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Show that the magnetic field B at a point in between the plates of a parallel-plate capacitor during charging is εε0μr2dEdt (symbols having usual meaning).

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प्रश्न

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).

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उत्तर

Let us assume Id be the displacement current in the region between two plates of parallel plate capacitor, in the figure.


The magnetic field at a point between two plates of capacitor  at a perpendicular distance r from the axis of plates is given by

B = `(mu_0 2I_d)/(4pir) = mu_0/(2pir) I_d = mu_0/(2pir) xx ε_r (dphi_E)/(dt)`  ......`[because I_d = (E_0dphi_E)/(dt)]`

⇒ B = `(mu_0ε_r)/(2pir) d/(dt) (Epir^2) = (mu_0ε_r)/(2pir) pir^2 (dF)/(dt)`

⇒ B = `(mu_0ε_r)/2 (dE)/(dt)`  .....`[because phi_E = Epir^2]`

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पाठ 8: Electromagnetic Waves - MCQ I [पृष्ठ ५१]

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एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 12
पाठ 8 Electromagnetic Waves
MCQ I | Q 8.21 | पृष्ठ ५१

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

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.


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  1. Calculate the displacement current density inside the cable.
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