Self Inductance And Energy In Magnetic Field Electromagnet L-3 Self Inductance and Energy in Magnetic Field Electromagnet
→ \rightarrow → → \rightarrow → Self Inductance and Energy in Magnetic Field Electromagnet → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Mutual Inductance
→ \rightarrow → → \rightarrow → Mutual Inductance → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Self Inductance
Self Inductance and Energy in Magnetic Field Electromagnet → \rightarrow → → \rightarrow → Self Inductance → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Toroid
Mean radius = r
Area of cross section = A
Assume : Magnetic field is uniform.
∮ B ⃗ . d l ⃗ = μ 0 I e n c \oint \vec B . \vec {dl} = \mu_0 I_{enc} ∮ B . d l = μ 0 I e n c
2 π r B = μ 0 N t I 2 \pi r B = \mu _0 N_t I 2 π r B = μ 0 N t I
B = μ 0 N t 2 π r . I \frac {\mu_0 N_t}{2 \pi r} . I 2 π r μ 0 N t . I
Mutual Inductance → \rightarrow → → \rightarrow → Toroid → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Magnetic Flux
Magnetic flux through each turn = B.A = μ 0 N t A 2 π r . I \frac {\mu_0 N_t A}{2 \pi r}.I 2 π r μ 0 N t A . I
Total magnetic flux = μ 0 N t 2 A 2 π r . I \frac {\mu_0 N_{t}^{2} A}{2 \pi r}.I 2 π r μ 0 N t 2 A . I
Self Inductances, L = μ 0 N t 2 A 2 π r \frac {\mu_0 N_{t}^{2} A}{2 \pi r} 2 π r μ 0 N t 2 A
Self Inductance → \rightarrow → → \rightarrow → Magnetic Flux → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Example
L = μ 0 N t 2 A 2 π r L=\frac{\mu_0 N_t^2 A}{2 \pi r} L = 2 π r μ 0 N t 2 A
N t = 200 N_t=200 N t = 200
A = 5 c m 2 = 5 × 10 − 4 m 2 A=5 \mathrm{cm}^2=5 \times 10^{-4} \mathrm{m}^2 A = 5 cm 2 = 5 × 1 0 − 4 m 2
r = 10 cm = 0.1m
L = 4 π × 10 − 7 × 4 × 10 4 × 5 × 10 − 4 2 π × 0.1 L = \frac{4 \pi \times 10^{-7} \times 4 \times 10^4 \times 5 \times 10^{-4}}{2 \pi \times 0.1} L = 2 π × 0.1 4 π × 1 0 − 7 × 4 × 1 0 4 × 5 × 1 0 − 4
= 40 × 10 − 6 =40 \times 10^{-6} = 40 × 1 0 − 6 μ \mu μ
Toroid → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Example
d I d t = 5 A 10 × 10 − 6 s \frac{d I}{d t}=\frac{5 A}{10 \times 10^{-6} s} d t d I = 10 × 1 0 − 6 s 5 A
=5 × 10 5 A / s 5 \times 10^5 A/s 5 × 1 0 5 A / s
Induces EMF
E = − L d I d t = − 40 × 10 − 6 × 5 × 10 5 E =-L \frac{d I}{d t} =-40 \times 10^{-6} \times 5 \times 10^5 E = − L d t d I = − 40 × 1 0 − 6 × 5 × 1 0 5
Magnetic Flux → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Energy in Magnetic Fields
Coil with self inductances: L
For current changing with time, induced emf
E = -L d I d t \frac {dI}{dt} d t d I
E = Work done in moving a unit charge through the circuit
Work done by the external agent = -E
Example → \rightarrow → → \rightarrow → Energy in Magnetic Fields → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Energy in Magnetic Fields
I : represent total charge energy the circuit per unit time.
Work done per unit time.
d w d t = − E I = − L I d I d t {dw}{dt} = -E I = - L I \frac{dI}{dt} d w d t = − E I = − L I d t d I
Total work done in increasing the current from 0 to I
W = L ∫ 0 I . I d I L \int_{0}^{I} . IdI L ∫ 0 I . I d I
W = 1 2 L I 2 \frac{1}{2} L I^2 2 1 L I 2
Example → \rightarrow → → \rightarrow → Energy in Magnetic Fields → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Solenoid
Closely bound
Very long
Magnetic field is uniform within the solenoid. B = μ 0 N I \mu_0 N I μ 0 N I
Self Inductances. L =μ 0 N 2 π r 2 l \mu_0 N^2 \pi r^2 l μ 0 N 2 π r 2 l
N : Number of turns per unit length.
Energy in Magnetic Fields → \rightarrow → → \rightarrow → Solenoid → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Energy Stores
Energy stores = 1 2 L I 2 \frac{1}{2} L I^2 2 1 L I 2
= 1 2 μ 0 N 2 π r 2 l ⋅ I 2 =\frac{1}{2} \mu_0 N^2 \pi r^2 l \cdot I^2 = 2 1 μ 0 N 2 π r 2 l ⋅ I 2
= 1 2 μ 0 N 2 I 2 π r 2 l =\frac{1}{2} \mu_0 N^2 I^2 \pi r^2 l = 2 1 μ 0 N 2 I 2 π r 2 l
= 1 2 μ 0 ( μ 0 N I ) 2 ( π r 2 l ) =\frac{1}{2 \mu_0}\left(\mu_0 N I\right)^2\left(\pi r^2 l\right) = 2 μ 0 1 ( μ 0 N I ) 2 ( π r 2 l )
= 1 2 μ 0 B 2 × v o l u m e =\frac{1}{2 \mu_0} B^{2} \times volume = 2 μ 0 1 B 2 × v o l u m e
Energy in Magnetic Fields → \rightarrow → → \rightarrow → Energy Stores → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Energy Density
Energy density = Energy per unit volume
U B = 1 2 μ 0 B 2 U_B = \frac{1}{2\mu_0}B^2 U B = 2 μ 0 1 B 2
Electrostatic energy stores per unit volume.
U E = 1 2 ϵ 0 E 2 U_E = \frac{1}{2}\epsilon_0 E^2 U E = 2 1 ϵ 0 E 2
Solenoid → \rightarrow → → \rightarrow → Energy Density → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Example
Magnetic fields , B = I T B=I T B = I T
Energy density = 1 2 μ 0 B 2 =\frac{1}{2 \mu_0} B^2 = 2 μ 0 1 B 2
= 1 2 × 4 π × 10 − 7 × 1 =\frac{1}{2 \times 4\pi \times 10^{-7}} \times 1 = 2 × 4 π × 1 0 − 7 1 × 1
= 1 8 π × 10 7 J / m 3 =\frac{1}{8 \pi} \times 10^7 \mathrm{J} / \mathrm{m}^3 = 8 π 1 × 1 0 7 J / m 3
Energy Stores → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Solenoid
N = 1000 turns/m
I = 1 A
B =μ 0 N I \mu_0 N I μ 0 N I
=4 π × 10 − 7 × 10 3 4 \pi \times 10^{-7} \times 10^3 4 π × 1 0 − 7 × 1 0 3 4 π × 10 − 4 T 4 \pi \times 10^{-4} T 4 π × 1 0 − 4 T
U B = 1 2 μ 0 B 2 = 1 2 μ μ 0 2 N 2 I 2 U_B =\frac{1}{2 \mu_0} B^2 = \frac{1}{2 \mu} \mu_0^2 N^2 I^2 U B = 2 μ 0 1 B 2 = 2 μ 1 μ 0 2 N 2 I 2
= μ 0 N 2 I 2 2 = 4 π × 10 − 7 × 10 6 × 1 2 =\frac{\mu_0 N^2 I^2}{2} = \frac{4 \pi \times 10^{-7} \times 10^6 \times 1}{2} = 2 μ 0 N 2 I 2 = 2 4 π × 1 0 − 7 × 1 0 6 × 1
= 2 π × 10 − 1 J / m 3 =2 \pi \times 10^{-1} \mathrm{~J} / \mathrm{m}^3 = 2 π × 1 0 − 1 J / m 3
Energy Density → \rightarrow → → \rightarrow → Solenoid → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Example
Non uniform magnetic field
∮ B ⃗ ⋅ d L → = μ 0 I e n c \oint \vec{B} \cdot \overrightarrow{d L}=\mu_0 I_{enc} ∮ B ⋅ d L = μ 0 I e n c
2 π r ⋅ B = μ 0 I 2 \pi r \cdot B=\mu_0 I 2 π r ⋅ B = μ 0 I
B = μ 0 I 2 π r B =\frac{\mu_0 I}{2 \pi r} B = 2 π r μ 0 I
a < r < b
Example → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Energy Density of Magnetic Field
U B = 1 2 μ 0 B 2 U_B =\frac{1}{2 \mu_0} B^2 U B = 2 μ 0 1 B 2
= 1 2 μ 0 ( μ 0 I 2 π r ) 2 =\frac{1}{2 \mu_0}\left(\frac{\mu_0 I}{2 \pi r}\right)^2 = 2 μ 0 1 ( 2 π r μ 0 I ) 2
= μ 0 I 2 8 π 2 r 2 =\frac{\mu_0 I^2}{8 \pi^2 r^2} = 8 π 2 r 2 μ 0 I 2
Energy in a length l
Solenoid → \rightarrow → → \rightarrow → Energy Density of Magnetic Field → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Total Magnetic Energy
Elementary volume = 2 π . r . d r . l 2 \pi .r .d r .l 2 π . r . d r . l
Total magnetic energy =∫ a b U B 2 π . r . d r . l \int_a^b U_B 2 \pi. r. d r. l ∫ a b U B 2 π . r . d r . l
= μ 0 I 2 8 π 2 2 π ∫ a b r . d r r 2 l =\frac{\mu_0 I^2}{8 \pi^2} 2 \pi \int_a^b \frac{r .d r}{r^2} l = 8 π 2 μ 0 I 2 2 π ∫ a b r 2 r . d r l
= μ 0 I 2 4 π l ∫ a b d r r =\frac{\mu_0 I^2}{4 \pi} l \int_a^b \frac{d r}{r} = 4 π μ 0 I 2 l ∫ a b r d r
Example → \rightarrow → → \rightarrow → Total Magnetic Energy → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Total Magnetic Energy
Total magnetic energy =μ 0 l 4 π ln ( b a ) I 2 = 1 2 L I 2 \frac{\mu_0 l}{4 \pi} \ln \left(\frac{b}{a}\right) I^2 = \frac{1}{2} L I^2 4 π μ 0 l ln ( a b ) I 2 = 2 1 L I 2
Self Inductance
L = μ 0 l 2 π l n b a \frac{\mu_0 l}{2 \pi} ln \frac{b}{a} 2 π μ 0 l l n a b
Self inductance per unit length =μ 0 2 π ln ( b a ) \frac{\mu_0}{2 \pi} \ln \left(\frac{b}{a}\right) 2 π μ 0 ln ( a b )
Energy Density of Magnetic Field → \rightarrow → → \rightarrow → Total Magnetic Energy → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Coaxial Cable
Total Magnetic Energy → \rightarrow → → \rightarrow → Coaxial Cable → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Inductance
Φ B = ∫ B ⃗ ⋅ d A ⃗ \Phi_B =\int \vec{B} \cdot \vec{dA} Φ B = ∫ B ⋅ d A
= μ 0 I 2 π l . ln ( b a ) =\frac{\mu_0 I}{2 \pi} l .\ln (\frac{b}{a}) = 2 π μ 0 I l . ln ( a b )
L =μ 0 L 2 π ln ( b a ) \frac{\mu_0 L }{2 \pi} \ln(\frac{b}{a}) 2 π μ 0 L ln ( a b )
Total Magnetic Energy → \rightarrow → → \rightarrow → Inductance → \rightarrow → → \rightarrow →
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Thank You
Coaxial Cable → \rightarrow → → \rightarrow → Thank You → \rightarrow → → \rightarrow →
Resume presentation
Self Inductance And Energy In Magnetic Field Electromagnet L-3 Self Inductance and Energy in Magnetic Field Electromagnet $\rightarrow$ $\rightarrow$ Self Inductance and Energy in Magnetic Field Electromagnet $\rightarrow$ Mutual Inductance $\rightarrow$ Self Inductance
