Electromagnetic radiation exists because changing magnetic fields induce changing electric fields and vice versa. This fact often appears inconsistent with the way some physics textbooks solve particular problems using Faraday’s law. These types of problems often ask students to find the induced electric field given a current that does not vary linearly with time. A typical example involves a long solenoid carrying a sinusoidal current. This problem is usually solved as an example or assigned as a homework exercise. The solution offered by many textbooks uses the approximation that the induced, changing electric field produces a negligible magnetic field, which is only valid at low frequencies. If this approximation is not explicitly acknowledged, then the solution appears inconsistent with the description of electromagnetic radiation. In other cases, when the problem is solved without this approximation, the electric and magnetic fields are derived from the vector potential. We present a detailed calculation of the electric and magnetic fields inside and outside the long solenoid without using the vector potential. We then offer a comparison of our solution and a solution given in an introductory textbook.

Topics
Faraday's law, Electromagnetic radiation, Magnetic fields, Electromagnetic coils, Educational aids
1.
See R. A. Serway and R. J. Biechner, Physics for Scientists and Engineers, 5th ed. (Harcourt College Publishers, Orlando, FL, 2000), pp. 993–994 as an example of how many introductory texts perform this calculation.
2.
See D. J. Griffiths, Introduction to Electrodynamics, 3rd ed. (Prentice Hall, Upper Saddle River, NJ, 1999), p. 308. Griffiths explains the quasistatic approximation.
Also see R. L. Reese, University Physics (Brooks/Cole Publishing Co., Pacific Grove, CA, 2000), pp. 895–952 for a consistent approach; Reese introduces Faraday’s law concurrently with Maxwell’s equations and electromagnetic radiation.
Also see R. Wolfson and J. M. Pasachoff, Physics, 2nd ed. (Harper Collins College Publishers, New York, 1995), p. 799 for an acknowledgement of the approximation.
3.
See Ref. 1, pp. 979–1013. The fact that changing electric fields induce magnetic fields is only briefly mentioned on pp. 1000; the approximation is not recognized anywhere else in the chapter.
Also see P. A. Tipler, Physics, 4th ed. (W. H. Freeman and Co., New York, 1999), pp. 927–958;
V. D. Barger and M. G. Olsson, Classical Electricity and Magnetism (Allyn and Bacon, Inc., Newton, MA, 1987), pp. 240–280. The approximation is not mentioned in either book in the chapter covering Faraday’s law.
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5.
See also integral derivation of differential equations in the Appendix. Introductory texts generally rely on integrals.
6.
To be more rigorous, integrate Eq. (6) from Rs−ε to Rs+ε, and take the limit as ε→0. Since the electric field is continuous, the last term is zero, and you are left with the boundary condition.
7.
Equations (3) and (4) are also consistent with Maxwell’s other two laws. Namely, ∇⋅B=0 and ∇⋅E=ρ/ε0 where, for our problem, the charge density is 0.
8.
Equations (8) and (9) are actually the Green’s function for Eq. (7) with an arbitrary source term.
9.
I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products (Academic, Orlando, FL, 1980), p. 967.
10.
Reference 8, p. 969.
11.
Reference 8, p. 961. Note the 1/r asymptotic dependence. This is consistent with the power radiated, which is proportional to E2+B2. To satisfy energy conservation, a point source, which has spherical symmetry, should have the power radiated per unit area as 1/r2; for an infinite solenoid with cylindrical symmetry, the power radiated per unit area should fall off as 1/r.
12.
This makes sense in two other ways: (a) As solutions to the differential equations, we could have chosen the Hankel functions instead of the Bessel and Neumann functions. Now, since H(1) and H(2) are equivalent to J+iN and J−iN, respectively, they correspond to incoming and outgoing waves (for very large r, cos x+i sin x and cos x−i sin x)⁠. Thus we would need to get rid of H(1) if we only want outgoing waves. (b) In light of the vector potential, when the z integral is performed in Ref. 5, the retarded Green’s function (for outgoing waves) gives H(2). Had the advanced Green’s function been chosen, which corresponds to incoming waves, H(1) would have been the solution.
13.
The calculation of the self-capacitance of a finite solenoid is not the main focus of this paper. The interested reader can see R. G. Medhurst, Resistance and Self-Capacitance of Single-Layer Solenoids, Wireless Engineer, February 1947, pp. 35–43 and March 1947, pp. 80–92.
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© 2003 American Association of Physics Teachers.
2003
American Association of Physics Teachers
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