Magnetic Energy

Magnetic Energy

AU : Dec.-09, 10, 11, 13, 14, May-04, 13, 18

• Similar to capacitor, studied previously, inductor is also an energy storing element. In a capacitor, the energy is stored in the electrostatic field, while in an inductor energy is stored in the magnetic field. The energy stored by an inductor is given by,

• Consider a differential volume in a magnetic field  as shown in the Fig. 8.12.1.

• Consider that at the top and bottom surfaces of a differential volume, conducting sheets with current ΔI are present.

• From the definition of an inductance, we can write inductance, ΔL of a differential volume as,

where ΔS = Differential surface area = Δx Δz

• Now the differential current AI can be expressed interms of the magnetic field intensity H. The current flowing through the conducting sheets present at the top and bottom, is in y direction.

ΔI = (H) (Δy) ... (8.12.3)

• The energy stored in the inductance of a different volume is given by,

ΔW_m = 1/2 ΔL ΔI²

• Putting values of AL and AI from equations (8.12.2) and (8.12.3),

• But the differential volume can be written as,

Δv = Δx Δy Δz

• Hence energy stored in an inductor of a differential volume is given by,

ΔWm = 1/2 µ H² Δv ..... (8.12.4)

• The magnetostatic energy density function is defined as,

• An energy density function is expressed in joule / m³ (J / m³).

• The magnetostatic energy density can be expressed in different forms as

• In a linear medium, the energy in a magnetostatic field is given by,

Ex. 8.12.1 A very long solenoid with 2 × 2 cm cross-section has an iron core (µ r = 1000) and 4000 turns/meter. If it carries current of 500 mA, find

a) Its self inductance per meter

b) The energy per meter stored in field.

AU : Dec.-14, Marks 6

Sol. : a) The inductance per unit length of a solenoid is given by,

L' = µN²A N/m, where A = Area of corss-section,

N = Number of turns

L = (µ₀µ_r) N²A

= (1000 × 4 × π × 10^-7) (4000)² (2 × 10^-2 × 2 × 10^-2)

= 8.042 H/m

b) The energy stored by inductor is given by,

Wm = 1/2  L I² = 1/ 2 (8.042) (500 × 10^-3)²

= 1.005 J/m

Ex. 8.12.2 An iron ring, 0.2 m in diameter and 10 cm² sectional area of the core, is uniformly wound with 250 turns of wire. The wire carries a current of 4 A. The relative permeability of iron is 500. Determine the value of self-inductance and the stored energy.

AU : May-18, Marks 7

Sol. :

Examples for Practice

Ex. 8.12.3 A coil has 1000 turns and carries a magnetic flux of 10 mWb. The resistance of the coil is 4 Ω. If it is connected to a 40 V d.c. supply, estimate the energy stored in the magnetic field when the current attains its final steady state value.

AU : May-04, 13

[Ans.: 50 W]

Ex. 8.12.4 A air core toroid has a mean radius of 40 mm and is wound with 400 turns of wire. The circular cross-section of the toroid has a radius of 4 mm. A current of 10 A is passed in the wire. Find the inductance and the energy stored.

[Ans.: 40.2123 µH, 2.0106 mW]

Review Questions

1. Derive expression for the energy density in the magnetic field.

AU : Dec.-11, Marks 16

2. Give short note on the following : Magnetic energy density.

AU : Dec.-09, Marks 4

3. Obtain the expression for energy stored in the magnetic field and also derive the expression for magnetic energy density.

AU : Dec.-10, 13, Marks 16; Dec.-14, Marks 8