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As a parallel-plate capacitor with circular plates \(20 \mathrm{~cm}\) in diameter is being charged, the current density of the displacement current in the region between the plates is uniform and has a magnitude of \(20 \mathrm{~A} / \mathrm{m}^{2}\). (a) Calculate the magnitude \(B\) of the magnetic field at a distance \(r=50 \mathrm{~mm}\) from the axis of symmetry of this region. (b) Calculate \(d E / d t\) in this region.

Short Answer

Expert verified
(a) \(B = 2 \times 10^{-7} \text{ T}\); (b) \(dE/dt = 2.26 \times 10^{11} \text{ V/m/s}\).

Step by step solution

01

Determine the Area of the Circular Plates

Given the diameter of the plates as \(20\) cm, the radius \(R\) can be calculated as \(R = \frac{20}{2} = 10\) cm = 0.1 m. The area \(A\) of the circular plates can be found using the formula for the area of a circle \(A = \pi R^2\), which gives \[A = \pi (0.1)^2 = 0.01\pi \ m^2.\]
02

Calculate the Displacement Current

The displacement current \(I_d\) is given by the product of the current density \(J_d\) and the area \(A\). Given \(J_d = 20\) A/m², \[I_d = J_d \cdot A = 20 \cdot 0.01\pi = 0.2\pi \text{ A}.\]
03

Calculate the Magnetic Field Using Ampère’s Law

At a distance \(r = 50\) mm = 0.05 m from the axis, apply Ampère's Law for the magnetic field of a displacement current, which states that \(B \cdot 2\pi r = \mu_0 I_d \cdot \frac{r^2}{R^2}\). Thus, \[B = \frac{\mu_0 I_d r}{2\pi R^2}.\] Substitute \(\mu_0 = 4\pi \times 10^{-7}\) Tm/A, \(I_d = 0.2\pi\) A, \(r = 0.05\) m, and \(R = 0.1\) m:\[B = \frac{4\pi \times 10^{-7} \cdot 0.2\pi \cdot 0.05}{2\pi \times (0.1)^2} = 2\times 10^{-7} \text{ T}.\]
04

Calculate the Rate of Change of Electric Field (dE/dt)

The displacement current is related to the rate of change of the electric field \(dE/dt\) by the equation \(I_d = \varepsilon_0 \frac{dE}{dt} A\), where \(\varepsilon_0 = 8.85 \times 10^{-12}\) F/m. Rearrange to find \[\frac{dE}{dt} = \frac{I_d}{\varepsilon_0 A}.\] Substitute \(I_d = 0.2\pi\) A, \(\varepsilon_0 = 8.85 \times 10^{-12}\) F/m, and \(A = 0.01\pi\) m²:\[\frac{dE}{dt} = \frac{0.2\pi}{8.85 \times 10^{-12} \cdot 0.01\pi} = 2.26 \times 10^{11} \text{ V/m/s}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ampère's Law
Let's dive into the essence of Ampère's Law, a core principle in magnetism. It helps us understand the relationship between electric currents and magnetic fields. Ampère's Law states that the integral of the magnetic field \( B \) around a closed loop is equal to the permeability of free space \( \mu_0 \) times the electric current encased by the loop. This can be expressed as: \[ \oint B \cdot dl = \mu_0 I \]
In simpler terms, it shows how currents create magnetic fields. For the case of our problem, we're interested in how a changing electric field (between capacitor plates) can also contribute to a magnetic field. This is where the concept of displacement current comes into play, extending Ampère's Law to accommodate time-varying electric fields. Remember, when a current passes through a wire loop, it generates a magnetic field, and similarly, changing electric fields can mimic this effect even in the absence of a traditional current.
Displacement Current
The idea of displacement current sneaks into our understanding of magnetism via Ampère's extended Law. It addresses a gap left by simply considering conduction currents (ones that flow in a wire) by introducing a way to account for the effects of changing electric fields.
Displacement current is not a real current in the sense of moving charges, but rather a term used to describe the changing electric field within a capacitor that gives rise to a magnetic field. Mathematically, it is represented as: \[ I_d = \varepsilon_0 \frac{dE}{dt} A \]
Here, \( I_d \) represents the displacement current, \( \varepsilon_0 \) is the permittivity of free space, \( \frac{dE}{dt} \) is the rate of change of the electric field, and \( A \) is the area through which the field is changing. This concept allows us to use Ampère's Law even in dynamic situations where the electric field changes over time, such as in a charging capacitor.
Electric Field
At its core, the electric field is a representation of how charges can exert force on one another. It is defined as a vector field around charged particles and surfaces, and it describes the amount of force a positive test charge would experience at any given point in space.
Mathematically, the electric field \( E \) is given by the equation: \[ E = \frac{F}{q} \]
where \( F \) is the force experienced by a small positive test charge \( q \).
In our exercise context with the parallel-plate capacitor, the electric field is crucial as it changes while the capacitor charges. Understanding the rate of change of this field \( \frac{dE}{dt} \) is essential for calculating the displacement current, which subsequently allows us to determine the magnetic field around the capacitor plates. Thus, electric fields not only signify direct interactions between charges but also affect magnetic interactions when they vary over time.

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Most popular questions from this chapter

A magnetic compass has its needle, of mass \(0.050 \mathrm{~kg}\) and length \(4.0 \mathrm{~cm}\), aligned with the horizontal component of Earth's magnetic field at a place where that component has the value \(B_{h}=\) \(16 \mu \mathrm{T}\). After the compass is given a momentary gentle shake, the needle oscillates with angular frequency \(\omega=45 \mathrm{rad} / \mathrm{s}\). Assuming that the needle is a uniform thin rod mounted at its center, find the magnitude of its magnetic dipole moment.

In New Hampshire the average horizontal component of Earth's magnetic field in 1912 was \(16 \mu \mathrm{T}\), and the average inclination or "dip" was \(73^{\circ} .\) What was the corresponding magnitude of Earth's magnetic field?

The magnetic flux through each of five faces of a die (singular of "dice") is given by \(\Phi_{B}=\pm N \mathrm{~Wb}\), where \(N(=1\) to 5\()\) is the number of spots on the face. The flux is positive (outward) for \(N\) even and negative (inward) for \(N\) odd. What is the flux through the sixth face of the die?

Earth has a magnetic dipole moment of \(8.0 \times 10^{22} \mathrm{~J} / \mathrm{T}\). (a) What current would have to be produced in a single turn of wire extending around Earth at its geomagnetic equator if we wished to set up such a dipole? Could such an arrangement be used to cancel out Earth's magnetism (b) at points in space well above Earth's surface or (c) on Earth's surface?

A \(0.50 \mathrm{~T}\) magnetic field is applied to a paramagnetic gas whose atoms have an intrinsic magnetic dipole moment of \(1.0 \times\) \(10^{-23} \mathrm{~J} / \mathrm{T}\). At what temperature will the mean kinetic energy of translation of the atoms equal the energy required to reverse such a dipole end for end in this magnetic field?

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