Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

If you have some experience in electromagnetism, you could do the following problem concerning the curious situation illustrated in Figure 1.8. The electric and magnetic fields at a point r1 due to a charge q2 at r2 moving with constant velocity v2 (with v2c ) are 15 E(r1)=14πϵoq2s2s^ and B(r1)=μ04πq2s2v2×s^ where s=r1r2 is the vector pointing from r2 to r1. (The first of these you should recognize as Coulomb's law.) If F12el and F12mag  denote the electric and magnetic forces on a charge q1 at r1 with velocity v1, show that F12mag(v1v2/c2)F12el. This shows that in the non-relativistic domain it is legitimate to ignore the magnetic force between two moving charges.

Short Answer

Expert verified
The magnetic force is negligible compared to the electric force for non-relativistic speeds.

Step by step solution

01

Write Down Expressions for Forces

The electric force on charge q1 at r1 is given by Coulomb's law:F12el=q1E(r1)=14πϵoq1q2s2s^.The magnetic force is given by the Lorentz force law:F12mag=q1(v1×B(r1)).Substitute B(r1) into the expression for F12mag:F12mag=q1μ04πq2s2(v1×(v2×s^)).
02

Simplify Magnetic Force Expression

Use the vector triple product identity: (a×(b×c))=(ac)b(ab)c.Substitute into the magnetic force expression:(v1s^)v2(v1v2)s^.Thus,F12mag=q1q2μ04πs2((v1s^)v2(v1v2)s^).
03

Calculate Magnitude of Magnetic Force

The magnitude of F12mag is the sum of the magnitudes of the component vectors:|F12mag|=q1q2μ04πs2|(v1s^)v2(v1v2)s^|.Apply the inequality |ab||a|+|b| to the magnitudes:q1q2μ04πs2(|v1s^||v2|+|v1v2|).
04

Connect Terms and Apply Approximation

Using μ0ϵ0c2=1, substitute μ0=1ϵ0c2 into the magnetic force expression:|F12mag|q1q24πϵ0s2v2c2(v1+v2).Notice that the electric force is F12el=q1q24πϵ0s2s^:|F12mag|(v1v2c2+v22c2)|F12el|,where v22/c2 is negligible compared to v1v2/c2.
05

Conclusion and Result

Because v2c, the term v22c2 can be ignored, leaving only:|F12mag|v1v2c2|F12el|.This shows the magnetic force is negligible compared to the electric force in the non-relativistic domain.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Electric Field
The electric field is a fundamental concept in electromagnetism. It refers to the region around a charged particle where an electric force can be felt by another charge. In simple terms, imagine a charged object creating an invisible field around it. If another charge enters this field, it experiences a force that is proportional to its own charge and inversely proportional to the square of the distance from the source charge.

The formula for the electric field E, as in this exercise, is given by:
  • E(r1)=14πϵoq2s2s^
This formula illustrates how the electric field depends on the charge creating it, the distance from the charge, and the permittivity of free space, ϵ0.

Essentially, the electric field is a vector that points away from positive charges and towards negative charges. This aligns with the idea of forces in physics where like charges repel and unlike charges attract, following Coulomb's Law.
Magnetic Force
Magnetic force arises from the motion of electric charges. It is a fundamental force in nature, responsible for actions such as the operation of electric motors and the navigation of compasses.

A moving charge, like a current in a wire, produces a magnetic field around it. If another charge moves within this field, it experiences a magnetic force, not aligned with the electric field, but rather perpendicular to both the velocity of the moving charge and the direction of the field.

For this exercise, the magnetic field B(r1) produced by a moving charge q2 is expressed as:
  • B(r1)=μ04πq2s2v2×s^
Here, μ0 represents the magnetic constant or permeability of free space. The cross product v2×s^ indicates the force's direction is perpendicular to the plane formed by v2 and s^. Magnetic forces are unique because they require motion; a static charge does not experience a magnetic force.
Lorentz Force
The Lorentz force equation is crucial in understanding how charged particles move in electric and magnetic fields. It combines both electric and magnetic forces acting on a charge. By understanding this, we gain insight into the motion of electrons in fields, a principle that powers many modern technologies like cathode ray tubes and the function of cyclotrons.

The total force F on a moving charge q in external electric E and magnetic fields B is given by:
  • F=q(E+v×B)
This equation shows the superposition of the electric and magnetic forces. The electric part of the Lorentz force is straightforward, simply qE. The magnetic part, q(v×B), signifies how motion through the magnetic field affects the charge.

The forces interact in such a way that if a charged particle moves through both fields, its path can curve due to the perpendicular magnetic forces, much like in the example problem where motion at speeds much less than light allows simpler approximations.
Coulomb's Law
Coulomb's Law is as fundamental to electromagnetism as Newton's Law of Gravitation is to gravity. It calculates the electric force between two point charges. This foundational law helps us quantify how charges interact and the resultant forces they exert on each other.

The formula for Coulomb's Law in its common form is:
  • F12el=14πϵoq1q2s2s^
Here, F12el is the force between charges q1 and q2, s is the distance separating them, and s^ is the unit vector pointing from one charge to the other.

This law reveals that the force is inversely proportional to the square of the distance between charges, meaning it rapidly decreases as they move apart. This inverse-square law also helps explain why electric forces can be strong at close range and almost negligible at great distances.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

[Computer] The differential equation (1.51) for the skateboard of Example 1.2 cannot be solved in terms of elementary functions, but is easily solved numerically. (a) If you have access to software, such as Mathematica, Maple, or Matlab, that can solve differential equations numerically, solve the differential equation for the case that the board is released from ϕo=20 degrees, using the values R=5m and g=9.8m/s2. Make a plot of ϕ against time for two or three periods. ( b ) On the same picture, plot the approximate solution (1.57) with the same ϕo=20 Comment on your two graphs. Note: If you haven't used the numerical solver before, you will need to learn the necessary syntax. For example, in Mathematica you will need to learn the syntax for "NDSolve”and how to plot the solution that it provides. This takes a bit of time, but is something that is very well worth learning.

Prove that if v(t) is any vector that depends on time (for example the velocity of a moving particle) but which has constant magnitude, then v˙(t) is orthogonal to v(t). Prove the converse that if v˙(t) is orthogonal to v(t), then |v(t)| is constant. [Hint: Consider the derivative of v2.] This is a very handy result. It explains why, in two- dimensional polars, dr^/dt has to be in the direction of ϕ^ and vice versa. It also shows that the speed of a charged particle in a magnetic field is constant, since the acceleration is perpendicular to the velocity.

Find the angle between a body diagonal of a cube and any one of its face diagonals. [Hint: Choose a cube with side 1 and with one corner at O and the opposite corner at the point (1,1,1) . Write down the vector that represents a body diagonal and another that represents a face diagonal, and then find the angle between them as in Problem 1.4.]

A student kicks a frictionless puck with initial speed vo, so that it slides straight up a plane that is inclined at an angle θ above the horizontal. (a) Write down Newton's second law for the puck and solve to give its position as a function of time. (b) How long will the puck take to return to its starting point?

An astronaut in gravity-free space is twirling a mass m on the end of a string of length R in a circle, with constant angular velocity ω Write down Newton's second law (1.48) in polar coordinates and find the tension in the string.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free