Problem 16.7
หัวข้อ: คลื่นแม่เหล็กไฟฟ้าแบบระนาบ (Plane Electromagnetic Waves)
\(\therefore \vec{B}(x,t) = (2.50 \mu T)\cos ((0.838 rad/m)x - (2.51\times 10^{9} rad/s)t) \;\blacksquare\)
ดูวิธีทำ
ĉîĵ
กำหนดให้ \(= , E_{\max}= 750 N/C, f = 40.0 MHz, \vec{E}(0,0) = E_{\max}\)
a)จาก\(T=\frac{1}{f}=25.0ns\;\blacksquare\)
b) จาก \(c = fλ\)
\(\lambda =\frac{c}{f}=7.50m\;\blacksquare\)
c) จาก \(k = \frac{2\pi}{\lambda}= 0.838 rad/m\)
จาก \(ω = 2πf = 2.51\times 10^{9} rad/s\)
จาก \(E(x,t) = E_{\max}\cos (kx - \omega t + φ)\)
ĵ
\(\vec{E}(x,t) = E_{\max}\cos (kx - \omega t + φ)\)
ĵĵ
\(E_{\max}= E_{\max}\cos (k(0) - \omega (0) + φ)\)
\(1 = \cos (φ)\)
\(φ = 0 rad\)
ĵ
\(\therefore \vec{E}(x,t) = (750 N/C)\cos ((0.838 rad/m)x - (2.51\times 10^{9} rad/s)t) \;\blacksquare\)
d) จาก \(c = \frac{E_{\max}}{B_{\max}}\)
\(B_{\max}=\frac{E_{\max}}{c}= 2.50 \mu T\)
ĉÊ B̂
จาก \(= \times \)
îĵ B̂
\(= \times\)
B̂k̂
\(\therefore =\)
จาก \(B(x,t) = B_{\max}\cos (kx - \omega t + φ)\)
k̂
\(\therefore \vec{B}(x,t) = (2.50 \mu T)\cos ((0.838 rad/m)x - (2.51\times 10^{9} rad/s)t) \;\blacksquare\)