Inside the wire, the pictured horizontal Ampere loop encloses no
current. By symmetry, any magnetic field induced inside would depend
only on radius s, and would point in the 
direction,
since each vertical bit of current induces a 
field
circulating about it in a righthanded sense, which would add to give a
lefthanded circular field in the center. Thus 
is constant along the Ampere loop, with Ampere's law giving
Outside the wire, by symmetry, any magnetic field induced depends
only on radius s, and points in the 
direction,
since each vertical bit of current induces a field
circulating about it in a righthanded sense, which adds to give a
righthanded circular field outside the wire. Here the current enclosed by our horizontal Ampere loop is exactly the wire's total current I. Thus Ampere's law gives
Thus the wire with current density J uniformly distributed over its outside surface induces magnetic field
Inside the wire (for 
), the pictured horizontal Ampere loop
encloses only some of the wire's current. The current enclosed is
We can set the proportionality constant k by requiring the total current carried by the entire wire to be I:
Thus the current enclosed at s;SPMlt;a is
Again by symmetry, any magnetic field induced inside depends only on
radius s and points in the direction. Thus is constant along the Ampere loop, with Ampere's law
giving
Outside the wire, again by symmetry, any magnetic field induced depends
only on radius s, and points in the direction. Here the current enclosed by our horizontal Ampere loop is exactly the wire's total current I. Thus Ampere's law gives
Thus the wire with current density J = ks induces magnetic field
Also note that these two cylindrically symmetric examples, which differ only in how they distribute current within a wire, necessarily lead to the same magnetic field outside the wire; they can differ only in the magnetic field induce inside the wire.
While we could derive this by Ampere's law, we've already derived the
magnetic field induced by a solenoid using Ampere's law in the
examples: vanishes outside the solenoid, and inside is
, where 
is found from the
tangential solenoid current via a right hand rule (it is the axis the
tangential current flows counterclockwise -- or righthanded -- about). In
our case, the magnetic field induced by the inner solenoid is
while that induced by the outer solenoid is
where 
points to the right on the page.
We get the magnetic field induced by the two solenoids together by superposition,
Labeling the bar's distance from the left edge of the circuit by x, the current loop contains magnetic flux
since is perpendicular to the
plane of the current loop. The induced emf is just
and since this induced emf is that voltage source for the induced current,
The minus
sign means that the current flows opposite the righthanded sense
determined by the current loop's normal. We used into the page as the
normal (the direction for 
). This determines a positive
current as clockwise on the page (righthanded about the axis into the
page), and a negative current as counterclockwise on the page. Thus our current is
Note that this counterclockwise induced current also agrees with Lenz' law: the growing current loop has growing magnetic flux into the page, so the induced current flows counterclockwise, to induce opposing magnetic flux out of the page.
where 
points along the current, which flows up through the bar. up, crossed with into the page, determines a Lorentz force
This induced magnetic force also makes sense via Lenz' law: the motion of the bar to the right increases magnetic flux into the page. The induced current creates a magnetic force on the bar which opposes its rightward motion. Without an external pulling force in this situation, nature really would put an end to the changing magnetic flux, by slowing and eventually stopping the bar.
Note that the current loop determines a normal 
which is
initially aligned with , but then rotates counterclockwise to
point at an angle 
, relative to . Thus the flux through the current loop at time t is
since 
is constant over the current loop's area of 
.
The induced emf is just
where is the axis that the solenoid's current flows
counterclockwise about. Since this magnetic field is changing, a current loop placed in the field contains changing magnetic flux and thus has an induced emf given by Faraday's law:
To find this induced electric field we consider circular current loops in the xy-plane, first inside, then outside, the solenoid. In either case, by symmetry, the 
-field points tangentially, in the -direction, and depends only on radius s, remaining constant along the loop integral. Thus in either case
Inside the solenoid, the flux enclosed by the current loop is
since the loop is perpendicular to the constant magnetic field.
Since 
is changing, we have
Thus inside the solenoid, Faraday's law gives
Recalling our assertion 
, this means that inside the solenoid,
Outside the solenoid, the flux enclosed by a circular current loop is
since the constant -field only exists inside the solenoid, over an area 
, where a is the solenoid radius. Again , so
Thus outside the solenoid, Faraday's law gives
or restoring directions,
Thus a solenoid of radius a, with changing current I(t) in the direction, induces an electric field
As shown in problem 5, the induced emf outside the solenoid is
inducing a current through the exterior current loop of
The minus sign means that this current flows in the direction, where points tangentially along
the solenoid current loops. Thus current flows opposite the sense of
the solenoid current loops, or to the right (or out of the page,
depending on how you look at it) through the resistor.
Between the plates we have electric field
directed as shown. Using cylindrical coordinates, 
. This field is increasing as more and more charge is deposited on the capacitor's plates, inducing a tangential -field by Maxwell's displacement current:
For circular Ampere loop at radius s this gives
since B is constant and tangential along the loop integral, and 
is constant and perpendicular to the surface enclosed by the loop. The current loop encloses area 
, and
so Maxwell's displacement current law says
Restoring directions, then,
between the capacitor plates, for s;SPMlt;a.
where 
is the wave vector (pointing in the
direction of propagation; when quantized 
will give the plane
wave's momentum); the frequency 
; is the polarization vector, giving the plane of oscillation of the -field; and 
is a phase angle showing where in its oscillation cycle the field starts, at t=0.
Here we are given the frequency 
, the amplitude 
, and the
phase angle 
, for an electromagnetic wave traveling in the
direction (so 
) and polarized in the
z-direction (so 
). Using
we find
where we have used the evenness of the cosine function. This electromagnetic (light) wave is sketched below:
