Electromagnetism - EM Radiation

A charged particle, A, moving at a speed much less than c, decelerates uniformly. A second particle, B, has one-half the mass, twice the charge, three times the velocity, and four times the acceleration of particle A. According to classical electrodynamics, the ratio P_B/P_A of the powers radiated is

A. 16
B. 32
C. 48
D. 64
E. 72

(GR9677 #70)

Solution:

Larmor formula:
Power radiated by an accelerating non-relativistic point charge particle,
P = q²a² ⁄ 6πɛ₀c³

P does not depend on m and v.

P_B/P_A =  (2q)² (4a)² ⁄q²a²  =  64

Answer: D

Electromagnetism - EM Radiation

An electron oscillates back and forth along the + and − x-axes, consequently emitting electromagnetic radiation. Which of the following statements concerning the radiation is NOT true?

1. The total rate of radiation of energy into all directions is proportional to the square of the electron's acceleration.
2. The total rate of radiation of energy into all directions is proportional to the square of the electron's charge.
3. Far from the electron, the rate at which radiated energy crosses a perpendicular unit area decreases as the inverse square of the distance from the electron
4. Far from the electron, the rate at which radiated energy crosses a perpendicular unit area is a maximum when the unit area is located on the + or − x-axes
5. Far from the electron, the radiated energy is carried equally by the transverse electric and the transverse magnetic fields.

(GR9277 #84)

Solution:

Power radiated, P = ∮ S ∙ dA
S is poynting vector, where S = (1/μ₀) E × B  = (1/cμ₀) |E|²
Electric field, E = − (kq/r)(a_⊥/c²)

(A) TRUE
P ∝ a²  since |E|² ∝  a_⊥²

(B) TRUE,
P ∝ q² since |E|² ∝ q²

(C) TRUE
P ∝ 1/r² since |E|² ∝ 1/r²

(D) FALSE
The electron oscillates back and forth along the + and − x-axes. Thus, the direction of acceleration is along the + and − x axes. P_max is when the area of electric flux perpendicular, ⊥ to the acceleration of the particle (not located on the + or − x-axes).
P_max ∝ a_⊥² since |E|² ∝  a_⊥²

(E) TRUE, 
Power radiated, P = ∮ S ∙ dA
S is poynting vector, where S = (1/μ₀) E × B
P is carried equally by E and B field.

Answer: D

Electromagnetism - EM Radiation

A charge particle oscillates harmonically along the x-axis as shown above. The radiation from the particle is detected at a distant point P which lies in the xy-plane. The electric field at P is in the

A. ±z direction and has a maximum amplitude at θ = 90^o
B. ±z direction and has a minimum amplitude at θ = 90^o
C. xy-plane and has a maximum amplitude at θ = 90^o
D. xy-plane and has a minimum amplitude at θ = 90^o
E. xy-plane and has a maximum amplitude at θ = 45^o

(GR8677 #53)

Solution:

Since point P lies in the xy-plane, the electric field at P also lies in the xy-plane, and its maximum amplitude is at θ = 90^o (perpendicular to the acceleration of particle).

Answer: C