This problem leads you through a study of electric and magnetic fields as viewed by two different observers, that is, in two different reference frames. Suppose a charge q₁ is moving with a constant velocity v₁(vector) (Fig. P20.94) and that its trajectory takes it to within a distance L of stationary charge q₂.
(a) What are the magnitude and direction of the resulting electric current?
(b) Equation 20.19 gives the magnetic field produced by a current-carrying wire. Assume this relation can also be applied here (it can!) and use it to find the magnetic field produced by the current in part (a). What is the resulting magnetic field a distance L from the path of q₁?
(c) Now assume charge q₂ has a velocity v₂(vector) parallel to v₁(vector). Use the result from part (b) to show that the magnitude of the magnetic force on q₂ when the two charges are at their closest is F = μ₀q₁q₂v₁v₂ /(2πL).
(d) What is the direction of the force in part (c) if q₁ and q₂ are both positive charges? If one is positive and one is negative? An observer watching q1 and q2 travel by would say that the force found in part (c) is a magnetic force. An observer in a moving reference frame would see things differently. If an observer is moving along with q1 at velocity v S 1, he or she would say that q1 is not moving, so the current and hence the magnetic field and the magnetic force are zero. However, this observer would still find that there is a force between the two charges as given in part (c) and would attribute it to an extra electric field E_new.
(e) Find the magnitude and direction of E_new. The Coulomb’s law force produced by E_new must be equal to the force found in part (c). This electric field is a relativistic effect, and we’ll discuss the ideas behind it in Chapter 27.