The induced magnetic field at radial distance 6 . 0 mm from the central axis of a circular parallel-plate capacitor is 1 . 2 × 10 - 7 T . The plates have radius 4 . 0 mm . At what rate d E d t is the electric field between the plates changing?

Given that Radial distance r = 6 . 00 mm Induced magnetic field B = 1 . 2 × 10 - 7 T Radius of plate of capacitor R = 4 . 0 mm Changing electric flux induces a magnetic field. Consider a circle of radius r Between the plates, with its centre on the axis of the capacitor. Since there is no current between the capacitor plates, the Ampere Maxwell’s law reduces to ∮ B → · d l → = μ 0 i d ; Here displacement current i d = ε 0 d Φ E d t ⇒ ∮ B → · d l → = μ 0 ε 0 d Φ E d t . . . 1 where B → is the magnetic field at points on the circle, and Φ E is the electric flux through the circle. Since, the B → field on the circle is in the tangential direction, and Φ E = A E = π R 2 E , where R is the radius of the capacitor. From equation 1 We have 2 π r B = μ 0 ε 0 π R 2 d E d t ⇒ B = μ 0 ε 0 R 2 2 r d E d t ( r ≥ R ) . Solving for d E d t , we obtain d E d t = 2 B r μ 0 ε 0 R 2 ⇒ d E d t = 2 1 . 2 × 10 - 7 T 6 . 0 × 10 - 3 m 4 π × 10 - 7 T · m - 1 A - 1 8 . 85 × 10 - 12 C 2 N - 1 · m 2 4 . 0 × 10 - 3 m 2 ⇒ d E d t = 8 . 1 × 10 12 V m - 1 s - 1 .

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The electron drift speed is small, and the charge of the electron is also small, but still, we obtain large current in a conductor. This is due to

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Electromagnetic Waves>Exercise>Q 14

If the electric current through an electric bulb is

3.2 A

, the number of electrons flowing through it in one second is

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Electromagnetic Waves>Exercise>Q 15

The charge on a parallel plate capacitor is varying as

q = q_{0}

\sin (2 π n t)

. The plates are very large and close together. Neglecting the edge effects, the displacement current through the capacitor is:

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Electromagnetic Waves>Exercise>Q 16

The figure shows a capacitor made of two circular plates each of radius $12 cm$ , and separated by $5 cm$ . The capacitor is being charged by an external source. The charging current is constant and is equal to $0.15 A$ , obtain the displacement current across the plates?

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Electromagnetic Waves>Exercise>Q 17

Sea water at frequency

ν = 4 \times 10^{8} Hz

has permittivity

ε \approx 80 ε_{o}

, permeability

µ \approx µ_{o}

and resistivity

ρ = 0.25 Ω - m

. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source

V (t) = V_{o} \sin (2 π ν t)

. What fraction of the conduction current density is the displacement current density?

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Electromagnetic Waves>Exercise>Q 18

Explain the need of displacement current.

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Electromagnetic Waves>Exercise>Q 19

The induced magnetic field at radial distance

6.0 mm

from the central axis of a circular parallel-plate capacitor is

1.2 \times 10^{- 7} T

. The plates have radius

4.0 mm

. At what rate

\frac{d E}{d t}

is the electric field between the plates changing?

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Electromagnetic Waves>Exercise>Q 20

Non-uniform displacement current. Figure shows a circular region of radius $R = 3.00 cm$ in which a displacement current $i_{d}$ is directed out of the figure. The magnitude of the displacement current is $i_{d} = (3.00 A) (r / R),$ where $r$ is the radial distance $(r \leq R)$ from the center. What is the magnitude of the magnetic field due to $i_{d}$ at radial distances
(a) $2.00 cm$ and
(b) $6.00 cm$ ?

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Electromagnetic Waves>Exercise>Q 21

One requires

11 eV

of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in

The electron drift speed is small, and the charge of the electron is also small, but still, we obtain large current in a conductor. This is due to

Important Questions on Electromagnetic Waves

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