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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?

Short Answer

Expert verified
The magnetic field at the tip of the cone is given by \(\mu_0 \sigma \omega \tan(\theta) L / 4 \cos(\theta)\).

Step by step solution

01

Calculate the Charge

The first step is to calculate the total charge of the surface of the cone, denoted as \(Q\). It can be obtained by multiplying the surface charge density \(\sigma\) by the surface area of the open cone, which is \(\pi L^2 \tan(\theta)\). Hence, \(Q = \sigma \cdot \pi L^2 \tan(\theta)\).
02

Calculate the Current

The next step is to calculate the current \(I\), which is produced by the rotating charge. The charge passes a given point once per revolution, i.e., once per period of the rotation \(T = 2 \pi / \omega\). Hence \(I = Q / T = \sigma \cdot \pi L^2 \tan(\theta) / (2 \pi / \omega) = \sigma \cdot L^2 \omega \tan(\theta) / 2\).
03

Calculate the Magnetic Field

Now, to find the magnetic field \(B\) at the tip of the cone, employ Ampere's circuital law in magnetism considering a circular loop of radius \(r\) at the tip as the path. For a current loop \(I\), this magnetic field at a distance \(r\) on the axis is given by \(B = \mu_0 I / (2 r)\). Here, \(r = L \cos(\theta)\), the distance from the tip to the center of the conical surface. So, \(B = \mu_0 \sigma \omega \tan(\theta) L / 4 \cos(\theta)\)

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Key Concepts

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

Ampere's Circuital Law
Ampere's Circuital Law is a fundamental principle in electromagnetism that relates the magnetic field circulating around a closed loop to the electric current passing through that loop. This concept is key when determining the magnetic fields for symmetrical and steady currents.
\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I \]
This equation states that the integral of the magnetic field \( \mathbf{B} \) along a closed path is equal to the permeability of free space \( \mu_0 \), multiplied by the total electric current \( I \) that passes through the enclosed area of the path.

Using Ampere’s circuital law helps calculate the magnetic field in cases like a hollow cone that spins, generating a continuous current due to its rotating charge. Knowing the amount of current \( I \) and the geometry of the loop can help solve for the magnetic field at specific points. Creating a path that coincides with the rotational symmetry—like a ring at the cone’s tip—maximizes symmetry and simplifies calculations.
Surface Charge Density
Surface charge density \( \sigma \) refers to the amount of electric charge per unit area on a surface. It influences how much charge an object can hold over its area and is expressed in coulombs per square meter \( \text{C/m}^2 \).

To calculate the total charge on a hollow cone, one would multiply the surface charge density by the cone's surface area. For the cone discussed, its lateral surface area is \( \pi L^2 \tan(\theta) \). Therefore, the total charge \( Q \) is given by \( \sigma \times \text{surface area} \).
  • If the surface charge density increases, the total charge on the cone also increases.
  • This, in turn, affects the generated magnetic and electric fields around the cone.

Understanding surface charge density is crucial for analyzing the distribution of electric charge across various surfaces, particularly in electrostatic and electromagnetic applications.
Angular Frequency
Angular frequency \( \omega \) is a measure of how quickly an object rotates or oscillates. It is the rate of change of the angle with which an object covers a certain arc and is given in radians per second.
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the period of rotation.

In systems where an object like a hollow cone spins around a given axis, the angular frequency is a crucial parameter. It directly affects the magnetic field generated by the rotation of charged surfaces. The faster the rotational rate, the greater the current produced and, hence, a stronger magnetic field.
  • It helps determine how often the charge on the cone completes a full cycle.
  • Aligns with harmonic motions seen in pendulums and circular motions.

By increasing the angular frequency, you can enhance the characteristics of the resultant magnetic field.
Hollow Cone
A hollow cone is a three-dimensional object resembling a cone with no base. It is often used in physics problems due to its symmetrical structure which simplifies the calculation of fields and forces.
  • Identified by its vertex angle \(2\theta\), slant height \(L\), and open surface area.
  • In this scenario, acts as a perfect model for observing charge distributions over a 2D surface in 3D space.

The hollow cone spins around its symmetry axis, which can create interesting electromagnetic effects due to induced motion of surface charges.

When considering fields at its tip, factors like its rotational symmetry allow for leveraging laws of symmetry with electromagnetic principles, like Ampere’s law, making complex calculations more manageable.

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Most popular questions from this chapter

Solenoid field, inside and outside *** Consider an infinite solenoid with circular cross section. The current is \(I\), and there are \(n\) turns per unit length. Show that the magnetic field is zero outside and \(B=\mu_{0} n I\) (in the longitudinal direction) everywhere inside. Do this in three steps as follows. (a) Show that the field has only a longitudinal component. Hint: Consider the contributions to the field from rings that are symmetrically located with respect to a given point. (b) Use Ampère's law to show that the field has a uniform value outside and a uniform value inside, and that these two values differ by \(\mu_{0} n I\). (c) Show that \(B \rightarrow 0\) as \(r \rightarrow \infty .\) There are various ways to do this. One is to obtain an upper bound on the field contribution due to a given ring by unwrapping the ring into a straight wire segment, and then finding the field due to this straight segment.

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, \(A_{0}\) ?

Hall voltage \(* *\) A Hall probe for measuring magnetic fields is made from arsenic-doped silicon, which has \(2 \cdot 10^{21}\) conduction electrons per \(\mathrm{m}^{3}\) and a resistivity of \(0.016 \mathrm{ohm}-\mathrm{m}\). The Hall voltage is measured across a ribbon of this \(n\)-type silicon that is \(0.2 \mathrm{~cm}\) wide, \(0.005\) \(\mathrm{cm}\) thick, and \(0.5 \mathrm{~cm}\) long between thicker ends at which it is connected into a \(1 \mathrm{~V}\) battery circuit. What voltage will be measured across the \(0.2 \mathrm{~cm}\) dimension of the ribbon when the probe is inserted into a field of 1 kilogauss?

Zero field outside a solenoid \(*\) * We showed in the solution to Problem \(6.19\) that the magnetic field is zero outside an infinite solenoid with arbitrary (uniform) crosssectional shape. We can demonstrate this fact in another way, similar in spirit to Problem 1.17.

Constant magnitude of \(B\) ** How should the current density inside a thick cylindrical wire depend on \(r\) so that the magnetic field has constant magnitude inside the wire?

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