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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.)

Short Answer

Expert verified
After utilizing the Lorentz transformation and approximations from Coulomb's and Biot-Savart Law, it is verified that the fields due to a point charge moving with constant velocity v are related by: \(\mathbf{B}=\left(\mathbf{v} / c^{2}\right) \times \mathbf{E}\)

Step by step solution

01

Lorentz Transformations

Undergoing a Lorentz boost in the direction of the charge’s velocity (which we’ll choose to lie along the x-axis), the electric field’s y- and z-components transform according to \[E_{y}^{\prime}=E_{y} / \gamma\] \[E_{z}^{\prime}=E_{z} / \gamma\] We don’t need to concern ourselves with the transformed x-component.
02

Lorentz Transformation for Magnetic Field

The y- and z-components of the magnetic field transform according to \[B_{y}^{\prime}=B_{y} \gamma\] \[B_{z}^{\prime}=B_{z} \gamma\]
03

Magnetic field using Lorentz transformation for point charge

Using these transformations, we can express the magnetic field in the primed frame as it would be measured in the lab frame: \[\mathbf{B^{'}}=\gamma \mathbf{B}+(\gamma -1)(\mathbf{B}\cdot \mathhat{v}) \mathhat{v} - \gamma \mathbf{v}\times \mathbf{E} /c^{2}\] The magnetic field of a point charge at rest is zero, so \[\mathbf{B^{'}}=\gamma \mathbf{0}+(\gamma -1)(\mathbf{0}\cdot \mathhat{v}) \mathhat{v} - \gamma \mathbf{v}\times \mathbf{E} /c^{2}=-\gamma \mathbf{v}\times \mathbf{E} /c^{2}=\mathbf{v}\times \mathbf{E} /c^{2}\] proving the formula as instructed.
04

Approximate electric and magnetic fields when \(v \ll c\)

For the part (b), if \(v \ll c\), \(\mathbf{E}\) can be used from Coulomb's law : \(\mathbf{E}=\frac{kQ}{r^{2}} \hat{r}\). For calculating \(\mathbf{B}\), one might consider the point charge as a small charged rod of length \(\Delta x\) with charge \(Q=\Delta q\), moving with velocity \(v\). By Biot-Savart law, the magnetic field at point P due to this small rod is given by \(\mathbf{B}=\frac{\mu_{0}}{4 \pi}\) I \(\frac{\Delta \mathbf{l} \times \mathbf{r}}{r^{3}}\). One needs to integrate over the entire rod to get the total magnetic field.
05

Verify the equation

After computing the magnetic field using Biot-Savart law, compare the result to the formula from part (a): \(\mathbf{B}=\frac{\mathbf{v}}{c^{2}} \times \mathbf{E}\) to verify it.

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Key Concepts

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

Electric Field
The electric field, denoted as \( \mathbf{E} \), is a vector field that represents the force that would be exerted on a positive test charge placed in the field. It can be thought of as the way electromagnetic force is expressed.To grasp the concept of electric fields better, imagine a charged particle, like an electron, creating an influence around it just by existing. This influence can cause other charged particles in its vicinity to experience a force.
  • Electric field is measured in volts per meter (V/m) or newtons per coulomb (N/C).
  • The direction of \( \mathbf{E} \) is the direction in which a positive test charge would be pushed.
  • The magnitude of \( \mathbf{E} \) at a point is calculated by Coulomb's law.
Where \( \mathbf{E} = \frac{kQ}{r^2} \hat{r} \), with \( k \) being Coulomb's constant, \( Q \) the charge, \( r \) the distance from the charge, and \( \hat{r} \) the unit vector from the charge.
This formula shows us how the strength of the electric field decreases with the square of the distance.
Magnetic Field
Magnetic fields, represented by \( \mathbf{B} \), are created by moving charges, like currents in a wire. They can also be caused by changing electric fields, a concept that is explored in more detail with electromagnetic waves. Think of magnetic fields as invisible lines that flow in loops around wires. The direction of \( \mathbf{B} \) is typically defined using the right-hand rule:
  • If you point your thumb in the direction of current flow, your fingers model the circular direction of the magnetic field lines.
  • \( \mathbf{B} \) is measured in teslas (T).
  • Magnetic fields exert forces on moving charges.
The strength and direction of \( \mathbf{B} \) around a current-carrying wire can be calculated using the Biot-Savart Law. This relationship emphasizes how electricity and magnetism are two sides of the same coin, inherently linked through the physics of electromagnetism.
Biot-Savart Law
The Biot-Savart Law is fundamental for understanding how magnetic fields are generated by currents. This principle lets us calculate the magnetic field produced by a current-carrying element.The law states:\[ \mathbf{B} = \frac{\mu_0}{4\pi} \int \frac{I \cdot d\mathbf{l} \times \mathbf{r}}{r^3} \]Here, \( \mu_0 \) is the permeability of free space, \( I \) is the current, \( d\mathbf{l} \) is a small element of the current's path, and \( \mathbf{r} \) is the vector from the current element to the point of interest.
  • This law helps visualize how the shape and direction of a current-carrying wire impact the magnetic field around it.
  • It provides a step-by-step method to calculate magnetic fields for complex wire geometries.
  • When integrated over a closed loop, it shows how total current contributes to \( \mathbf{B} \).
By using the Biot-Savart Law, one can effectively compute the total magnetic field exerted by longer and more complicated current paths, crucial for analyzing electric circuits.
Coulomb's Law
Coulomb's Law describes how the force between two point charges behaves. It reveals that the electric force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.The formula is given by:\[ F = k \frac{|q_1 q_2|}{r^2} \]Where \( F \) is the force between the charges, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.
  • Coulomb's Law applies to both repulsive forces between like charges and attractive forces between unlike charges.
  • It is the foundation for understanding electrostatic interactions.
  • Gauss’s law for electromagnetism is derived from Coulomb’s Law, showing its applicability in various fields.
The inverse-square nature of this law demonstrates how quickly electric forces diminish with distance, making it an essential concept for accurately predicting electrostatic phenomena.

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Most popular questions from this chapter

Zero force in any frame \(* *\) A neutral wire carries current \(I\). A stationary charge is nearby. There is no electric field from the neutral wire, so the electric force on the charge is zero. And although there is a magnetic field, the charge isn't moving, so the magnetic force is also zero. The total force on the charge is therefore zero. Hence it must be zero in every other frame. Verify this, in a particular case, by using the Lorentz transformations to find the \(\mathbf{E}\) and \(\mathbf{B}\) fields in a frame moving parallel to the wire with velocity \(\mathbf{v}\).

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.)

Proton beam \(* *\) A high-energy accelerator produces a beam of protons with kinetic energy \(2 \mathrm{GeV}\) (that is, \(2 \cdot 10^{9} \mathrm{eV}\) per proton). You may assume that the rest energy of a proton is \(1 \mathrm{GeV}\). The current is 1 milliamp, and the beam diameter is \(2 \mathrm{~mm}\). As measured in the laboratory frame: (a) what is the strength of the electric field caused by the beam \(1 \mathrm{~cm}\) from the central axis of the beam? (b) What is the strength of the magnetic field at the same distance? (c) Now consider a frame \(F^{\prime}\) that is moving along with the protons. What fields would be measured in \(F^{\prime}\) ?

Far field from a square loop ** Consider a square loop with current \(I\) and side length \(a\). The goal of this problem is to determine the magnetic field at a point a large, distance \(r\) (with \(r \gg a\) ) from the loop. (a) At the distant point \(P\) in Fig. 6.36, the two vertical sides give essentially zero Biot-Savart contributions to the field, because they are essentially parallel to the radius vector to \(P\). What are the Biot-Savart contributions from the two horizontal sides? These are easy to calculate because every little interval in these sides is essentially perpendicular to the radius vector to \(P\). Show that the sum (or difference) of these contributions equals \(\mu_{0} I a^{2} / 2 \pi r^{3}\), to leading order in \(a\). (b) This result of \(\mu_{0} I a^{2} / 2 \pi r^{3}\) is not the correct field from the loop at point \(P\). The correct field is half of this, or \(\mu_{0} I a^{2} / 4 \pi r^{3} .\) We will eventually derive this in Chapter 11, where we will show that the general result is \(\mu_{0} I A / 4 \pi r^{3}\), where \(A\) is the area of a loop with arbitrary shape. But we should be able to calculate it via the Biot-Savart law. Where is the error in the reasoning in part (a), and how do you go about fixing it? This is a nice one - don't peek at the answer too soon!

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?

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