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A rotating solid cylinder ** (a) A very long cylinder with radius \(R\) and uniform volume charge density \(\rho\) spins with frequency \(\omega\) around its axis. What is the magnetic field at a point on the axis? (b) How would your answer change if all the charge were concentrated on the surface?

Short Answer

Expert verified
The magnetic field on the axis inside the cylinder is \(B = \frac{\mu_0 \omega \rho r}{2}\), outside the cylinder it is \(B = \mu_0 \omega \rho R^2/(2r)\). If all charge was concentrated on the surface, there wouldn’t be any magnetic field inside the cylinder, and for outside the cylinder, the magnetic field remains the same as in the first part of the problem.

Step by step solution

01

Determining the Current

The current I in the cylinder can be predicted by taking the volume integral of the charge density over the given ring of thickness \(dr\), which rotates around \(r\). This integral can be defined as \(I = \int dI = \int_{0}^{R} \omega \rho 2\pi r dr \). The volume integral results in \(I = \omega \rho \pi R^2\).
02

Applying Ampere’s Law

The magnetic field can be found by applying the Ampere’s law, which states that on integrating B around any closed loop, the result is \(\mu_0\) times the net current enclosed by the loop. In this case, the loop integral looks like \(B 2\pi r = \mu_0 I\). In the consideration of field points inside the cylinder, \(I = \omega \rho \pi r^2 \). This leads to giving us the magnetic field \(B = \frac{\mu_0 \omega \rho r}{2}\). Outside the cylinder, the field decreases linearly with \(1/r\), given that \(B = \mu_0 \omega \rho R^2/(2r)\).
03

Considering all charge concentrated on the surface

If all the charge is concentrated on the surface, the magnetic field inside the cylinder would be \(B = 0\), because now there is no enclosed current. In contrast, for the cases outside the cylinder, things will remain as per part (a) because the enclosed current remains the same.

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Key Concepts

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

Ampere's Law
Ampere's Law is a foundational concept in electromagnetism that relates the circulating magnetic field in space to the current that generates it. It is pivotal when calculating the magnetic field produced by a steady current. In mathematical terms, Ampere's Law is expressed by the equation \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} \), where \( \oint \vec{B} \cdot d\vec{l} \) is the magnetic field \( \vec{B} \) integrated over a closed loop, \( \mu_0 \) is the permeability of free space, and \( I_{\text{enc}} \) is the net current enclosed by the loop. This law is essential for solving the problem at hand as it allows for the determination of the magnetic field generated by a rotating solid cylinder with a given current distribution.

In our educational exercise, we apply Ampere's Law by creating an imaginary loop inside the solid cylinder. This approach simplifies the problem, as the symmetry tells us that the magnetic field at a certain distance from the axis is uniform in magnitude and direction along our loop. Consequently, Ampere's Law enables us to link the enclosed current to the magnetic field strength with great efficiency.
Volume Charge Density
Volume charge density, denoted as \( \rho \), is a measure of how much electric charge is distributed within a given volume of a material. It is a crucial parameter in calculating the current inside the rotating cylinder for our problem. The volume charge density is defined as the quantity of charge \( Q \) per unit volume \( V \), expressed mathematically as \( \rho = \frac{Q}{V} \).

Understanding the concept of volume charge density is important for determining how the charges within the solid cylinder contribute to the magnetic field. In the context of our exercise, we use the uniform volume charge density to calculate the current within the cylinder, \( I = \omega \rho \pi R^2 \) by integrating the charge density over the volume of the rotating cylindrical shell. This current, as a source of the magnetic field, is then applied to Ampere's Law to find the magnetic field at any point within the cylinder.
Magnetic Field Inside a Solid Cylinder
The magnetic field inside a solid cylinder that carries a current can be visualized using the right-hand rule, where the thumb points in the direction of the current and the curled fingers show the direction of the magnetic field. For a rotating solid cylinder with a uniform volume charge density, the generated magnetic field lines form concentric circles around the axis of rotation.

To calculate the magnetic field inside the cylinder, we use Ampere's Law and consider a circular path concentric with the cylinder's axis. Inside the cylinder, the magnetic field is directly proportional to the distance from the axis \( r \) as \( B = \frac{\mu_0 \omega \rho r}{2} \). This relationship indicates a linear increase in the magnetic field strength from the axis outwards until reaching the surface of the cylinder. The simplicity in the behavior of the magnetic field inside the solid cylinder comes from the uniform volume charge density and the symmetry of the setup, making it easier to grasp for students.
Surface Charge Distribution
Surface charge distribution involves the distribution of electric charges over the surface of a conductor. In the case of the rotating cylinder, if the charge is assumed to be uniformly distributed over its surface rather than throughout its volume, it affects the resulting magnetic field significantly. Instead of a volume charge density \( \rho \) throughout the material, we have a surface charge density \( \sigma \), where \( \sigma = \frac{Q}{A} \) with \( A \) being the surface area.

When all the charge is on the surface of the cylinder, the magnetic field inside the cylinder is zero, \( B = 0 \), because Ampere's Law would enclose no current within a loop inside the surface. The existence of the magnetic field is directly associated with the flow of charges, and in this scenario, there's no charge flow inside the cylinder to generate a magnetic field. However, outside the cylinder, the magnetic field is as calculated previously because the total current intercepted by an Amperian loop is the same as before. This concept is a crucial educational example of how charge distribution plays a significant role in the behavior of the magnetic field.

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Most popular questions from this chapter

Equal magnitudes Suppose we have a situation in which the component of the magnetic field parallel to the plane of a sheet has the same magnitude on both sides, but changes direction by \(90^{\circ}\) in going through the sheet. What is going on here? Would there be a force on the sheet? Should our formula for the force on a current sheet apply to cases like this?

Fields from two rings \(*\) A ring with radius \(r\) and linear charge density \(\lambda\) spins with frequency \(\omega\). A second ring with radius \(2 r\) has the same density \(\lambda\) and frequency \(\omega\). Each ring produces a magnetic field at its center. How do the magnitudes of these fields compare?

Maximum field in a cyclotron ** For some purposes it is useful to accelerate negative hydrogen ions in a cyclotron. A negative hydrogen ion, \(\mathrm{H}^{-}\), is a hydrogen atom to which an extra electron has become attached. The attachment is fairly weak; an electric field of only \(4.5 \cdot 10^{8} \mathrm{~V} / \mathrm{m}\) in the frame of the ion (a rather small field by atomic standards) will pull an electron loose, leaving a hydrogen atom. If we want to accelerate \(\mathrm{H}^{-}\)ions up to a kinetic energy of \(1 \mathrm{GeV}\left(10^{9} \mathrm{eV}\right)\), what is the highest magnetic field we dare use to keep them on a circular orbit up to final energy? (To find \(\gamma\) for this problem you only need the rest energy of the \(\mathrm{H}^{-}\)ion, which is of course practically the same as that of the proton, approximately \(1 \mathrm{GeV}\).)

Force in three frames *** A charge \(q\) moves with speed \(v\) parallel to a wire with linear charge density \(\lambda\) (as measured in the lab frame). The charges in the wire move with speed \(u\) in the opposite direction, as shown in Fig. 6.38. If the charge \(q\) is a distance \(r\) from the wire, find the force on it in (a) the given lab frame, (b) its own rest frame, (c) the rest frame of the charges in the wire. Do this by calculating the electric and magnetic forces in the various frames. Then check that the force in the charge's rest frame relates properly to the forces in the other two frames. You can use the fact that the \(\gamma\) factor associated with the relativistic addition of \(u\) and \(v\) is \(\gamma_{u} \gamma_{v}\left(1+\beta_{u} \beta_{v}\right)\)

The retarded potential \(* * *\) A point charge \(q\) moves with speed \(v\) along the line \(y=r\) in the \(x y\) plane. We want to find the magnetic field at the origin at the moment the charge crosses the \(y\) axis. (a) Starting with the electric field in the charge's frame, use the Lorentz transformation to show that, in the lab frame, the magnitude of the magnetic field at the origin (at the moment the charge crosses the \(y\) axis) equals \(B=\left(\mu_{0} / 4 \pi\right)\left(\gamma q v / r^{2}\right)\) (b) Use the Biot-Savart law to calculate the magnetic field at the origin. For the purposes of obtaining the current, you may assume that the "point" charge takes the shape of a very short stick. You should obtain an incorrect answer, lacking the \(\gamma\) factor in the above correct answer. (c) The Biot-Savart method is invalid because the Biot-Savart law holds for steady currents (or slowly changing ones, but see Footnote 8 ). But the current due to the point charge is certainly not steady. At a given location along the line of the charge's motion, the current is zero, then nonzero, then zero again. For non-steady currents, the validity of the Biot-Savart law can be restored if we use the so-called "retarded time." \(^{\prime 11}\) The basic idea with the retarded time is that, since information can travel no faster than the speed of light, the magnetic field at the origin, at the moment the charge crosses the \(y\) axis, must be related to what the charge was doing at an earlier time. More precisely, this earlier time (the "retarded time") is the time such that if a light signal were emitted from the charge at this time, then it would reach the origin at the same instant the charge crosses the \(y\) axis. Said in another way, if someone standing at the origin takes a photograph of the surroundings at the moment the charge crosses the \(y\) axis, then the position of the charge in the photograph (which will not be on the \(y\) axis) is the charge's location we are concerned with. \(^{12}\)

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