University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

95

Pages:

5
Author:

Jabin F.

# Radiation:-~ EM Waves are radiation. # Sources of radiation:- If any charge particle is moving with constant velocity would it produce E.M radiation? (Ans. No) #Any accelerated or deaccelerated point charge can produce E.M. radiation. #Any charge particle performing circular motion (Yes) Radiation (accelerated motion) Classical E.D- Classical physics was not able to explain Atomic stability. we will explain #Three Sources of EM Radiation Oscillating electric dipole Oscillating magnetic dipole Accelerated point charge Page 1: (P(t) = P_0 \log(\omega t)) Ideally find (E \rightarrow B \rightarrow I \rightarrow \langle P \rangle) (\langle P \rangle = \frac{\mu_0 \langle (\vec{p})^2 \rangle}{6 \pi c}) (\dot{p} = P_0 \omega \cos(\omega t)) (\langle (\dot{p})^2 \rangle = \frac{P_0^2 \omega^2}{2}) (\langle P \rangle = \frac{\mu_0 P_0^2 \omega^4}{12 \pi c}) (P = \frac{\mu_0 P_0^2 \omega^4}{6 \pi c} \cos^2(\omega t)) [Instantaneous power] Page 2: (x(t) = A \sin(\omega t)) (p(t) = 2q x(t)) (p(t) = 2A q \sin(\omega t)) (x(t) = A \sin(\omega t)) (p(t) = 2q x(t)) (p(t) = 2A q \sin(\omega t)) (\dot{p}(t) = -2A q \omega^2 \sin(\omega t)) (\langle P \rangle = \frac{\mu_0 \langle (\dot{p})^2 \rangle}{6 \pi c}) (\langle P \rangle = \frac{\mu_0 (2A)^2 q^2 \omega^4}{12 \pi c}) Power pattern along x-axis (p(t) = 2q x(t)) (\dot{p}(t) = 2q \dot{x}(t) = 2q v) (p(t) = 2q x(t) = 2q v = q a) (a = acceleration) (P = \frac{\mu_0 q^2 a^2}{6 \pi c}) (Larmor formula)