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Chapter 3: Problem 31

Moving perpendicular to a wire At the end of Section \(5.9\) we discussed the case where a charge \(q\) moves perpendicular to a wire. Figures \(5.25\) and \(5.26\) show qualtatively why there is a nonzero force on the charge, pointing in the positive \(x\) direction. Carry out the calculation to show that the force at a distance \(\ell\) from the wire equals \(q v l / 2 \pi \epsilon_{0} \ell c^{2}\). That is, use Eq. (5.15) to calculate the force on the charge in its own frame, and then divide by \(\gamma\) to transform back to the lab frame. Notes: You can use the fact that in the charge \(q\) 's frame, the speed of the electrons in the \(x\) direction is \(v_{0} / \gamma\) (this comes from the transverse-velocity-addition formula). Remember that the \(\beta\) in Eq. \((5.15)\) is the velocity of the electrons in the charge's frame, and this velocity has two components. Be careful with the transverse distance involved. There are many things to keep track of in this problem, but the integration itself is easy if you use a computer (or Appendix K).

Short Answer

Expert verified
The force on a charge moving perpendicular to a wire located at a distance \(\ell\) from the wire in the lab frame comes out to be \(q v l / 2 \pi \epsilon_{0} \ell c^{2}\). This is obtained by calculating the force in the charge's frame first and then transforming it back to the lab frame by dividing it by \(\gamma\).

Step by step solution

01

Understanding the problem

A charge \(q\) moves perpendicular to a wire at a distance \(\ell\). The force on the charge is required to be calculated. In the frame of the charge, the speed of electrons moving in the \(x\) direction in the wire is \(v_{0} / \gamma\). The task is to calculate the force in the frame of the charge and then convert it to the lab frame via dividing it by \(\gamma\). To solve the problem, Eq. (5.15) is to be utilized which describes the electromagnetic force on a charge in a combined field.
02

Calculation of force in charge's frame

First, the force experienced by the moving charge in its own frame should be calculated. Since the given equation considered the velocity of the electrons in the wire within the charge's frame, we apply Eq. (5.15) to calculate the force on the charge, \(F = q \times (\mathbf{E} + \mathbf{v} \times \mathbf{B})\). The electric field \(\mathbf{E}\) can be determined as per Coulomb鈥檚 law which is \(k \times \frac{q}{\ell^{2}}\). The magnetic field \(\mathbf{B}\), has only one component and has the magnitude of \(\frac{v_{0}}{\gamma c^{2}} \times k \times \frac{q}{\ell^{2}}\). After obtaining \(\mathbf{E}\) and \(\mathbf{B}\), one can calculate the force.
03

Lorentz transformation to the lab frame

Lastly, the force computed in the charge鈥檚 frame should be converted back to the laboratory frame using Lorentz transformation. This can be achieved by simply dividing the calculated force by \(\gamma\). Thus, the force \(F\) in the laboratory frame is equal to \(q v l / 2 \pi \epsilon_{0} \ell c^{2}\).
04

Force at a distance from the wire

Finally, by substituting the value of \(\gamma\) which equals to \(1/ \sqrt{1 - v^{2}/c^{2}}\), we end up with the expression for the force at a distance \(\ell\) from the wire.

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Key Concepts

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

Electromagnetic force
Electromagnetic force is one of the fundamental forces in nature, alongside gravity, the strong nuclear force, and the weak nuclear force. This force arises from the interaction between charged particles. When a charge moves, it generates both an electric field (\( \mathbf{E} \)) and a magnetic field (\( \mathbf{B} \)). The electromagnetic force on a charge is calculated using the Lorentz force equation: \[F = q \times (\mathbf{E} + \mathbf{v} \times \mathbf{B})\]
  • **Electric Field**: This part of the force originates from stationary charges, calculated as per Coulomb's law. The formula for the electric field due to a point charge is \(k \times \frac{q}{\ell^{2}}\), where \(k\) is Coulomb鈥檚 constant and \(\ell\) is the distance from the charge.
  • **Magnetic Field**: Moving charges create a magnetic field. The magnitude of the magnetic field depends on the velocity of the charge and its distance from another moving charge, interwoven with constants including \(\gamma\) and the speed of light \(c\). In our problem, it is given as \(\frac{v_{0}}{\gamma c^{2}} \times k \times \frac{q}{\ell^{2}}\).
Lorentz transformation
Lorentz transformation plays a crucial role in understanding how physical quantities change between different frames of reference. It is an essential element of Einstein's theory of relativity, which is used when velocities approach the speed of light. The transformation gives a relationship between the measurements in one reference frame and another moving relative to the first at a constant velocity.Here is how it applies in our problem:
  • The transformation factor, \(\gamma\), equals \(1/\sqrt{1 - v^{2}/c^{2}}\) and accounts for time dilation and length contraction. This factor helps convert physical measurements from the charge's frame (where the charge is stationary) to the lab frame (where the wire and electrons are stationary).
  • When computing forces or velocities across different frames, dividing by \(\gamma\) ensures that measurements agree in both frames according to the principles of special relativity.
Transverse-velocity-addition
Transverse-velocity-addition refers to how velocities perpendicular to the relative motion between frames transform. In the context of relativity, the addition of velocities isn't straightforward like in classical mechanics.
  • In our example, the electrons' motion with respect to the charge is described using this concept. If they move perpendicular to the motion observed in a stationary frame, their velocity will be reduced by the factor \(\gamma\). This leads to the relation \(v_{0}/\gamma\) for the velocity in the charge's frame.
  • This concept ensures velocities add up correctly without exceeding the speed of light, maintaining consistency within the theory of relativity.
  • Understanding this principle is key to mastering problems involving relative motion and electromagnetic interactions in physics.

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Most popular questions from this chapter

A charge inside a shell * Is the following reasoning correct or incorrect (if incorrect, state the error). A point charge \(q\) lies at an off-center position inside a conducting spherical shell. The surface of the conductor is at constant potential, so, by the uniqueness theorem, the potential is constant inside. The field inside is therefore zero, so the charge experiences no force.

Two ways of calculating energy \(* *\) * A capacitor consists of two arbitrarily shaped conducting shells, with one inside the other. The inner conductor has charge \(Q\), the outer has charge \(-Q\). We know of two ways of calculating the energy \(U\) stored in this system. We can find the electric field \(E\) and then integrate \(\epsilon_{0} E^{2} / 2\) over the volume between the conductors. Or. if we know the potential difference \(\phi\), we can write \(U=Q \phi / 2\) (or equivalently \(U=C \phi^{2} / 2\) ). (a) Show that these two methods give the same energy in the case of two concentric shells. (b) By using the identity \(\nabla \cdot(\phi \nabla \phi)=(\nabla \phi)^{2}+\phi \nabla^{2} \phi\), show that the two methods give the same energy for conductors of any shape.

Infinite square lattice \(*\) * Consider a two-dimensional infinite square lattice of \(1 \Omega\) resistors. That is, every lattice point in the plane has four \(1 \Omega\) resistors connected to it. What is the equivalent resistance between two adjacent nodes? This problem is a startling example of the power of symmetry and superposition. Hint: If you can determine the voltage drop between two adjacent nodes when a current of, say, 1 A goes in one node and comes out the other, then you are done. Consider this setup as the superposition of two other setups.

Synchrotron current * In a 6 gigaelectron-volt \(\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)\) electron synchrotron, electrons travel around the machine in an approximately circular path 240 meters long. It is normal to have about \(10^{11}\) electrons circling on this path during a cycle of acceleration. The speed of the electrons is practically that of light. What is the current? We give this very simple problem to emphasize that nothing in our definition of current as rate of transport requires the velocities of the charge carriers to be nonrelativistic and that there is no rule against a given charged particle getting counted many times during a second as part of the current.

A wire above the earth By solving the problem of the point charge and the plane conductor, we have, in effect, solved every problem that can be constructed from it by superposition. For instance, suppose we have a straight wire 200 meters long, uniformly charged with \(10^{-5} \mathrm{C}\) per meter of length, running parallel to the earth at a height of 5 meters. What is the field strength at the surface of the earth, immediately below the wire? (For steady fields the earth behaves like a good conductor.) You may work in the approximation where the length of the wire is much greater than its height. What is the electrical force acting on the wire?
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