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Chapter 6: Problem 49

Field at the center of a disk * A disk with radius \(R\) and surface charge density \(\sigma\) spins with angular frequency \(\omega\). What is the magnetic field at the center?

Short Answer

Expert verified
The magnetic field at the center of the disk is \(B = \frac{\mu_0 \sigma \omega R^2}{4}\).

Step by step solution

01

Set up the Biot-Savart Law

We first recognize that the rotating disk of charge can be viewed as circular current loops, with the current \(I\) for a loop of radius \(r\) and thickness \(dr\) given by \(I = \sigma \omega r dr\). Considering a point on this loop, the current element \(Idl\) generates a small magnetic field \(dB\) at the center of the disk: \(dB = \frac{\mu_0 Idl \sin\theta}{4\pi r^2}\) where \(\theta\) is the angle between \(dl\) and the position vector for the center of the disk, and \(\mu_0\) is the permeability of free space. Given that the disk is planar, \(dl\) and the position vector are perpendicular, hence \(\sin\theta = 1\). So, \(dB = \frac{\mu_0 Idl}{4\pi r^2}\). It's worth noting that the direction of \(dB\) is into the plane of the page (due to the right-hand rule).
02

Integrate the Contributions

Applying the Biot-Savart Law gives us the contribution to the total magnetic field at the center from a small element of the disk. The next step is to add up or integrate these contributions over the entire disk. First consider the contribution from a small ring with radius \(r\) and thickness \(dr\). The length \(dl\) of this ring is the circumference \(2\pi r\). Hence, \(dB = \frac{\mu_0 I 2\pi r}{4\pi r^2} = \frac{\mu_0 I}{2r}\). Substitute the expression for \(I\) to find \(dB = \frac{\mu_0 \sigma \omega r dr}{2}\). Now we can integrate \(dB\) over the entire disk, from \(r = 0\) to \(r = R\), to find the total magnetic field \(B\): \(B = \int_0^R dB = \frac{\mu_0 \sigma \omega}{2} \int_0^R r dr\).
03

Solve the Integral and Obtain the Magnetic Field

Solving the integral, we find \(\int_0^R r dr = \frac{R^2}{2}\). Substituting this into the previous equation gives the total magnetic field at the center of the disk: \(B = \frac{\mu_0 \sigma \omega R^2}{4}\).

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Key Concepts

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

Magnetic Field
The magnetic field is a region around a magnetic material or a moving electric charge within which the force of magnetism acts. In physics, a moving charge generates a magnetic field which can affect other charges in motion. This field is typically represented by field lines that show the direction and strength of the magnetic force.

For a system like a rotating disk with a surface charge, understanding the magnetic field is essential. When charges move, their movement creates loops of current, which in turn generates a magnetic field. The magnitude and direction of this field can be established through laws such as Biot-Savart Law. In the context of the given exercise, we calculate the magnetic field at the disk's center using the contributions from each infinitesimal part of the disk.
Rotating Disk
A rotating disk refers to a flat circular object that spins around a central axis. When a disk rotates, any charge on its surface also moves, effectively creating a circulating current.

The speed at which parts of the disk move is determined by its angular velocity. The further away from the center of the disk, the higher the linear velocity of the rotating charge due to its larger path. This motion creates loops of current which result in a magnetic field at the center of the disk. Understanding the dynamics of a rotating disk is crucial when calculating physical phenomena like the magnetic field generated by its motion.
Surface Charge Density
Surface charge density, denoted often as \(\sigma\), is the amount of electric charge per unit area on a surface. In scenarios involving rotating disks, it's a fundamental parameter because it influences the current that flows over the surface.

Mathematically, it is expressed as the ratio of charge \(Q\) to the surface area \(A\): \(\sigma = \frac{Q}{A}\). On a rotating disk, knowing the surface charge density helps in determining the total amount of current because, as the disk rotates, the charge traverses a path that behaves like a current-carrying loop. By understanding \(\sigma\), we can calculate the induced magnetic field using Biot-Savart Law by considering the charge movement as a current flow.
Angular Frequency
Angular frequency (\(\omega\)), often used in rotational dynamics, describes how quickly a rotating object spins. It is defined as the rate of change of the angle (\(\theta\)) with which an object rotates around a particular axis, measured in radians per second.

In the context of the rotating disk, the angular frequency explains how quickly the charge at a particular radius of the disk moves along its circular path. It directly affects the current formed by the moving charge, as a higher angular frequency means the charges move faster, resulting in a stronger current and thus a larger magnetic field at the center of the disk. Understanding angular frequency is essential to solve problems related to magnetic fields generated by rotating charged systems.

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Most popular questions from this chapter

Copper solenoid \(* *\) A solenoid is made by winding two layers of No. 14 copper wire on a cylindrical form \(8 \mathrm{~cm}\) in diameter. There are four turns per centimeter in each layer, and the length of the solenoid is \(32 \mathrm{~cm}\). From the wire tables we find that No. 14 copper wire, which has a diameter of \(0.163 \mathrm{~cm}\), has a resistance of \(0.010 \mathrm{ohm} / \mathrm{m}\) at \(75^{\circ} \mathrm{C}\). (The coil will run hot!) If the solenoid is connected to a \(50 \mathrm{~V}\) generator, what will be the magnetic field strength at the center of the solenoid in gauss, and what is the power dissipation in watts?

Scaled-up ring \(*\) Consider two circular rings of copper wire. One ring is a scaled-up version of the other, twice as large in all regards (radius, crosssectional radius). If currents around the rings are driven by equal voltage sources, how do the magnetic fields at the centers compare?

Rings with opposite currents Two parallel rings have the same axis and are separated by a small distance \(\epsilon\). They have the same radius \(a\), and they carry the same current \(I\) but in opposite directions. Consider the magnetic field at points on the axis of the rings. The field is zero midway between the rings, because the contributions from the rings cancel. And the field is zero very far away. So it must reach a maximum value at some point in between. Find this point. Work in the approximation where \(\epsilon \ll a\)

A slab and a sheet \(* *\) A volume current density \(\mathbf{J}=J \hat{\mathbf{z}}\) exists in a slab between the infinite planes at \(x=-b\) and \(x=b\). (So the current is coming out of the page in Fig. 6.37.) Additionally, a surface current density \(\mathcal{J}=2 b J\) points in the \(-\hat{\mathbf{z}}\) direction on the plane at \(x=b\) (a) Find the magnetic field as a function of \(x\), both inside and outside the slab. (b) Verify that \(\nabla \times \mathbf{B}=\mu_{0} \mathbf{J}\) inside the slab. (Don't worry about the boundaries.)

Motion in a B field \(*\) A particle of charge \(q\) and rest mass \(m\) is moving with velocity \(\mathbf{v}\) where the magnetic field is \(\mathbf{B}\). Here \(\mathbf{B}\) is perpendicular to \(\mathbf{v}\), and there is no electric field. Show that the path of the particle is a curve with radius of curvature \(R\) given by \(R=p / q B\), where \(p\) is the momentum of the particle, \(\gamma m v\). (Hint: Note that the force \(q \mathbf{v} \times \mathbf{B}\) can only change the direction of the momentum, not the magnitude. By what angle \(\Delta \theta\) is the direction of \(\mathbf{p}\) changed in a short time \(\Delta t ?\) ) If \(\mathbf{B}\) is the same everywhere, the particle will follow a circular path. Find the time required to complete one revolution.
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