1. Lecture 3-7 Magnetic Energy (pg. 43 – 42)
ECT1026 Field Theory
Lecture 3-7
Magnetic Energy
(pg. 43 – 42)

2. An inductor with inductance L is connected to a current source.
3.7 Magnetic Energy
An inductor with inductance L is connected to a current source.
Current I through the inductor increases from zero to final value I1.
Voltage V across the inductor is given by
Power =
The energy is equal to the time integral of power.
Total energy expended in building up the current is:

3. 3.7 Magnetic Energy
Establishing a current I through a loop of wire, a magnetic field is generated in the space around the loop.
The energy expended in establishing the magnetic field is stored in the magnetic field itself.
This energy is called magnetic energy (Wm) and magnitude is LI2/2.

4. Energy Stored in a Solenoid:
3.7 Magnetic Energy
Energy Stored in a Solenoid:
Cross-sectional area: A
Number: N
Length: l
Recall: Inductance of a solenoid is
B inside the solenoid is
(Example 3.5-1) 
Al = volume of solenoid
Magnetic Energy Density
(Joules/m3)

5. Energy Stored in an Inductor: wm =
3.7 Magnetic Energy
Energy Stored in an Inductor:
wm =
This equation valid for any medium with magnetic field H.
Given Magnetic field H, the total magnetic energy stored in a volume is: 1
m0H2 (Joules/m3) 2 1
 m0H2 dv(Joules)
Wm = 2 v

6. Example 3.7-1: 3.7 Magnetic Energy
Calculate the energy stored in the magnetic field of a solenoid (length = 30cm, diameter = 3cm) wound with 1000 turns of wire when carrying a current of 10A.
Solutions:
Recall: For solenoid, L =
Magnetic Energy = LI2 =
m0N2A l 1 1
4p10-7
(10001000)
(p0.0152m2)
(10A)2 2 2
0.3m
= J

7. Rectangular cross-section
3.7 Magnetic Energy
Example 3.7-2:
Use the expression for energy density of the magnetic field to calculate the self-inductance of a toroid with rectangular cross-section:
Inside the core
mN I
B =
2pr
Toroid height: h
Radius: r
Thickness: dr
Toroid showing
Rectangular cross-section

8.  dWm=  Solutions: 3.7 Magnetic Energy
Differential volume dV = 2prh dr, and assume m = m0 (air filled core)
So, energy stored in this elemental volume is
dWm =
Thus, Wm=
Inside the core
mN I
Toroid height: h
Radius: r
Thickness: dr
B =
2pr
1 B2
m0 N2 I2 h dr
dV =
2 m0 4p r b
 dWm=
m0 N2 I2 h dr
m0 N2 I2 h b  =
ln( ) 4p r 4p a a

9.  dWm=  Solutions: 3.7 Magnetic Energy Thus, Wm = Inside the core
mN I
Toroid height: h
Radius: r
Thickness: dr
B =
2pr b
m0 N2 I2 h dr
m0 N2 I2 h b
 dWm=  =
ln( ) 4p r 4p a a 1
Recall: Wm = LI2
So, L = 2 b a
m0 N2 h 2p
ln( )