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Chapter 9: Problem 2

Spherically symmetric current A spherically symmetric (and constant) current density flows radially inward to a spherical shell, causing the charge on the shell to increase at the constant rate \(d Q / d t\). Verify that Maxwell's equation, \(\nabla \times \mathbf{B}=\mu_{0} \mathbf{J}+\mu_{0} \epsilon_{0} \partial \mathbf{E} / \partial t\), is satisfied at points outside the shell.

Short Answer

Expert verified
Outside the shell, considering there is no current and the rate of charge does not vary with time, Maxwell's equation \(\nabla \times \mathbf{B}=\mu_{0} \mathbf{J}+\mu_{0} \epsilon_{0} \partial \mathbf{E} / \partial t\) simplifies to \(\nabla \times \mathbf{B} = \mu_{0} \epsilon_{0} \partial \mathbf{E} / \partial t\), which further simplifies to \(0=0\) as there is no change of electric field with time. Hence, Maxwell's equation is satisfied outside the spherical shell.

Step by step solution

01

Consider Parameters

Consider the given current density \(J\), magnetic field \(B\), the vacuum permeability \(\mu_0\), vacuum permittivity \(\epsilon_0\) and the electric field \(E\). Also, note that outside the sphere where there's no current, \(\mathbf{J} = 0\), as no current flows outside the spherical shell.
02

Apply Conditions

Now, the charge density \(\rho\) can be found from charge conservation, i.e., \(\nabla \cdot \mathbf{J} = -\partial\rho / \partial t \). But here, since \(\partial\rho / \partial t = dQ/dt\) which is given, the equation becomes \(0 = - dQ/dt\). This implies that \(dQ/dt\) does not change with time. Therefore, \(dQ/dt = 0\) outside the sphere.
03

Using Maxwell's Equation

We now have to consider the Maxwell's equation \(\nabla \times \mathbf{B}=\mu_{0} \mathbf{J}+\mu_{0} \epsilon_{0} \partial \mathbf{E} / \partial t\). Outside the sphere where \(J = 0\), the equation simplifies to \(\nabla \times \mathbf{B} = \mu_{0} \epsilon_{0} \partial \mathbf{E} / \partial t\). Since both sides of this equation are zero, Maxwell's equation is satisfied outside the spherical shell.

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Key Concepts

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

Spherically Symmetric Current
A spherically symmetric current exists when current flows uniformly in all directions towards a center point, in this case, a spherical shell. The uniformity of the current ensures symmetry around the sphere. Imagine streams of current moving radially inward, much like spokes on a wheel going towards the hub.
This type of current is crucial for certain physical models because it maintains consistency in how fields behave around the sphere.
  • It helps simplify equations due to its symmetry.
  • Ensures equal distribution of charges.
  • Makes it easier to apply Gauss's Law.
The inward flow also means charges build up on the shell over time, which is pivotal for analyzing fields outside the shell.
Vacuum Permeability
Vacuum permeability, denoted by \( \mu_0 \), is a fundamental physical constant. It quantifies the ability of a vacuum to support the formation of magnetic fields, reflecting a relationship between magnetic field strength and magnetic force.
Think of it as a measure of how a magnetic field influences and is influenced by a vacuum.
  • The value for \( \mu_0 \) is approximately \( 4 \pi \times 10^{-7} \) Tm/A (Tesla meter per Ampere).
  • It's a key part of Maxwell's equations which describe electromagnetism.
In scenarios outside a spherical shell, \( \mu_0 \) helps determine how magnetic fields react to the charged shell.
Charge Conservation
Charge conservation is a fundamental principle stating that the total charge in an isolated system remains constant over time. This means that charge cannot be created or destroyed, only transferred.
In this context, the continuity equation \( abla \cdot \mathbf{J} = -\partial\rho / \partial t \) is critical. It connects the current density \( \mathbf{J} \) and the rate of change of charge density \( \rho \).
  • For a spherically symmetric current, changes occur as charges accumulate on the shell.
  • Outside the shell, where \( \mathbf{J} = 0 \), charge doesn't change.
Understanding this ensures that Maxwell's equations hold true, verifying that the physics outside the shell remains consistent.
Electric Field Outside a Sphere
The electric field outside a sphere can be intriguing. It's the electric field generated by charges on the sphere and is related to how those charges are distributed.
Key here is that outside the sphere, the system behaves like a point charge located at the center of the sphere. The electric field, \( \mathbf{E} \), reduces with the square of the distance from the center due to spherical symmetry.
  • The formula: \( \mathbf{E}(r) = \frac{kQ}{r^2} \), where \( k \) is Coulomb's constant and \( Q \) is the total charge on the sphere.
  • This implies that outside, it resembles a dipole field, losing intensity with distance.
For validating Maxwell's equations, it's important as it simplifies the conditions needed in the magnetic field equations outside the sphere.

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Most popular questions from this chapter

A charge and a half-infinite wire A half-infinite wire carries current \(I\) from negative infinity to the origin, where it builds up at a point charge with increasing \(q\) (so \(d q / d t=I\) ). Consider the circle shown in Fig. 9.12, which has radius \(b\) and subtends an angle \(2 \theta\) with respect to the charge. Calculate the integral \(\int \mathbf{B} \cdot d \mathbf{s}\) around this circle. Do this in three ways. (a) Find the B field at a given point on the circle by using the BiotSavart law to add up the contributions from the different parts of the wire. (b) Use the integrated form of Maxwell's equation (that is, the generalized form of Amp猫re's law including the displacement current), $$ \int_{C} \mathbf{B} \cdot d \mathbf{s}=\mu_{0} I+\mu_{0} \epsilon_{0} \int_{S} \frac{\partial \mathbf{E}}{\partial t} \cdot d \mathbf{a} $$ with \(S\) chosen to be a surface that is bounded by the circle and doesn't intersect the wire, but is otherwise arbitrary. (You can invoke the result from Problem 1.15.) (c) Use the same strategy as in (b), but now let \(S\) intersect the wire.

Microwave background radiation ** Of all the electromagnetic energy in the universe, by far the largest amount is in the form of waves with wavelengths in the millimeter range. This is the cosmic microwave background radiation discovered by Penzias and Wilson in \(1965 .\) It apparently fills all space, including the vast space between galaxies, with an energy density of \(4 \cdot 10^{-14}\) joule \(/ \mathrm{m}^{3} .\) Calculate the rms electric field strength in this radiation, in volts/m. Roughly how far away from a 1 kilowatt radio transmitter would you find a comparable electromagnetic wave intensity?

Oscillating field in a solenoid *** A solenoid with radius \(R\) has \(n\) turns per unit length. The current varies with time according to \(I(t)=I_{0} \cos \omega t\). The magnetic field inside the solenoid, \(B(t)=\mu_{0} n I(t)\), therefore changes with time. In this problem you will need to make use of wisely chosen Faraday/Amp猫re loops. (a) Changing \(B\) fields cause \(E\) fields. Assuming that the \(B\) field is given by \(B_{0}(t) \equiv \mu_{0} n I_{0} \cos \omega t\), find the electric field at radius \(r\) inside the solenoid. (b) Changing \(E\) fields cause \(B\) fields. Find the \(B\) field (at radius \(r\) inside the solenoid) caused by the changing \(E\) field that you just found. More precisely, find the difference between the \(B\) at radius \(r\) and the \(B\) on the axis. Label this difference as \(\Delta B(r, t)\). (c) The total \(B\) field does not equal \(\mu_{0} n I_{0} \cos \omega t\) throughout the solenoid, due to the \(\Delta B(r, t)\) difference you just found. \({ }^{3}\) What is the ratio \(\Delta B(r, t) / B_{0}(t) ?\) Explain why we are justified in making the statement, "The magnetic field inside the solenoid is essentially equal to the naive \(\mu_{0} n I_{0} \cos \omega t\) value, provided that the changes in the current occur on a time scale that is long compared with the time it takes light to travel across the width of the solenoid." (This time is very short, so for an "everyday" value of \(\omega\), the field is essentially equal to \(\mu_{0} n I_{0} \cos \omega t\). )

Momentum in an electromagnetic field ** We know from Section \(9.6\) that traveling electromagnetic waves carry energy. But the theory of relativity tells us that anything that transports energy must also transport momentum. Since light may be considered to be made of massless particles (photons), the relation \(p=E / c\) must hold; see Eq. (G.19). In terms of \(\mathbf{E}\) and \(\mathbf{B}\), find the momentum density of a traveling electromagnetic wave. That is, find the quantity that, when integrated over a given volume, yields the momentum contained in the wave in that volume. Although we won't prove it here, the result that you just found for traveling waves is a special case of the more general result that the momentum density equals \(1 / c^{2}\) times the energy flow per area per time. This holds for any type of energy flow (matter or field). In particular, it holds for any type of electromagnetic field; even a static field with a nonzero \(\mathbf{E} \times \mathbf{B} / \mu_{0}\) Poynting vector carries momentum. For a nice example of this, see Problem 9.12.

Poynting vector and resistance heating ** A longitudinal \(\mathbf{E}\) field inside a wire causes a current via \(\mathbf{J}=\sigma \mathbf{E}\). And since the curl of \(\mathbf{E}\) is zero, this same longitudinal \(\mathbf{E}\) component must also exist right outside the surface of the wire. Show that the Poynting vector flux through a cylinder right outside the wire accounts for the \(I V\) resistance heating.
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