I had some very bright AP physics classes during my 35 years as a high school physics teacher. Because those students had already completed my PSSC-style first-year class, I was able to help them discover the basics of electricity and magnetism by using lab experiments and Socratic questioning. (My complete class worksheets can be downloaded from the URL in Ref. 1.) For classes with the ability to master the basics, I chose to have my students make a simple discovery that does not appear in their textbooks. I sent a carefully selected volunteer^{2} to the blackboard to solve the following problem with the help of classmates:

Imagine a radio wave passing through a vacuum. Imagine a small region at the crest of that wave, small enough so that the electric and magnetic fields within it are nearly uniform.

*Write formulas for the electric and magnetic field energies contained in that small region, in terms of its volume, field strengths, and constants.*[The energy densities are Eqs. (3) and (4), below. Multiply each by the volume.]

*Show how the total energy in the region can be expressed in terms of the magnetic field strength, the region’s volume, and constants. (By eliminating “E,” we avoid the temptation to confuse “**E**” for electric field with “E” for energy.) Be sure to detect and correct algebraic errors by checking units before continuing.*[Eq. (9) gives the total energy density. Multiply that total by the volume of the region.]

*Do the same for the momentum contained in the region in terms of B and constants.*[Use Eq. (10), below.]

*Write the formula for wave speed in terms of properties of the vacuum.*[Eq. (8), below]

*Use those momentum and speed formulas to discover a formula for the apparent mass of the region in terms of magnetic field strength and constants. Again, check units before continuing. Beware of confusing “C” (for “Coulombs”) with “c” representing the speed of light.*(Divide the momentum by the speed.)

*m*= (*ε*_{0}*cB*^{2})(vol.)/*c*= (*ε*_{0}*B*^{2})(vol.)*How does the energy in that small region depend on its apparent mass and speed?*(Divide the energy by the mass.)

Energy/mass = (

*B*^{2}/*μ*_{0})/(*ε*_{0}*B*^{2}) = 1/(*μ*_{0}*ε*_{0}) =*c*^{2}→*E*=*mc*^{2}.

Most students are thrilled to discover that final equation because it is usually associated with Einstein. It certainly does appear in his special theory of relativity, but it was lurking all along in Maxwell’s equations! Discovering Einstein’s most famous equation in this way is something that students will never forget, even if the steps we used are not perfectly rigorous.

The equations mentioned above can all be found in textbooks, but with careful guidance most of them can be discovered by motivated high school students. Coulomb’s law is the exception, but the old black-and-white film by Eric Rogers solves that problem brilliantly.^{3}

Students need to be told that an “electric field” is a condition in space that causes a force to act on a charged particle, and that the definition of “electric field strength” is used with Coulomb’s law to create a formula for the strength of the electric field produced by a charged particle, just as the analogous formula was created for gravitation.

After students have discovered how to find the area of a sphere, it is easy for them to discover Gauss’s law. The proportionality constant in that law (1/4*πk* in terms of Coulomb’s constant) is cumbersome, so we replace it with a new constant “*ε*_{0}” called “permittivity of a vacuum.”

Students need to learn how to integrate the electric fields produced by a straight row of particles to find the electric field produced by a long, straight charged wire, and how to use that result to describe the fields produced by uniformly charged planes. That leads to the parallel-plate capacitance formula, which is as simple as it could possibly be, with the proportionality constant turning out to be *ε*_{0}. It is important for students to use both Gauss’s law and the capacitance formula to figure out what SI units *ε*_{0} must have.

Students need to be told that a “magnetic field” is a condition in space that causes equal and opposite forces to act on the two ends of a compass needle. Then with tangent galvanometer experiments using concentric and coplanar circular coils (fine insulated wire with many turns is best, and the leads must be twisted together), students can make two important discoveries:

The magnetic field generated at the center of a circular coil is proportional to the coil current and the number of turns.

If the larger coil has twice the radius and twice as many turns as the smaller one and they are connected in series so that one coil current is clockwise and the other is counterclockwise, we find that their magnetic fields cancel at the center. We conclude that the magnetic field generated by a circular loop of current is inversely proportional to the loop’s radius.

*d*

**” contributed by one segment must be tangent to an imaginary circle that is centered on the segment and that lies in a plane perpendicular to the segment. Since the outer coil has four times as many segments as the inner one and its segments are twice as far from the center, we must conclude that**

*B**d*

**is proportional to the segment’s current and inversely proportional to the square of the segment’s distance from the center**

*B*^{4}:

**represents a unit vector in the**

*r̂***direction.**

*r**μ*

_{0}.” The result is as simple as it possibly could be:

The current-segment law can also be used to discover a formula for the magnetic field generated by a long straight current-carrying wire, attributed to Biot and Savart. That leads to the integral form of Ampere’s law, where the same constant appears. Those derivations can be found in most calculus-level physics textbooks, so they need not be repeated here. Again, it is important for students to *discover* them with careful guidance, and to use one or both of those equations to figure out what SI units *μ*_{0} must have.

Again, the steps leading to those discoveries are explained clearly in most textbooks, but it is best to lead students through those steps by the Socratic method, rather than simply copying them from a book.

**” in a direction perpendicular to the uniform electric field “**

*v***” between its plates, i.e., parallel to the planes of the plates. An imaginary rectangle that is stationary relative to the observer lies in a plane parallel to the plates and between them. In the truck’s frame of reference, that rectangle moves backward. As it enters the uniform electric field, we find an increasing electric flux penetrating the plane of the rectangle. In the observer’s frame of reference, that flux’s increase rate multiplied by**

*E**ε*

_{0}constitutes a “pseudocurrent,” also known as a “displacement current.” With Ampere’s law, we conclude that in the observer’s frame of reference, an induced magnetic field must exist between the plates, and that field must be perpendicular to both the electric field and the velocity of the truck. We conclude that a moving electric field creates an induced magnetic field, and its strength and direction can be predicted with a simple vector equation involving

*ε*

_{0}and the cross product of the velocity and electric field:

**,” is perpendicular to**

*B***, and is moving in a direction perpendicular to itself and**

*B***so that it cuts field lines like a blade cutting grass. Since we have already discovered that a magnetic force acts on a charged particle moving through a magnetic field (by using a magnet to move the dot on an oscilloscope screen and then by using a current balance to measure the magnetic force on a current-carrying wire), it is not difficult to discover a formula for the magnetic force on a moving electron:**

*B***and**

*v***:**

*B**E*/

*c*, so the energy densities of the electric and magnetic fields in such a wave are equal. A formula for their sum is easy to create:

Students are then asked to imagine an electromagnetic wave encountering a conducting wall in a plane perpendicular to the wave velocity. The electric component of the wave causes electrons in the wall to accelerate, and the magnetic component exerts a force on those moving electrons. Students figure out the directions of those forces with the usual hand rules. It turns out that the magnetic force pushes the electrons in the direction of the wave velocity. Since the electrons are trapped in the wall, we conclude that the wave pushes the wall. Therefore, the wave must have momentum.

(It’s true that this “hand-waving” argument ignores stray fields, but that problem can be avoided by using a Rowland ring instead of a box.)

The steps above require many weeks, and some go beyond what is expected in preparation for the Advanced Placement exam. For classes who could handle it, I chose to have students make the simple discovery described at the beginning of this article. Discovering Einstein’s most famous equation in this way is something that students will never forget, even if the steps described above are not all perfectly rigorous. It is also a stepping stone to the special theory of relativity.

## REFERENCES

*Am. J. Phys.*

**Arthur Hovey** *earned a BA in physics and mathematics (1964) and an MA in teaching from Yale University in 1966, and retired from physics teaching in 2000*.