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Chapter 7: Q7.21P (page 321)
Imagine a uniform magnetic field, pointing in the direction and filling all space . A positive charge is at rest, at the origin. Now somebody turns off the magnetic field, thereby inducing an electric field. In what direction does the charge move?
The direction of the charge is indetermined.
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Question: The preceding problem was an artificial model for the charging capacitor, designed to avoid complications associated with the current spreading out over the surface of the plates. For a more realistic model, imagine thin wires that connect to the centers of the plates (Fig. 7.46a). Again, the current I is constant, the radius of the capacitor is a, and the separation of the plates is w << a. Assume that the current flows out over the plates in such a way that the surface charge is uniform, at any given time, and is zero at t = 0.
(a) Find the electric field between the plates, as a function of t.
(b) Find the displacement current through a circle of radius in the plane mid-way between the plates. Using this circle as your "Amperian loop," and the flat surface that spans it, find the magnetic field at a distance s from the axis.
Figure 7.46
(c) Repeat part (b), but this time uses the cylindrical surface in Fig. 7.46(b), which is open at the right end and extends to the left through the plate and terminates outside the capacitor. Notice that the displacement current through this surface is zero, and there are two contributions to Ienc.
A long cylindrical shell of radius carries a uniform surface charge on the upper half and an opposite charge on the lower half (Fig. 3.40). Find the electric potential inside and outside the cylinder.
As a lecture demonstration a short cylindrical bar magnet is dropped down a vertical aluminum pipe of slightly larger diameter, about 2 meters long. It takes several seconds to emerge at the bottom, whereas an otherwise identical piece of unmagnetized iron makes the trip in a fraction of a second. Explain why the magnet falls more slowly.
A transformer (Prob. 7.57) takes an input AC voltage of amplitude , and delivers an output voltage of amplitude , which is determined by the turns ratio . If , the output voltage is greater than the input voltage. Why doesn't this violate conservation of energy? Answer: Power is the product of voltage and current; if the voltage goes up, the current must come down. The purpose of this problem is to see exactly how this works out, in a simplified model.
(a) In an ideal transformer, the same flux passes through all turns of the primary and of the secondary. Show that in this case , where is the mutual inductance of the coils, and , are their individual self-inductances.
(b) Suppose the primary is driven with AC voltage , and the secondary is connected to a resistor, . Show that the two currents satisfy the relations
(c) Using the result in (a), solve these equations for localid="1658292112247" and . (Assume has no DC component.)
(d) Show that the output voltage divided by the input voltage is equal to the turns ratio: .
(e) Calculate the input power localid="1658292395855" and the output power , and show that their averages over a full cycle are equal.
An alternating current flows down a long straight wire, and returns along a coaxial conducting tube of radius a.
(a) In what direction does the induced electric field point (radial, circumferential, or longitudinal)?
(b) Assuming that the field goes to zero as , find
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