physics : class12 Oscillating Dipole Radiation : Radio Waves : Electromagnetic Waves MCQ Questions and Answers
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Oscillating Dipole Radiation
MCQ questions and answers.
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MCQ Questions
1. An electric dipole has moment p(t) = p0 cos(ωt) z-hat. In the radiation zone, at which polar angle θ from the z-axis is the time-averaged radiated intensity maximum?
A.
θ = 0
B.
θ = π/4
C.
θ = π/2
D.
The intensity is independent of θ
Correct Answer:
θ = π/2
Explanation:
The radiation-zone electric field of an oscillating electric dipole is proportional to sin θ, so the time-averaged intensity is proportional to sin²θ. It is therefore maximum in the plane perpendicular to the dipole axis, at θ = π/2, and zero along the dipole axis.
2. For a harmonically oscillating electric dipole with fixed amplitude p0, the total time-averaged radiated power varies with angular frequency ω as
A.
ω
B.
ω²
C.
ω³
D.
ω⁴
Correct Answer:
ω⁴
Explanation:
The average power radiated by an electric dipole is proportional to the square of the second time derivative of its dipole moment. For p(t) = p0 cos(ωt), the second derivative has amplitude ω²p0, so the average radiated power is proportional to ω⁴p0².
3. Two identical short oscillating electric dipoles radiate at the same frequency and have the same dipole-moment amplitude. Dipole 2 has twice the length of dipole 1 and carries half the current amplitude. What is the ratio of their total average radiated powers, P2/P1?
A.
1/4
B.
1/2
C.
1
D.
4
Correct Answer:
1
Explanation:
For a short dipole, the dipole moment amplitude is proportional to current amplitude times dipole length divided by angular frequency. The product I0l is unchanged because the length doubles while the current halves. Since radiated power is proportional to (I0l)² at fixed frequency, the two powers are equal.
4. A sinusoidally oscillating electric dipole radiates an electric field in the far zone. At a fixed observation point, which phase relation holds between the radiation electric field and the dipole moment p(t) = p0 cos(ωt) z-hat, apart from the propagation delay?
A.
The radiation field is in phase with p(t)
B.
The radiation field is 90° ahead of p(t)
C.
The radiation field is 180° out of phase with p(t)
D.
The radiation field is unrelated in phase to p(t)
Correct Answer:
The radiation field is 180° out of phase with p(t)
Explanation:
The far-zone radiation electric field is proportional to the second time derivative of the dipole moment. For harmonic motion, d²p/dt² = -ω²p, producing a phase reversal of 180° relative to p(t), in addition to the retarded propagation phase.
5. For an oscillating electric dipole, the radiation-zone magnetic field at a point is related to the electric field by which statement?
A.
The magnetic field is parallel to the electric field and has equal magnitude
B.
The magnetic field is perpendicular to both the electric field and the propagation direction, with B = E/c
C.
The magnetic field is perpendicular to the electric field, with B = cE
D.
The magnetic field vanishes because the source is electric
Correct Answer:
The magnetic field is perpendicular to both the electric field and the propagation direction, with B = E/c
Explanation:
In the radiation zone, the dipole field is locally a transverse electromagnetic wave. The electric field, magnetic field, and radial propagation direction are mutually perpendicular, and their magnitudes satisfy E = cB.
6. At a fixed distance in the far field of an oscillating electric dipole, the time-averaged intensity at θ = 30° is what fraction of the intensity at θ = 90°?
A.
1/4
B.
1/2
C.
√3/2
D.
3/4
Correct Answer:
1/4
Explanation:
The dipole radiation intensity varies as sin²θ. Therefore, the ratio is I(30°)/I(90°) = sin²30°/sin²90° = (1/2)² = 1/4.
7. An oscillating electric dipole radiates power P at frequency f. If its dipole-moment amplitude is doubled and its frequency is reduced to f/2, what is the new total average radiated power?
A.
P/16
B.
P/4
C.
P
D.
4P
Correct Answer:
P/4
Explanation:
The dipole radiation power is proportional to p0²ω⁴. Doubling p0 contributes a factor of 4, while replacing ω by ω/2 contributes a factor of 1/16. The combined factor is 4/16 = 1/4, so the new power is P/4.
8. A short electric dipole is oriented along the x-axis and oscillates harmonically. In the far field, the radiation observed along the z-axis is
A.
zero because the observation direction is perpendicular to the dipole
B.
maximum and linearly polarized along the x-axis
C.
maximum and circularly polarized
D.
independent of the dipole orientation
Correct Answer:
maximum and linearly polarized along the x-axis
Explanation:
Dipole radiation vanishes along the dipole axis and is maximum in directions perpendicular to it. The z-axis is perpendicular to the x-oriented dipole, so the radiation is maximum there. The radiated electric field lies along the component of the dipole transverse to the propagation direction, hence it is linearly polarized along x.
9. A localized oscillating electric dipole produces a radiation-zone field whose amplitude at distance r is E0. At distance 3r, in the same direction and at the same time dependence, what are the field amplitude and time-averaged intensity?
A.
E0/3 and I0/3
B.
E0/3 and I0/9
C.
E0/9 and I0/3
D.
E0/9 and I0/81
Correct Answer:
E0/3 and I0/9
Explanation:
The radiation field of an oscillating dipole decreases as 1/r in the far zone. Thus, at 3r its amplitude is E0/3. Since intensity is proportional to the square of the field amplitude, it becomes I0/9.
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