Explain how energy can be stored in a magnetic field
Derive the equation for energy stored in a coaxial cable given the magnetic energy density
The energy of a capacitor is stored in the electric field between its plates. Similarly, an inductor has the capability to store energy, but in its magnetic field. This energy can be found by integrating the
magnetic energy density ,
over the appropriate volume. To understand where this formula comes from, let’s consider the long, cylindrical solenoid of the previous section. Again using the infinite solenoid approximation, we can assume that the magnetic field is essentially constant and given by
everywhere inside the solenoid. Thus, the energy stored in a solenoid or the magnetic energy density times volume is equivalent to
Although derived for a special case, this equation gives the energy stored in the magnetic field of
any inductor. We can see this by considering an arbitrary inductor through which a changing current is passing. At any instant, the magnitude of the induced emf is
so the power absorbed by the inductor is
The total energy stored in the magnetic field when the current increases from 0 to
I in a time interval from 0 to
t can be determined by integrating this expression:
Self-inductance of a coaxial cable
[link] shows two long, concentric cylindrical shells of radii
and
As discussed in
Capacitance on capacitance, this configuration is a simplified representation of a
coaxial cable . The capacitance per unit length of the cable has already been calculated. Now (a) determine the magnetic energy stored per unit length of the coaxial cable and (b) use this result to find the self-inductance per unit length of the cable.
Strategy
The magnetic field both inside and outside the coaxial cable is determined by Ampère’s law. Based on this magnetic field, we can use
[link] to calculate the energy density of the magnetic field. The magnetic energy is calculated by an integral of the magnetic energy density times the differential volume over the cylindrical shell. After the integration is carried out, we have a closed-form solution for part (a). The self-inductance per unit length is determined based on this result and
[link] .
Solution
We determine the magnetic field between the conductors by applying Ampère’s law to the dashed circular path shown in
[link] (b). Because of the cylindrical symmetry,
is constant along the path, and
This gives us
In the region outside the cable, a similar application of Ampère’s law shows that
, since no net current crosses the area bounded by a circular path where
This argument also holds when
that is, in the region within the inner cylinder. All the magnetic energy of the cable is therefore stored between the two conductors. Since the energy density of the magnetic field is
the energy stored in a cylindrical shell of inner radius
r , outer radius
and length
l (see part (c) of the figure) is
Thus, the total energy of the magnetic field in a length
l of the cable is
where
L is the self-inductance of a length
l of the coaxial cable. Equating the previous two equations, we find that the self-inductance per unit length of the cable is
Significance
The inductance per unit length depends only on the inner and outer radii as seen in the result. To increase the inductance, we could either increase the outer radius
or decrease the inner radius
. In the limit as the two radii become equal, the inductance goes to zero. In this limit, there is no coaxial cable. Also, the magnetic energy per unit length from part (a) is proportional to the square of the current.
Check Your Understanding How much energy is stored in the inductor of
[link] after the current reaches its maximum value?
At the instant a current of 0.20 A is flowing through a coil of wire, the energy stored in its magnetic field is
What is the self-inductance of the coil?
Suppose that a rectangular toroid has 2000 windings and a self-inductance of 0.040 H. If
, what is the current flowing through a rectangular toroid when the energy in its magnetic field is
Solenoid
A is tightly wound while solenoid
B has windings that are evenly spaced with a gap equal to the diameter of the wire. The solenoids are otherwise identical. Determine the ratio of the energies stored per unit length of these solenoids when the same current flows through each.
A 10-H inductor carries a current of 20 A. How much ice at
could be melted by the energy stored in the magnetic field of the inductor? (
Hint : Use the value
for ice.)
A coil with a self-inductance of 3.0 H and a resistance of
carries a steady current of 2.0 A. (a) What is the energy stored in the magnetic field of the coil? (b) What is the energy per second dissipated in the resistance of the coil?
A current of 1.2 A is flowing in a coaxial cable whose outer radius is five times its inner radius. What is the magnetic field energy stored in a 3.0-m length of the cable?
A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?