Electromagnetic theory miscellaneous
- A plane electromagnetic wave in free space is specified by the electric field
E = ax [20 cos (ωt – βz) + 5 cos (ωt + βz)] V/m.
The associated magnetic field is—
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E and H are related by η
i.e. E/ H = η,
the characteristic impedance of the medium which is 120 π for free space. Also, phase of H in the reflected component of the wave has to reverse so that travelling of the reflected wave is correctly given by the direction of E × H.
Of course, E × H are perpendicular to each other in the incident component or the forward travelling component of the wave. Keeping in view the above facts and the given expression for E.
H = ay (1/ 120 π) 20 cos (ωt – βz) – 5 cos (ωt + βz) A /m.Correct Option: C
E and H are related by η
i.e. E/ H = η,
the characteristic impedance of the medium which is 120 π for free space. Also, phase of H in the reflected component of the wave has to reverse so that travelling of the reflected wave is correctly given by the direction of E × H.
Of course, E × H are perpendicular to each other in the incident component or the forward travelling component of the wave. Keeping in view the above facts and the given expression for E.
H = ay (1/ 120 π) 20 cos (ωt – βz) – 5 cos (ωt + βz) A /m.
- The maximum usable frequency of an ionospheric layer at 60º incidance and with 8 MHz. The critical frequency is—
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Maximum usable frequency, muf = fc sec θ
where, given θ i = 60º
fc = 8 MHz
Now, muf = 8 × 106 × sec 60º
= 16 MHzCorrect Option: A
Maximum usable frequency, muf = fc sec θ
where, given θ i = 60º
fc = 8 MHz
Now, muf = 8 × 106 × sec 60º
= 16 MHz
- The potential difference between the forces A and B of a uniformly polarized infinite slab shown in the figure is—
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D = ∈0 E + P also D = ∈0∈r E
∴∈0∈r E = ∈0 E + P or E = P ∈0(∈r − 1)
∴Potential difference V = Ed = Pd ∈0(∈r − 1) Correct Option: A
D = ∈0 E + P also D = ∈0∈r E
∴∈0∈r E = ∈0 E + P or E = P ∈0(∈r − 1)
∴Potential difference V = Ed = Pd ∈0(∈r − 1)
- Three concentric conducting spherical surfaces of radii R1, R2 and R3 (R1 < R2 < R3) carry charges of –1, –2 and 4 coulombs respectively. The charges on the inner and outer surfaces of the outermost sphere will be respectively (in coulombs)—
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The total charge on spheres 1 and 2 taken together is – 1 + (–2) = – 3C. This charge will induced + 3C on inner surface and – 3C on the outer surface of sphere 3. Thus the total charge on the inner surface of sphere 3 is 3C and on the outer surface – 3 + 4 = 1 C.
Correct Option: B
The total charge on spheres 1 and 2 taken together is – 1 + (–2) = – 3C. This charge will induced + 3C on inner surface and – 3C on the outer surface of sphere 3. Thus the total charge on the inner surface of sphere 3 is 3C and on the outer surface – 3 + 4 = 1 C.
- In the figure shown below the force acting on the conductor PQ is in the direction of—
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The direction of flux density produced by all the elements of two veritcal sections of the current shown, at points on PQ is to the plane of the paper and outward. This can be checked with IB: Idl × IR. Then the force on an element of PQ is found from dF = Idl × B. Keeping in view the direction of dl which is the as that of l (at points on PQ, we find direction of force on PQ is along – n.)
Correct Option: C
The direction of flux density produced by all the elements of two veritcal sections of the current shown, at points on PQ is to the plane of the paper and outward. This can be checked with IB: Idl × IR. Then the force on an element of PQ is found from dF = Idl × B. Keeping in view the direction of dl which is the as that of l (at points on PQ, we find direction of force on PQ is along – n.)
