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Chapter 6: Problem 66

Fields in a new frame \(*\) * In the neighborhood of the origin in the coordinate system \(x, y\), \(z\), there is an electric field \(\mathbf{E}\) of magnitude \(100 \mathrm{~V} / \mathrm{m}\), pointing in a direction that makes angles of \(30^{\circ}\) with the \(x\) axis, \(60^{\circ}\) with the \(y\) axis. The frame \(F^{\prime}\) has its axes parallel to those just described, but is moving, relative to the first frame, with a speed \(0.6 c\) in the positive \(y\) direction. Find the direction and magnitude of the electric field that will be reported by an observer in the frame \(F^{\prime}\). What magnetic field does this observer report?

Short Answer

Expert verified
The observer in the moving frame will measure the electric and magnetic fields with the calculated magnitudes and directions, which can be derived from the Lorentz transformations and the component computations. The fields will be different from what are measured in the stationary frame due to special relativity and the relative motion.

Step by step solution

01

Compute Electric Field Components

The electric field components can be computed via spherical to Cartesian transformation as follows: \(E_x = E sin(30), E_y = E sin(60), E_z = E cos(60)\). Given \(E=100v/m\), compute these quantities to get the electric field in Cartesian coordinates.
02

Apply Lorentz Transformation

Apply Lorentz transformation which describes how two observers moving relative to each other will measure electric field. The transformed fields in \(F'\) frame are given by \(E'x = E_x, E'y = E_y/\gamma + \beta B_z, E'z = E_z/\gamma - \beta B_y\). Here \(\gamma = 1/\sqrt{1-\beta^2}\), where \(\beta=v/c\) is the relative velocity between the observers in terms of the speed of light. Substitute \(v=0.6c\) to calculate \(\gamma\) and then use the Lorentz transformation with \(B_y = B_z = 0\) (no magnetic field in \(F\) frame) to compute \(E'\) components.
03

Compute Resultant Field Magnitude & Direction in \(F'\) Frame

The magnitude of the electric field in the \(F'\) frame can be computed with the Euclidean norm of the field components: \(E' = \sqrt{{E'x}^2+{E'y}^2+{E'z}^2}\). The direction in terms of angles can be obtained by formulas: \(cos(\alpha') = E'x / E', cos(\beta') = E'y / E', cos(\gamma') = E'z / E'\)
04

Compute Magnetic Field Reported by Observer

In the \(F'\) frame, the observer measures a magnetic field due to the relative motion. The magnetic field components can be given by Lorentz Transformation as \(B_x = B_x, B_y = B_y/\gamma - \beta E_z, B_z = B_z/\gamma + \beta E_y\). Substitute the values of \(E_y, E_z, B_y = B_z = 0, \gamma, \beta\) and calculate the magnetic field components. As in step 3, the magnitude and direction of the magnetic field can be obtained.

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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 is a vector field surrounding an electric charge that exerts a force on other charges, influencing them remotely. In our exercise, an electric field of magnitude 100 V/m is positioned at the origin of our coordinate system. This field points in a specific direction that forms angles with the axes of our reference frame.

To completely describe the field in the initial frame, we must find its components along the x, y, and z axes. To do this, we use trigonometry based on the provided angles: for instance, for the x-component we have used \(E_x = E \sin(30^\circ)\), and similar expressions for the y and z components. Understanding these components provides a base for how the electric field is perceived in different inertial frames, laying the groundwork for our analysis through Lorentz transformation in the context of special relativity.
Magnetic Field
A magnetic field is another aspect of the unified electromagnetic force and is represented as a vector field that exerts a magnetic force on moving electric charges and magnetic dipoles. Magnetic fields originate from moving electrical charges and are intrinsically linked with electric fields under the electromagnetic phenomena umbrella.

In the exercise, we're initially given a static scenario where there鈥檚 only an electric field present. However, as we move to a different frame of reference, one moving relative to the first, a magnetic field arises as a result of this relative motion. Applying Lorentz transformation equations allow us to calculate the components of this magnetic field as observed in the moving frame. This is a classic result from special relativity, where an electric field in one frame may manifest as both an electric and a magnetic field in another.
Special Relativity
Special relativity is a fundamental theory about the structure of spacetime, formulated by Albert Einstein, which introduces the idea that the laws of physics are the same for all non-accelerating observers. One of the key precepts is that the speed of light in a vacuum is the same for all observers, no matter the speed at which an observer travels.

In terms of electromagnetism, special relativity profoundly affects the perception of electric and magnetic fields. The Lorentz transformation, derived from special relativity, provides the mathematical means by which we can calculate how the measurements of space and time, and thus of electric and magnetic fields, will differ between two observers moving at a constant velocity relative to one another. The problem at hand involves applying this transformative concept to adjust the observed electric field based on the relative motion between frames, exemplifying the intertwined nature of space, time, and electromagnetic forces.
Coordinate Transformation
Coordinate transformation is the process of converting a set of given coordinates in one system to another system. This concept is essential when we aim to describe the same physical situation from different points of view or reference frames. In classical physics, these transformations are often simple, but they become more complex and significantly more interesting under the framework of special relativity.

The Lorentz transformation is a specific type of coordinate transformation in special relativity. It includes time as a fourth dimension and involves the relative speed of the two reference frames as a key factor in the transformation equations. By using the Lorentz transformation, as performed in the exercise, we can accurately determine how the components of the electric field and any resulting magnetic field are observed in a frame moving at a significant fraction of the speed of light relative to the initial frame. Applying these transformations precisely gives us the ability to predict physical phenomena in the newly transformed frame.

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Most popular questions from this chapter

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.)

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?

Motion in a B field \(*\) A particle of charge \(q\) and rest mass \(m\) is moving with velocity \(\mathbf{v}\) where the magnetic field is \(\mathbf{B}\). Here \(\mathbf{B}\) is perpendicular to \(\mathbf{v}\), and there is no electric field. Show that the path of the particle is a curve with radius of curvature \(R\) given by \(R=p / q B\), where \(p\) is the momentum of the particle, \(\gamma m v\). (Hint: Note that the force \(q \mathbf{v} \times \mathbf{B}\) can only change the direction of the momentum, not the magnitude. By what angle \(\Delta \theta\) is the direction of \(\mathbf{p}\) changed in a short time \(\Delta t ?\) ) If \(\mathbf{B}\) is the same everywhere, the particle will follow a circular path. Find the time required to complete one revolution.

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?

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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