/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 A Charged Dielectric Disk. A thi... [FREE SOLUTION] | 91Ó°ÊÓ

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A Charged Dielectric Disk. A thin disk of dielectric material with radius \(a\) has a total charge \(+Q\) distributed uniformly over its surface. It rotates \(n\) times per second about an axis perpendicular to the surface of the disk and passing through its center. Find the magnetic field at the center of the disk. (Hint: Divide the disk into concentric rings of infinitesimal width.)

Short Answer

Expert verified
The magnetic field at the center of the disk is \(B = \frac{\mu_0 n Q}{a}\).

Step by step solution

01

Understand the Problem

The problem involves calculating the magnetic field at the center of a rotating charged dielectric disk. The disk has a charge +Q uniformly distributed over its surface and rotates at a frequency of n revolutions per second. The approach involves considering the disk as composed of concentric rings to simplify the integration.
02

Divide the Disk into Concentric Rings

Consider an infinitesimally small ring of radius r and thickness dr on the disk. The surface charge density \( \sigma \) is given by \( \sigma = \frac{Q}{\pi a^2} \). The charge on this infinitesimal ring, \( dQ \), is \( \sigma \, \times \, \text{area of the ring} = \sigma (2\pi r \, dr) \).
03

Calculate the Current Due to the Rotating Ring

As the disk rotates, the charges on the ring produce a current. The linear velocity at radius r is \( v = 2\pi r n \). The current \( dI \) due to the rotating charge is \( dI = \frac{dQ}{T} = \frac{dQ}{1/n} = n \, dQ \). Substituting for \( dQ \), we get \( dI = n \, \sigma \, (2\pi r \, dr) \).
04

Use the Biot-Savart Law for Magnetic Field

The magnetic field at the center due to an infinitesimal current element is given by the Biot-Savart Law. For a circular current loop, it gives a field \( dB = \frac{\mu_0 \, dI}{2r} \). Substituting \( dI \), \( dB = \frac{\mu_0 \, n \, \sigma \, (2\pi r \, dr)}{2r} = \mu_0 \, \pi \, n \, \sigma \, dr \).
05

Integrate Over the Entire Disk

Integrate the expression for \( dB \) from \( r = 0 \) to \( r = a \) to find the total magnetic field \( B \) at the center:\[ B = \int_0^a \mu_0 \pi n \sigma \, dr = \mu_0 \pi n \sigma \int_0^a dr = \mu_0 \pi n \sigma a = \mu_0 \pi n \frac{Q}{\pi a^2} a = \frac{\mu_0 n Q}{a}. \]
06

Final Result for the Magnetic Field

The magnetic field at the center of the disk is \( B = \frac{\mu_0 n Q}{a} \). This result shows that the field depends linearly on the number of rotations per second \( n \), the total charge \( Q \), and inversely on the radius \( a \) of the disk.

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Key Concepts

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

Dielectric Disk
A dielectric disk is a type of material that does not conduct electricity but can support electrostatic fields. In our exercise, we have a thin disk of this material with a specified radius and a total positive charge, denoted as \(+Q\), distributed evenly across its surface. This means that the surface of the disk is evenly covered in charged particles. When dealing with charge distributions, we assume that the disk is perfectly planar and uniform, which simplifies the physics involved. The disk in this exercise is rotating around an axis passing through its center, allowing us to study the effects of this motion on the magnetic field produced at the disk's center. By dividing the disk into smaller concentric rings, we can calculate the magnetic contributions from each part of the disk and then sum these to find the total effect.
Surface Charge Density
Surface charge density is a crucial concept when dealing with charged surfaces. It tells us how much charge is present per unit area on the surface. For the dielectric disk, the surface charge density \( \sigma \) is defined as \( \sigma = \frac{Q}{\pi a^2} \), where \(Q\) is the total charge and \(a\) is the radius of the disk. This formula derives from distributing the total charge evenly across the total area of the disk. Understanding surface charge density helps in calculating the charge present on infinitesimally small portions of the disk, like thin rings. By knowing \( \sigma \), you can determine the charge \( dQ \) on an infinitesimally small ring at a radius \( r \) with thickness \( dr \), as \( dQ = \sigma \times (2 \pi r dr) \). This step is fundamental in calculating the current and, subsequently, the magnetic field due to these rotating charges.
Biot-Savart Law
The Biot-Savart Law is a powerful tool in electromagnetism used for calculating the magnetic field generated by a current-carrying conductor. In the context of this problem, we apply it to find the magnetic field at the center of the rotating dielectric disk. It states that the magnetic field \( dB \) due to an infinitesimal current element \( dI \) is proportional to \( \frac{\mu_0 dI}{2r} \) for points along the axis of a circular current loop, where \( \mu_0 \) is the permeability of free space.Applying this law to each infinitesimal ring in the disk gives us the contribution of magnetic field from each ring. We add up (integrate) all these infinitesimal contributions to find the total magnetic field at the disk's center. This process highlights the importance of Biot-Savart Law in calculating fields from distributed sources.
Current Loop
In this exercise, rotating charges on the disk are treated like tiny current loops. Current loops are a classic configuration in which the Biot-Savart Law is often applied. To understand why, think of each rotating ring on the disk as generating a current due to its motion. This is similar to how a current in a wire loop generates a magnetic field.The linear velocity of charges in a ring at radius \( r \) is given by \( v = 2\pi r n \), where \( n \) is the number of rotations per second. This velocity, combined with charge density, allows us to determine the effective current \( dI \) in the ring as \( dI = n \times dQ \). Now, since each concentric ring can be considered an individual current loop, the total magnetic field is a result of integrating the contributions from all such loops across the whole disk. This approach demonstrates a powerful method to analyze magnetic fields resulting from complex charge distributions in motion.

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Most popular questions from this chapter

Two identical circular, wire loops 40.0 \(\mathrm{cm}\) in diameter each carry a current of 1.50 \(\mathrm{A}\) in the same direction. These loops are parallel to each other and are 25.0 \(\mathrm{cm}\) apart. Line \(a b\) is normal to the plane of the loops and passes through their centers. A proton is fired at 2400 \(\mathrm{km} / \mathrm{sperpendicular}\) to line \(a b\) from a point midway between the centers of the loops. Find the magnitude and direction of the magnetic force these loops exert on the proton just after it is fired.

A pair of long, rigid metal rods, each of length \(L,\) lie parallel to each other on a perfectly smooth table. Their ends are connected by identical, very light conducting springs of force constant \(k\) (Fig. 28.55 ) and negligible unstretched length. If a current \(I\) runs through this circuit, the springs will stretch. At what separation will the rods remain at rest? Assume that \(k\) is large enough so that the separation of the mods will be much less than \(L\) .

Two long, parallel wires are separated by a distance of \(2.50 \mathrm{cm} .\) The force per unit length that each wire exerts on the other is 4.00 \(\times 10^{-5} \mathrm{N} / \mathrm{m}\) , and the wires repel each other. The current in one wire is 0.600 \(\mathrm{A}\) . (a) What is the current in the second wire? (b) Are the two currents in the same direction or in opposite directions?

Helmholtz Coils. Fig. 28.59 is a sectional view of two circular coils with radius \(a\) , each wound with \(N\) turns of wire carrying a current \(I,\) circulating in the same direction in both coils. The coils are separated by a distance \(a\) equal to their radii. In this configuration the coils are called Helmholtz coils; they produce a very uniform magnetic field in the region between them. (a) Derive the expression for the magnitude \(B\) of the magnetic field at a point on the axis a distance \(x\) to the right of point \(P,\) which is midway between the coils. (b) Graph \(B\) versus \(x\) for \(x=0\) to \(x=a / 2\) . Compare this graph to one for the magnetic field due to the right-hand coil alone. (c) From part (a), obtain an expression for the magnitude of the magnetic field at point \(P .\) (d) Calculate the magnitude of the magnetic field at \(P\) if \(N=300\) turns, \(I=6.00 \mathrm{A},\) and \(a=8.00 \mathrm{cm} .\) ( e) Calculate \(d B / d x\) and \(d^{2} B / d x^{2}\) at \(P(x=0)\) . Discuss how your results show that the field is very uniform in the vicinity of \(P .\)

A long solenoid with 60 turns of wire per centimeter carries a current of 0.15 \(\mathrm{A}\) . The wire that makes up the solenoid is wrapped around a solid core of silicon steel \(\left(K_{\mathrm{m}}=5200\right) .\) (The wire of the solenoid is jacketed with an insulator so that none of the current flows into the core.) (a) For a point inside the core, find the magnitudes of (i) the magnetic field \(\overrightarrow{\boldsymbol{B}}_{0}\) due to the solenoid current; (ii) the magnetization \(\vec{M} ;\) (iii) the total magnetic field \(\overrightarrow{\boldsymbol{B}}\) . (b) In a sketch of the solenoid and core, show the directions of the vectors \(\overrightarrow{\boldsymbol{B}}, \overrightarrow{\boldsymbol{B}}_{0}\) , and \(\overrightarrow{\boldsymbol{M}}\) inside the core.

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