There is a current of 1.2 amps in a
coaxial cable whose outer radius is five times its inner radius. The copper in the coaxial cable has
a magnetic permeability 𝜇 equals 1.26 times 10 to the negative sixth henrys per
meter. What is the magnetic field energy
stored in a 3.0-meter length of the cable?
If we look at a cross-section of
our coaxial cable, we’re told that the inner radius, the radius of the inner
conducting core, is one-fifth as large as the outer radius of the cable. We know that a coaxial cable works
by having current move in one direction in the central core and then looping back to
move in the opposite direction in the outer conductor.
If we looked at this cross-section
end on, we would see current coming out of the page towards us in the core and
moving away from us into the page at the outer conductor. When it comes to the magnetic field
produced by a coaxial cable, that field is equal to zero inside the conducting
core. And it’s also equal to zero outside
the cable. The only place the magnetic field
is nonzero is in between the central conducting core and the outer conductor. It’s the energy of that field in a
3.0-meter length of the wire carrying 1.2 amps of current with a given magnetic
permeability that we want to calculate.
Energy stored within a magnetic
field, we’ve called it 𝐸 sub 𝐵, can be thought of as energy due to induction, 𝐸
sub 𝐿. In particular, the energy in a
magnetic field is equal to one-half the self-induction coefficient of the object
creating the field, multiplied by the current running through that object
squared. This equation tells us that if we
wanna calculate the energy stored in the magnetic field, and we do, then we’ll need
to solve for the self-induction coefficient of this coaxial cable.
Remember that the self-induction of
an object is its capacity for a changing magnetic field within itself to induce a
changing current. Or put another way, given a
time-varying magnetic field within an object, how likely is that field to induce an
EMF in the object? For a coaxial cable like the one we
have here, there’s a mathematical relationship for the self-induction coefficient
𝐿. For a coaxial cable, that
self-induction coefficient is equal to the magnetic permeability of the cable
multiplied by its length divided by two 𝜋, all multiplied by the natural logarithm
of the outer radius of the cable divided by its inner radius.
Now in our case, we’re told that
the length of the coaxial cable we’ll work with is 3.0 meters long and that its
magnetic permeability is 1.26 times 10 to the negative sixth henrys per meter. And we’re also told that the outer
radius of the coaxial cable divided by the inner radius is equal to five. That is, the outer radius is five
times the inner radius. All this means that we can write
the self-induction coefficient 𝐿 of our coaxial cable as 1.26 times 10 to the
negative sixth henrys per meter multiplied by 3.0 meters divided by two 𝜋, all
multiplied by the natural logarithm of five.
Before we calculate this value,
recall that it’s not our final answer. Our final answer will be the
magnetic field energy stored in this cable. If we take the self-induction
coefficient and multiply that by one-half and then multiplied that by the current in
the coaxial cable squared, then we’ll have the value we want, the magnetic field
energy. So rather than making two
calculations, we’ll take our expression for self-induction coefficient 𝐿 and insert
it into this equation for the magnetic field energy. Note that we’ve used the value of
1.2 amps for our current.
When we calculate this expression
for a magnetic field energy, to two significant figures, we find a result of 7.0
times 10 to the negative seventh joules. That’s how much energy is stored in
the magnetic field of a 3.0-meter length of this cable.