Resource Lesson
Ampere's Law
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An electric current is the flow of charge past a point in a circuit.
It is measured in a unit called an
ampere
, or amp, which is equivalent to a flow rate of one coulomb or charge each second. When discussing this rate at which charges flow, we need to mention that once an electric field is established across a wire, each individual charge actually drifts at a very slow (less than 1 mm/sec) speed along the wire. It is the very high number of charges moving at any one time that creates the "larger current".
1 amp = 6.25 x 10
^{18}
electrons/sec
Within a conductor, it is often useful to define current in terms of its
current density
, or
j
. The units on
j
are amps/m
^{2}
. Current is sometimes referred to as "the flux of current density" over an cross-sectional area.
Find the magnitude of the electric current flowing through a circular wire of radius 2 mm if it carries a uniform current density of 0.5 A/m
^{2}
.
Ampere's Law
No only do charged particles moving through an external magnetic field experience a magnetic force according to
they also create magnetic fields.
Gauss' Law
stated that the net flux from a closed surface was equal to the ratio of the enclosed charge divided by the permittivity of free space,
=8.85 x 10
^{-12}
C
^{2}
/Nm
^{2}
.
This statement reflects that positive charges are "sources" of flux lines and negative charges are "sinks" for flux lines allowing an unequal number of field lines to either originate or terminate inside of a closed surface.
Gauss' Law for Magnetism
states that the net magnetic flux through a closed surface is zero as there are no such things as monopoles. Since each magnet has both a north and a south pole, any magnet enclosed within a closed surface would have the same number of flux lines both exiting and entering the surface - the net flux would be zero.
Ampere's Law
says that if we replace the closed surface integral with a closed line integral then the magnetic field multiplied by the length of the curve will equal the sum of the enclosed currents times the permeability of free space, µ
_{o}
= 4pi x 10
^{-7}
N/A
^{2}
.
Remember that a dot product (
) reflects that we must look for the component of B along the path; that is, we are interested in choosing an Amperian path that flows in parallel to the magnetic field B so that theta equals either 0º or 180º.
Refer to the following information for the next question.
The Amperian loop circulates counterclockwise. Each current into the plane of the page has magnitude of 2A and each current out of the plane of the page has a magnitude of 3A.
Evaluate
.
Magnetic Fields Around a Current-Carrying Wire
Now let's use Ampere's Law to determine the strength of the magnetic field a distance
r
from a wire carrying a current
I
.
image courtesy of John Hopkins University
Physics Lecture Demonstrations
Suppose we needed to investigate the strength of the magnetic field both inside and outside of this wire.
Let's assume that the wire has a radius of
a
and a uniform current density
j
(shown in the diagram as a solid grey area with the current flowing in the +z direction), allowing us to express the current as
If we once again examine the magnetic field outside of the wire at a distance r > a, then we get the expression
If we now examine the magnetic field within the wire at a distance r < a, then we get the expression
Combining these two equations graphically shows us
two general "shape" combinations in this graph that are reminiscent to those for an electric field of a uniformly charged insulator as well as a gravitational field of a solid sphere having a uniform density (except that the later two are inverse square relationships when r > a).
Magnetic Fields Within a Solenoid
Recall that an ideal solenoid is a tightly turned coil of wire which produces a
uniform magnetic field
down its center. We will now use Ampere's Law to calculate the magnitude of this central magnetic field.
image courtesy of John Hopkins University
Physics Lecture Demonstrations
Beginning on the left side we will use the Amperian path ABCDA.
where B represents the strength of the central magnetic field and
represents the length of the line
CD
.
Completing our solution, we now have
where the ratio
represents the total number of loops in the solenoid divided by the length of the solenoid. This value is a constant represented by the variable
n
.
Magnetic Fields Within a Toroid
Now let's wrap a solenoid into a circle to form a toroid and examine its magnetic field.
image courtesy of Oberlin College
Demonstration Website
Once again, we will use Ampere's Law to calculate the magnetic field. To begin we will place an Amperian loop along the center of the toroid's coils so that its radius,
r
, lies between the inner and outer radii of the torus, a < r < b.
For r < a and r > b, B = 0. In the case of r < a, no currents are enclosed within the Amperian loop. For r > b, the currents flowing "in" (-z) and the currents flowing "out" (+z) cancel, giving a net enclosed current of zero.
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