*Notes to a video lecture on http://www.unizor.com*

__Magnetic Field Energy__

Our task is to determine the energy carried by a magnetic field. More precisely, a density of a potential energy as a function of a local characteristic of a magnetic field - the value of its

*intensity vector*

*.*

**B**First, we have to choose a device where a uniform magnetic field can be created and evaluate the total amount of energy we have to spent to create this uniform magnetic field. This energy will be stored in a certain volume of space inside this device occupied by a created magnetic field as its potential energy.

Then, dividing the total amount of energy of the magnetic field inside this space by the space volume, we will get a density of the potential energy of the magnetic field.

Recall the description of an infinitely long solenoid as a device that produces relatively uniform magnetic field. It was described in the lecture "Electromagnetism - Magnetism of Electric Current - Solenoid" of this course "Physics 4 Teens" on UNIZOR.COM.

The absolute value of an intensity of a uniform magnetic field

*inside such a solenoid, was expressed as*

**B**

**B = μ·n·I**where

*is*

**μ***permeability*of the medium inside a solenoid,

*is the density of the wire loops making up a solenoid and*

**n***is the electric current running through a solenoid.*

**I**As physicists usually do, we assume that our real solenoid, though not infinite, is long enough to use this formula for absolute value of a magnetic field intensity at all points inside a solenoid and it's zero outside.

While this methodology depends on the fact that the magnetic field we analyze is inside a solenoid, the final formula for energy density will depend only on the field's intensity and, as a field's local characteristic, will be the same, no matter what is the outside source of a magnetic field, whether it's a solenoid or a few permanent magnets, or the result of electromagnetic oscillations far from the source of these oscillations.

Consider a battery of voltage

*that we connect to a solenoid of resistance*

**V***to create a magnetic field inside it. Electric current*

**R***through a solenoid in the beginning equals to zero and, gradually, rises to some value*

**I(t)***(Ohm's Law).*

**I**_{max}=V/RSince the electric current varies with time from

*to*

**0***, the above formula for magnetic field intensity is time-dependent:*

**I**_{max}

**B(t) = μ·n·I(t)**Recall that changing electric current in a wire causes changing magnetic field around it that adversely affects the current, resisting its change. This is a self-induction effect described in the chapter "Electromagnetism - Self-Induction" of this course.

Therefore, the process of rising of an electric current through a solenoid requires some work performed by the battery against self-induction of the solenoid.

This work is converted into energy of the magnetic field inside a solenoid, and we will quantitatively evaluate it.

According to Faraday's Law of Induction, the electromotive force (EMF)

*generated against the increasing electric current in a wire loop is proportional to a rate of change of the magnetic flux*

**U***flowing through this single loop:*

**Φ**_{1}

**U(t) = −**d**Φ**d_{1}(t)/**t**Magnetic flux is related to a field intensity in a single loop of wire of area

*as*

**A**

**Φ**_{1}(t) = B(t)·AFor a solenoid with total number of wire loops

*the flux will be*

**N***times greater:*

**N**

**Φ**_{N}(t) = B(t)·A·NUsing the above stated dependency between field intensity

*and the electric current*

**B(t)***, we can express the magnetic flux through a solenoid as*

**I(t)**

**Φ**_{N}(t) = μ·n·I(t)·A·NObviously, if the length of a solenoid is

*and the density of the wire loops is*

**l***, the total number of wire loops is*

**n**

**N = n·l**Therefore,

**Φ**_{N}(t) = μ·n²·I(t)·A·lIncidentally, the expression

*is a volume of space inside a solenoid*

**A·l***, so we write the above formula as*

**V**_{space}

**Φ**_{N}(t) = μ·n²·I(t)·**V**_{space}The power

*at any moment of time*

**P(t)***needed to run the electric current (that is, the amount of work per unit of time) depends on the voltage, that is EMF*

**t***needed for it and the electric current at the time*

**U(t)***as*

**I(t)**

**P(t) = U(t)·I(t)**Using the expression above for the EMF, we derive an absolute value of power as a function of time

*[*

**P(t) =**d**Φ**d_{N}(t)/**t·I(t) =**

=d=

**/**d**t***]*

**μ·n²·I(t)·V**_{space}*[*

**·I(t) =**

= μ·n²·V= μ·n²·V

_{space}·*d*]

**I(t)/**d**t**

**·I(t)**The constants in front form a characteristic of a solenoid called

*inductance*

*, so we simplify the expression for power as*

**L=μ·n²·V**_{space}*[*

**P(t) = L·I(t)·***d*]

**I(t)/**d**t**During the time period from

*to*

**t***the electric current will rise by*

**t+**d**t***d*[

**I=***d*]

**I(t)/**d**t***and the work performed will be*

**·**d**t***d*.

**W=P(t)·**d**t**Therefore, to increase the electric current from

*to*

**I***we have to perform work equal to*

**I+**d**I***d*

**W = L·I·**d**I**The total amount of work

*we have to perform to rise the electric current from*

**W***to*

**0***can be obtained by integrating the above expression by*

**I**_{max}*:*

**I**

**W = ∫**_{[0,Imax]}

**L·I·**d**I = ½L·I²**_{max}When the electric current reaches

*, the magnetic intensity*

**I**_{max}*reaches its maximum value, and the potential energy accumulated in the magnetic field will be equal to work performed.*

**B**From the above formula for magnetic field intensity

*follows that*

**B=μ·n·I**

**I = B/(μ·n)**Substituting this into an expression for total work performed,

**W = ½L·B²/(μ·n)² =**

= ½·μ·n²·V

= ½·V= ½·μ·n²·V

_{space}·B²/(μ·n)² == ½·V

_{space}·B²/μDividing the total work performed

*(that is, the total potential energy accumulated by a solenoid during the process of rising the electric current) by the total volume inside a solenoid*

**W***, we obtain the potential energy density*

**V**_{space}*of a magnetic field as a function of a local field characteristic - its intensity*

**P**_{M}*:*

**B**

**P _{M} = B²/(2μ)**

*.*

**B**Even if the field is of a more complex kind (variable, non-uniform, infinite etc.), since only its intensity participates in the above expression for a potential energy density, the formula should be valid.

In particular, it's valid for a magnetic component of an oscillating electromagnetic field.

Combining the result for an electric component of electromagnetic field presented in the previous lecture with the formula above for magnetic component, we obtain a total potential energy density of an electromagnetic field

* P_{E+M} = ½*[

*]*

**ε·E² + B²/μ**Final comment about an

**oscillating electromagnetic field**.

Once formed at the source, variable electric field possess certain potential energy.

Where is it going if the source stopped producing it?

Since variable electric field produces variable magnetic field, this energy is transferred to a newly formed variable magnetic field into its potential energy.

In turn, this potential energy of a variable magnetic field is used to generate a new variable electric field with its potential energy, which it uses to generate a new variable magnetic field etc.

That's how energy is delivered by an electromagnetic oscillations to some final destinations, where it might be used to convert it to some other forms.

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