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A rotating cylinder \(*\) An infinite cylinder with radius \(R\) and surface charge density \(\sigma\) spins around its symmetry axis with angular frequency \(\omega\). Find the magnetic field inside the cylinder.

Short Answer

Expert verified
The magnetic field inside the spinning charged cylinder is \(B = \frac { \mu_0 \sigma \omega r}{2}\) where \(r < R\).

Step by step solution

01

Establish the relation between angular velocity and linear velocity

Start by converting the angular velocity to linear velocity. According to the equation \(v = \omega r\), the linear speed of the spinning surface of the cylinder is \(v = R \omega\) where \(R\) is the radius of the cylinder.
02

Calculate the equivalent current

Next, find the current which is equivalent to the spinning charge. Consider a narrow ring of width \(dx\) at a distance \(x\) from the axis. The current \(dI\) in this ring is given by \(dI = \sigma v dx\). Substituting the values we found in step 1 yields \(dI = \sigma R \omega dx\). Integrate this expression from 0 to \(R\) to get the total current \(I\) inside a radius \(r < R\): \(I = \int_0^r \sigma R \omega dx = \sigma R \omega r\).
03

Finding the Magnetic Field

Finally, use Ampere's law in the form \(B 2 \pi r = \mu_0 I\) to find the magnetic field \(B\) inside the cylinder. After putting the value of \(I\) from step 2 and solving the equation, we find \(B = \frac { \mu_0 \sigma \omega r}{2}\) for \(r < R\). Therefore, the magnetic field inside the cylinder is \(B = \frac { \mu_0 \sigma \omega r}{2}\).

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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 mathematical statement that relates the integrated magnetic field around a closed loop to the electric current passing through the loop. It's one of the four Maxwell's equations, which underpin all of classical electromagnetism. The law particularly says that for any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability of free space times the electric current enclosed by the loop.

Mathematically, Ampere's Law is written as \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}} \), where \(\mathbf{B}\) represents the magnetic field, \(d\mathbf{l}\) is an infinitesimal element of the loop, \(\mu_0\) is the permeability of free space, and \(I_{\text{enc}}\) is the enclosed current. In the given exercise, we apply Ampere's Law by taking a hypothetical loop inside the cylinder, which allows us to calculate the magnetic field generated by the rotating charged cylinder.

Understanding Ampere's Law is crucial not only in solving problems involving magnetic fields but also in grasping the foundational relationships between electric currents and magnetic fields in various geometries and configurations.
Surface Charge Density
Surface charge density is a measure of electric charge per unit area on a surface. Represented with the Greek letter \(\sigma\), it quantifies how much electric charge is distributed over a specific area. In physical terms, this concept tells us how 'packed' an area is with charge. It's an important concept for understanding the distribution of charge on objects and the resultant electric and magnetic fields.

Surface charge density is given by the formula \(\sigma = Q / A\), where \(Q\) is the total charge, and \(A\) is the area over which the charge is spread. In the context of the rotating cylinder problem, the surface charge density is uniform across the surface of the cylinder and is given as a known quantity. This uniformity allows us to reason that every small segment of the cylinder's surface contributes to the magnetic field inside the cylinder in a calculable and additive way.

The exercise showcases the direct involvement of surface charge density in determining the magnetic field inside the cylinder, as the rotating charged surface creates a moving current, which is essential in magnetic field generation according to electromagnetic theory.
Angular Velocity
Angular velocity is a vector quantity that represents the rate of rotation around an axis. It tells us how fast an object is spinning and in which direction. It’s often denoted with the Greek letter \(\omega\) and is measured in radians per second (rad/s).

For a spinning cylinder, as described in the textbook exercise, each part of the cylinder is moving at the same angular velocity, but the linear velocity is different at different distances from the axis. The linear velocity, \(v\), can be easily related to \(\omega\) by the equation \(v = \(\omega\) r\), where \(r\) is the distance from the axis of rotation. That means points on the surface of the cylinder are moving faster in terms of linear speed (\(v\)) than points closer to the axis, since their \(r\) is greater.

In the given problem, knowing the angular velocity is critical since it directly affects the 'equivalent current' created by the spinning charge. The higher the angular velocity, the greater the linear velocity at the surface, and hence, the stronger the magnetic field generated inside the cylinder. Understanding angular velocity is not just crucial for solving this exercise but is also fundamental for comprehending rotational motion in general, which has applications ranging from astrophysics to mechanical engineering.

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Most popular questions from this chapter

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 \(1 \mathrm{GeV}\).)

Vector potential inside a wire ** A round wire of radius \(r_{0}\) carries a current \(I\) distributed uniformly over the cross section of the wire. Let the axis of the wire be the \(z\) axis, with \(\hat{z}\) the direction of the current. Show that a vector potential of the form \(\mathbf{A}=A_{0} \hat{\mathbf{z}}\left(x^{2}+y^{2}\right)\) will correctly give the magnetic field \(\mathbf{B}\) of this current at all points inside the wire. What is the value of the constant, \(A_{0}\) ?

Field in the plane of a ring ** A ring with radius \(R\) carries a current \(I\). Show that the magnetic field due to the ring, at a point in the plane of the ring, a distance \(a\) from the center (either inside or outside the ring), is given by $$ B=2 \cdot \frac{\mu_{0} I}{4 \pi} \int_{0}^{\pi} \frac{(R-a \cos \theta) R d \theta}{\left(a^{2}+R^{2}-2 a R \cos \theta\right)^{3 / 2}} $$ Hint: The easiest way to handle the cross product in the BiotSavart law is to write the Cartesian coordinates of \(d l\) and \(\mathbf{r}\) in terms of an angle \(\theta\) in the ring. This integral can't be evaluated in closed form (except in terms of elliptic functions), but it can always be evaluated numerically if desired. For the special case of \(a=0\) at the center of the ring, the integral is easy to do; verify that it yields the result given in

Field at the tip of a cone \(* *\) A hollow cone (like a party hat) has vertex angle \(2 \theta\), slant height \(L\), and surface charge density \(\sigma\). It spins around its symmetry axis with angular frequency \(\omega\). What is the magnetic field at the tip?

E and B for a point charge ** (a) Use the Lorentz transformations to show that the \(\mathbf{E}\) and \(\mathbf{B}\) fields due to a point charge moving with constant velocity \(\mathbf{v}\) are related by \(\mathbf{B}=\left(\mathbf{v} / c^{2}\right) \times \mathbf{E}\) (b) If \(v \ll c\), then \(\mathbf{E}\) is essentially obtained from Coulomb's law, and \(\mathbf{B}\) can be calculated from the Biot-Savart law. Calculate \(\mathbf{B}\) this way, and then verify that it satisfies \(\mathbf{B}=\left(\mathbf{v} / c^{2}\right) \times\) E. (It may be helpful to think of the point charge as a tiny rod of charge, in order to get a handle on the \(d l\) in the BiotSavart law.)

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