Advertisements
Advertisements
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.
- What is the rms value of the conduction current?
- Is the conduction current equal to the displacement current?
- Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
Advertisements
उत्तर
Radius of each circular plate, R = 6.0 cm = 0.06 m
Capacitance of a parallel plate capacitor, C = 100 pF = 100 × 10−12 F
Supply voltage, V = 230 V
Angular frequency, ω = 300 rad s−1
(a) The rms value of conduction current, I = `"V"/"X"_"C"`
Where,
XC = Capacitive reactance
= `1/(ω"C")`
∴ I = V × ωC
= 230 × 300 × 100 × 10−12
= 6.9 × 10−6 A
= 6.9 μA
Hence, the rms value of the conduction current is 6.9 μA.
(b) Yes, conduction current is equal to displacement current.
(c) Magnetic field is given as:
B = `(μ_0"r")/(2pi"R"^2)"I"_0`
Where,
μ0 = Free space permeability = 4π × 10−7 N A−2
I0 = Maximum value of current = `sqrt2"I"`
r = Distance between the plates from the axis = 3.0 cm = 0.03 m
∴ B = `(4pi xx 10^-7 xx 0.03 xx sqrt2 xx 6.9 xx 10^-6)/(2pi xx (0.06)^2)`
= 1.63 × 10−11 T
Hence, the magnetic field at that point is 1.63 × 10−11 T.
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.
- Calculate the capacitance and the rate of charge of the potential difference between the plates.
- Obtain the displacement current across the plates.
- Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.
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?
A parallel-plate capacitor of plate-area A and plate separation d is joined to a battery of emf ε and internal resistance R at t = 0. Consider a plane surface of area A/2, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time.
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 ______.
The displacement of a particle from its mean position is given by x = 4 sin (10πt + 1.5π) cos (10πt + 1.5π). The motion of the particle is
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 ______
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 `B = 1/c hatk xx E = 1/ω (hatk xx E)`.
- The electromagnetic field can be written in terms of the associated magnetic field as `E = c(B xx hatk)`.
- `hatk.E = 0, hatk.B` = 0.
- `hatk xx E = 0, hatk xx B` = 0.
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).
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 long straight cable of length `l` is placed symmetrically along z-axis and has radius a(<< l). The cable consists of a thin wire and a co-axial conducting tube. An alternating current I(t) = I0 sin (2πνt) flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance s from the wire inside the cable is E(s,t) = µ0I0ν cos (2πνt) In `(s/a)hatk`.
- Calculate the displacement current density inside the cable.
- Integrate the displacement current density across the cross-section of the cable to find the total displacement current Id.
- Compare the conduction current I0 with the displacement current `I_0^d`.
Draw a neat labelled diagram of displacement current in the space between the plates of the capacitor.
A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its palates is increasing at a constant rate with time. The magnetic field arising due to displacement current is ______.
A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates, there is no conduction current:
