हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

An electromagnetic wave travelling along z-axis is given as: E = E0 cos (kz – ωt.). Choose the correct options from the following; The associated magnetic field is given as ωB=1ck^×E=1ω(k^×E).

Advertisements
Advertisements

प्रश्न

An electromagnetic wave travelling along z-axis is given as: E = E0 cos (kz – ωt.). Choose the correct options from the following;

  1. The associated magnetic field is given as `B = 1/c hatk xx E = 1/ω (hatk xx E)`.
  2. The electromagnetic field can be written in terms of the associated magnetic field as `E = c(B xx hatk)`.
  3. `hatk.E = 0, hatk.B` = 0.
  4. `hatk xx E = 0, hatk xx B` = 0.

विकल्प

  • a, b and c

  • a, c and d

  • b, c and d

  • b and d

MCQ
Advertisements

उत्तर

a, b and c

Explanation:

a. The direction of propagation of an electromagnetic wave is always along the direction of vector product `vecE xx vecB`. Refer to figure.


`vecB = Bhatj = B(hatk xx hati) = E/c (hatk xx hati)`

= `1/c [k xx Ehati] = 1/c [hatk xx vecE]`  .....`("as"  E/B = c)`

b. `vecE = Ehati = cB(hatj xx hatk) = c(Bhatj xx hatk) = c(vecB xx hatk)`

c. `hatk * vecE = hatk * (Ehati)` = 0, `veck * vecB = hatk * (Bhatj)` = 0

d. `hatk xx vecE = hatk xx (Ehati) = E(hatk xx hati) = Ehatj` and `hatk xx vecB = hatk xx (Bhatj) = B(hatk xx hatj) = - Bhati`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Electromagnetic Waves - MCQ I [पृष्ठ ४९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 12
अध्याय 8 Electromagnetic Waves
MCQ I | Q 8.09 | पृष्ठ ४९

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Figure shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A.

  1. Calculate the capacitance and the rate of charge of the potential difference between the plates.
  2. Obtain the displacement current across the plates.
  3. Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.


A parallel plate capacitor (Figure) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s−1.

  1. What is the rms value of the conduction current?
  2. Is the conduction current equal to the displacement current?
  3. Determine the amplitude of B at a point 3.0 cm from the axis between the plates.


A capacitor has been charged by a dc source. What are the magnitude of conduction and displacement current, when it is fully charged?


When an ideal capacitor is charged by a dc battery, no current flows. However, when an ac source is used, the current flows continuously. How does one explain this, based on the concept of displacement current?


Without the concept of displacement current it is not possible to correctly apply Ampere’s law on a path parallel to the plates of parallel plate capacitor having capacitance C in ______.


Displacement current is given by ______.


If the total energy of a particle executing SHM is E, then the potential energy V and the kinetic energy K of the particle in terms of E when its displacement is half of its amplitude will be ______.


A cylinder of radius R, length Land density p floats upright in a fluid of density p0. The cylinder is given a gentle downward push as a result of which there is a vertical displacement of size x; it is then released; the time period of resulting (undampe (D) oscillations is ______.


Displacement current goes through the gap between the plantes of a capacitors. When the charge of the capacitor:-


Which of the following is the unit of displacement current?


A capacitor of capacitance ‘C’, is connected across an ac source of voltage V, given by V = V0 sinωt The displacement current between the plates of the capacitor would then be given by ______.


A capacitor of capacitance ‘C’, is connected across an ac source of voltage V, given by V = V0 sinωt The displacement current between the plates of the capacitor would then be given by ______


The charge on a parallel plate capacitor varies as q = q0 cos 2πνt. The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through the capacitor?


Show that the magnetic field B at a point in between the plates of a parallel-plate capacitor during charging is `(ε_0mu_r)/2 (dE)/(dt)` (symbols having usual meaning).


Show that average value of radiant flux density ‘S’ over a single period ‘T’ is given by S = `1/(2cmu_0) E_0^2`.


Sea water at frequency ν = 4 × 108 Hz has permittivity ε ≈ 80 εo, permeability µ ≈ µo and resistivity ρ = 0.25 Ω–m. Imagine a parallel plate capacitor immersed in seawater and driven by an alternating voltage source V(t) = Vo sin (2πνt). What fraction of the conduction current density is the displacement current density?


A parallel plate capacitor is charged to 100 × 10-6 C. Due to radiations, falling from a radiating source, the plate loses charge at the rate of 2 × 10-7 Cs-1. The magnitude of displacement current is ______.


Draw a neat labelled diagram of displacement current in the space between the plates of the capacitor.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×