# Energy Stored in a Magnetic Field. Energy Density of a Magnetic Field. Mutual Induction

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16.13Energy Stored in a Magnetic Field. Energy Density of a Magnetic Field. Mutual Induction

In this Physics tutorial, you will learn:

• What is magnetic potential energy? Where is it stored?
• How is the law of energy conservation applied in magnetic fields?
• What is the rate of magnetic potential energy change?
• What is the energy density of magnetic field? How is it calculated?
• What is mutual induction? How is it related to the induced current and induced emf of the two respective circuits?

## Introduction

Do you do any work when moving a nail away from a magnet? What does this mean regarding the energy consumed? Is there any energy stored in the nail due to the magnetic field?

Is the magnetic field produced by a bar magnet equally strong everywhere around the magnet? How do you know this? What visual representation do you use to have a better understanding about the magnetic field strength in a certain region of space?

Do you think a RL circuit stores energy? Where do you base your answer?

Do you think two RL circuits affect the operation of each other when they are very close? Why?

In this tutorial, we will discuss more extensively about some properties of magnetic field such as energy stored in it and the density of this energy, especially in RL circuits, as the most flagrant example of interaction between electricity and magnetism. In addition, how two RL circuits placed near each other affect the operation of each other.

## Energy Stored in a Magnetic Field

In Section 14, we have seen that two opposite charges attract each other, so we must do work to prevent them from colliding. On the other hand, we must do work to move two like charges close to each other as well. This is because like charges repel each other and they react when we try to bring them closer. We explained that this occurs because the charges possess some energy stored in them (electric potential energy), which we must overcome by doing work from outside.

The same thing occurs when bringing two magnets near each other. If we bring the opposite poles of two magnets near each other, they are attracted and we must do some work to move them apart. This means the system of magnets is storing energy in the form of magnetic potential energy.

Unlike in electricity, when studying the magnetic potential energy, we deal with currents instead of electric charges. Therefore, it is more suitable to analyse the magnetic potential energy using electromagnets, rather than simple permanent magnets. More precisely, RL circuits are the best systems for this purpose as they involve both electricity (through resistors) and magnetism (through inductors). From the last tutorial "RL Circuits", we know that the equation that describes the relationship between the emf's (of the source and of inductor) and potential energy (of resistor), is

ε = L ∙ di/dt + i ∙ R

The above equation derives from the law of energy conservation, since any change in the above values is associated with the production or consumption of some energy (emf = work / charge).

When multiplying all terms of the above equation by i (to deal with power), we obtain

ε ∙ i = L ∙ i ∙ di/dt + i2 ∙ R (1)

Let's explain what does each term of the above equation represent.

1. The term ε ∙ i in the left side represents the total power delivered by the source, i.e. the total energy produced by the battery in the unit time. Thus, if a charge dQ passes through the battery in the time interval dt, the battery does the work ε ∙ dQ on the charge during the given time. The rate at which the battery is doing work (that is the total power delivered by the battery), is
Ptot = ε ∙ dQ/dt
In other words, the term on the left side of the equation (1) represents the rate at which the source delivers energy to the rest of the circuit.
2. The i2 ∙ R term represents the rate at which the energy appears (and is consumed) in the resistor in thermal form.
3. The first term due right of equality sign in equation (1) represents the rest from the total energy produced by the battery (that is not delivered as thermal energy) in the unit time. Obviously, this must be the magnetic potential energy in the unit time, which is stored in the magnetic field of the inductor. If we denote the magnetic potential energy by Wm, we obtain for the rate of magnetic potential energy stored in the magnetic field of the inductor:
dWM/dt = L ∙ i ∙ di/dt
Multiplying both sides of the last equation by dt, we obtain
dWM = L ∙ i ∙ di
Integrating both sides, we obtain
W M0d WM = i0L ∙ i ∙ di
Taking the inductance of solenoid as constant, we obtain
W M0d WM = L ∙ i0i ∙ di
WM = L ∙ i2/2
The last equation gives the (magnetic) potential energy stored in an inductor L when a current I flows through it. This equation is similar to that of the energy stored in a capacitor
Wc = Q2/2C
where the inductance L of inductor is analogue to the inverse of capacitance 1/C of capacitor and the current I flowing through the inductor is analogue to the charge Q stored in the capacitor.

### Example 1

A 400 mH inductor and a 30 Ω resistor are connected in series to a circuit supplied by a 24 V battery, as shown in the figure. 1. What is the energy stored in the magnetic field produced by the inductor after a long time of circuit's operation?
2. How long does it take to the resistor to dissipate in the form of heat an amount of energy equal to the energy stored in the magnetic field of inductor?

### Solution 1

Clues:

ε = 24 V
L = 400 mH = 0.4 H
R = 30 Ω
Wm = ?
t = ? (WR = Wm)

1. First, we calculate the current in in the circuit. Giving that after a long time of circuit's operation the inductor behaves simply as a conducting wire, we assume the total resistance in the circuit is only due to the resistor. Therefore, using the Ohm's Law yields
i = ε/R
= 24 V/30 Ω
= 0.8 A
The energy stored in the magnetic field produced by the inductor therefore is
WM = L ∙ i2/2
= (0.4 H) ∙ (0.8 A)2/2
= 0.128 J
2. We denote by WR (instead of Q) the heat energy dissipated by the resistor, in order to avoid confusion between symbols (as in this tutorial we have expressed the electric charge by Q). From the Joule's Law, we have
Wr = i2 ∙ R ∙ t
Since WR = Wm = 0.128 J, we obtain for the time t in which the total heat dissipated by the resistor equals the heat stored in the inductor:
t = WR/i2 ∙ R
= (0.128 J)/(0.8 A)2 ∙ (30 Ω)
= 0.00666s
This result means that most of heat produced by the source is dissipated in the form of heat energy by the resistor. Only a small amount of the total heat is stored in the magnetic field of the inductor.

## Energy Density of a Magnetic Field

Let's consider again a solenoid (inductor) connected to a source of electricity, similar to those discussed in the previous paragraphs.

From definition of inductance, we obtained in tutorial 16.9 "Inductance and Self-Induction," we found the expression for the self-inductance L produced in a solenoid of length L, and number of turns N:

L = μ0 ∙ N2 ∙ A/I

If we denote by n the number of turns per unit length, we obtain for the inductance of the solenoid:

L = μ0 ∙ n2 ∙ A ∙ l

Thus, the inductance per unit length near the middle of solenoid is

L/I = μ0 ∙ n2 ∙ A It is known that the magnetic field is stored only inside the inductor; we have seen in previous tutorials that the magnetic field outside a solenoid is close to zero. In addition, this magnetic field is uniformly distributed throughout the volume of solenoid, which corresponds to the volume of a cylinder (V = A ∙ l). Hence, we obtain for the energy per unit volume w stored in the solenoid

w = WM/V = WM/A ∙ l

Giving that

WM = L ∙ i2/2

we obtain for the energy per unit volume stored in the inductor, which represents the energy density of magnetic field:

w = L ∙ i2/2 ∙ A ∙ l
= 0 ∙ n2 ∙ A ∙ l) ∙ i2/2 ∙ A ∙ l
= μ0 ∙ n2 ∙ i2/2

In tutorial 16.2, we have explained that the magnetic field B inside a solenoid is

B = μ0 ∙ N ∙ i

Hence, we obtain for energy density of a magnetic field:

w = μ0 ∙ n2 ∙ i2/2
= μ02 ∙ n2 ∙ i2/2 ∙ μ0
= 0 ∙ N ∙ i)2/2 ∙ μ0

Thus,

w = B2/2 ∙ μ0

The above equation is true not only for solenoids but for all types of magnetic fields.

### Example 2

A 30 cm long solenoid containing 120 turns is connected to a 48V source in which is also connected a 6 Ω resistor. #### Calculate:

1. The magnetic field inside the solenoid
2. The energy density of magnetic field inside the solenoid

### Solution 2

Clues:

l = 30 cm = 0.30 m
N = 120
ε = 48 V
R = 6 Ω
0 = 4π × 10-7 N/A2)
B = ?
w = ?

1. First, we find two useful quantities, the number of turns in the solenoid per unit length N and the current I in the circuit. We have
n = N/I
= 120 turns/0.30 m
= 400turns/m
and
i = ε/R
= 48 V/6 Ω
= 8 A
The magnetic field B of the solenoid is
B = μ0 ∙ N ∙ i
= (4 ∙ 3.14 × 10-7 N/A2 ) ∙ (400 turns/m) ∙ (8 A)
= 4.0192 × 10-3 T
2. The energy density of magnetic field produced inside the solenoid is
w = B2/2 ∙ μ0
= (4.0192 × 10-3 T)2/2 ∙ 4 ∙ 3.14 × 10-7 N/A2
= 6.43 J/m3

## Mutual Induction

Let's recall the experiment explained in the tutorial 16.7 in which two loops are near each other and a current is induced in one of the loops when turning the switch ON or OFF in the circuit containing the other loop. In addition, the arrow symbol in the resistor means it is a rheostat (variable resistor). As such, the value of current in the circuit can changed not only by turning the switch ON or OFF but also by changing the values of resistance in the rheostat. In this way, we can change the magnitude not only of the induced current but of the corresponding magnetic field as well. The process of producing a current through a variable magnetic field is called induction, as explained in precedent tutorials. The induction by which electric current is produced in one coil by changing the magnetic field of the other coil requires the presence of two coils. If one coil is moved away, no current is induced in the other coil due to the long distance. Therefore, the current (and the resulting magnetic field) in one coil produced by this mutual interaction is known as mutual induction.

The mutual induction differs from the self-induction of an inductor, as in the case of inductor the presence of a single solenoid is enough to induce a magnetic field inside it.

When the variable resistor in the figure above is set in a fixed position R, the source produces a steady current I in the circuit. As a result, a magnetic field (and flux) are produced in the coil on the left. The mutual inductance of the coil 2 (due right) on the coil 1 (due left) is denoted by m21. It is calculated by

M21 = N2 ∙ Φ21/i1 (2)

In our figure we have N1 = N2 = 1 as both coils have one turn only, but if there are kore turns, we must write their number accordingly.

The above equation has the same form as the equation of inductance for a single coil

L = N ∙ Φ/i

explained in the tutorial 16.9.

We can write the equation (2) as

M21 ∙ i1 = N2 ∙ Φ21

If we slightly change the value of resistance R, we obtain a variation of current, so we can write

M21di1/dt = N221/dt

From the Faraday's Law, the right side of the above equation represents the emf induced in the coil 2 due to the change in current in the coil 1. Since it opposes the current I1, we obtain

ε2 = -N221/dt = -M21di1/dt

This reasoning can be used for the emf induced in the first coil as well (due to the law of conservation of energy). Therefore, we can write

ε1 = -N112/dt = -M12di2/dt

Experiments show that m12 = m21. Thus, we can write the mutual inductance simply by M. in this way, the two above equations become

ε1 = -M ∙ di2/dt

and

ε2 = -M ∙ di1/dt

### Example 3

A single loop is connected to a 24V battery and a rheostat. The position of rheostat shifts from 20Ω to 8Ω in 2 seconds. This brings an induced current in the second coil, which contains 400 turns. Each coil has an area of 12 cm2. Calculate:

1. The mutual inductance if the induced emf in the second coil at the end of process is 120V
2. The value shown by the ammeter at the end of process
3. The change of magnetic flux in the second coil
4. The change in the induced magnetic field in the second coil ### Solution 3

Clues:

ε1 = 24 V
Ri = 20 Ω
Rf = 8 Ω
dt = 2 s
A1 = A2 = A = 12 cm2 = 0.0012 m2
B1 = 400 mT = 0.4 T
N1 = 1
N2 = 400
ε2 = 200 V
a) M = ?
b) i2 = ?
c) ΔΦ2 = ?
d) ΔB2 = ?

1. First, we calculate the change in current in the first coil due to the change in resistance. We have
i1(initial) = ε/R1
= 24 V/20 Ω
= 1.2 A
and
i1(final) = ε/R2
= 24 V/8 Ω
= 3 A
Thus,
di1 = i1(final) - i1(initial)
= 3 A - 1.2 A
= 1.8 A
From theory, we know that
ε2 = -M ∙ di1/dt
Ignoring the negative sign and substituting the values, we obtain for the mutual inductance:
M = ε2 ∙ dt/di1
= (200 V) ∙ (2 s)/(1.8 A)
= 222 H
2. We use the equation
ε1 = -M ∙ di2/dt
to calculate the value shown by the ammeter (which is the value of i2, i.e. the value of current induced in the second coil), where M is the mutual inductance found at (a). Thus, ignoring again the negative sign, we obtain
di2 = ε1 ∙ dt/M
= (24 V) ∙ (2 s)/(222 H)
= 0.216 A
3. Giving that
ε2 = -N221/dt
we obtain for the change in the magnetic flux in the second coil (the magnitude only):
21 = ε2 ∙ dt/N2
= (200 V) ∙ (2 s)/400
= 1 Wb
4. Since for a fixed area of loop any change in magnetic flux is due to the change in the magnetic field, we have
∆Φ21 = ∆B2 ∙ A
∆B2 = ∆Φ21/A
= 1 Wb/0.0012 m2
= 833.3 T

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