ 20 x 10^-6 * 3 x 10^8 * 4 * 3.14 x 10^-7 ] ^1/2 = 8.681 x 10^-2 V/m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.