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Physics Lesson 16.13.2 - Energy Density of a Magnetic Field

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Welcome to our Physics lesson on Energy Density of a Magnetic Field, this is the second lesson of our suite of physics lessons covering the topic of Energy Stored in a Magnetic Field. Energy Density of a Magnetic Field. Mutual Induction, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

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
Physics Tutorials: This image provides visual information for the physics tutorial Energy Stored in a Magnetic Field. Energy Density of a Magnetic Field. Mutual Induction

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
= (μ0 ∙ 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
= (μ0 ∙ 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.

Physics Tutorials: This image provides visual information for the physics tutorial Energy Stored in a Magnetic Field. Energy Density of a Magnetic Field. Mutual Induction

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
(μ0 = 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

You have reached the end of Physics lesson 16.13.2 Energy Density of a Magnetic Field. There are 3 lessons in this physics tutorial covering Energy Stored in a Magnetic Field. Energy Density of a Magnetic Field. Mutual Induction, you can access all the lessons from this tutorial below.

More Energy Stored in a Magnetic Field. Energy Density of a Magnetic Field. Mutual Induction Lessons and Learning Resources

Magnetism Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial		Revision
Notes		Revision
Questions
16.13Energy Stored in a Magnetic Field. Energy Density of a Magnetic Field. Mutual Induction
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
16.13.1Energy Stored in a Magnetic Field
16.13.2Energy Density of a Magnetic Field
16.13.3Mutual Induction

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