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

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:

3.7 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.

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)

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

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 = 0.148 J

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

 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

 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( )