Inductancia: Energía Magnética

INDUCTANCIA ENERGÍA MAGNÉTICA

INDUCTOR 𝑈𝑛𝑎 𝑏𝑜𝑏𝑖𝑛𝑎 𝑜 𝒊𝒏𝒅𝒖𝒄𝒕𝒐𝒓 𝑒𝑠 𝑢𝑛 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑐𝑖𝑟𝑐𝑢𝑖𝑡𝑜 𝑞𝑢𝑒 𝑎𝑙𝑚𝑎𝑐𝑒𝑛𝑎 𝑒𝑛𝑒𝑟𝑔í𝑎 𝑒𝑛 𝑒𝑙 𝑐𝑎𝑚𝑝𝑜 𝑚𝑎𝑔𝑛é𝑡𝑖𝑐𝑜 𝑒𝑛 𝑒𝑙 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑑𝑒 𝑙𝑎 𝑏𝑜𝑏𝑖𝑛𝑎 𝑝𝑜𝑟 𝑙𝑎 𝑐𝑢𝑎𝑙 𝑐𝑖𝑟𝑐𝑢𝑙𝑎 𝑢𝑛𝑎 𝑐𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒.

INDUCTANCIA 𝑃𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑 𝑑𝑒 𝑙𝑜𝑠 𝑐𝑖𝑟𝑐𝑢𝑖𝑡𝑜𝑠 𝑒𝑙é𝑐𝑡𝑟𝑖𝑐𝑜𝑠 𝑝𝑜𝑟 𝑙𝑎 𝑐𝑢𝑎𝑙 𝑠𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑒 𝑢𝑛𝑎 𝑓𝑢𝑒𝑟𝑧𝑎 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑚𝑜𝑡𝑟𝑖𝑧 𝑐𝑢𝑎𝑛𝑑𝑜 𝑣𝑎𝑟í𝑎 𝑙𝑎 𝑐𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒 𝑞𝑢𝑒 𝑝𝑎𝑠𝑎.

INDUCTANCIA 𝐸𝑠𝑙𝑎𝑏𝑜𝑛𝑎𝑚𝑖𝑒𝑛𝑡𝑜 𝑑𝑒 𝑓𝑙𝑢𝑗𝑜: 𝞴 = 𝑁Φ 𝑆𝑖 𝑒𝑙 𝑚𝑒𝑑𝑖𝑜 𝑒𝑠 𝑙𝑖𝑛𝑒𝑎𝑙: 𝞴 ∝ 𝐼 𝞴 = 𝐿𝐼

𝞴 𝐿= 𝐼 𝑁𝛷 𝐿= 𝐼

FEM AUTOINDUCIDA 𝑁𝛷 𝐿= 𝑖 𝐿𝑖 = 𝑁𝛷 𝑑Φ 𝑑𝑖 𝑁 =𝐿 𝑑𝑡 𝑑𝑡 𝑑𝑖 𝜀 = −𝐿 𝑑𝑡

INDUCTANCIA MUTUA 𝐸𝑠 𝑙𝑎 𝑐𝑎𝑝𝑎𝑐𝑖𝑑𝑎𝑑 𝑑𝑒 𝑢𝑛 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑝𝑎𝑟𝑎 𝑖𝑛𝑑𝑢𝑐𝑖𝑟 𝑡𝑒𝑛𝑠𝑖ó𝑛 𝑒𝑛 𝑢𝑛 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑣𝑒𝑐𝑖𝑛𝑜.

INDUCTANCIA MUTUA 𝑀12

λ12 𝑁1 Φ12 = = 𝑖2 𝑖2

𝐷𝑜𝑛𝑑𝑒 Φ12 𝑒𝑠 𝑒𝑙 𝑓𝑙𝑢𝑗𝑜 𝑞𝑢𝑒 𝑝𝑎𝑠𝑎 𝑝𝑜𝑟 𝑒𝑙 𝑐𝑖𝑟𝑐𝑢𝑖𝑡𝑜 1 𝑑𝑒𝑏𝑖𝑑𝑜 𝑎 𝑙𝑎 𝑐𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒 𝑖2 𝑒𝑛 𝑒𝑙 𝑐𝑖𝑟𝑐𝑢𝑖𝑡𝑜 2.

𝑀21

λ21 𝑁2 Φ21 = = 𝑖1 𝑖1

𝑆𝑖𝑛 𝑒𝑚𝑏𝑎𝑟𝑔𝑜, 𝑀12 𝑠𝑖𝑒𝑚𝑝𝑟𝑒 𝑒𝑠 𝑖𝑔𝑢𝑎𝑙 𝑎 𝑀21, 𝑎ú𝑛 𝑐𝑢𝑎𝑛𝑑𝑜 𝑙𝑎𝑠 𝑑𝑜𝑠 𝑏𝑜𝑏𝑖𝑛𝑎𝑠 𝑛𝑜 𝑠𝑒𝑎𝑛 𝑠𝑖𝑚é𝑡𝑟𝑖𝑐𝑎𝑠. 𝐴 𝑒𝑠𝑡𝑒 𝑣𝑎𝑙𝑜𝑟 𝑐𝑜𝑚ú𝑛 𝑙𝑜 𝑙𝑙𝑎𝑚𝑎𝑚𝑜𝑠 𝑠𝑖𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑒 𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑖𝑎 𝑚𝑢𝑡𝑢𝑎.

𝑀12 = 𝑀21

EJEMPLO 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑙𝑎 𝑎𝑢𝑡𝑜𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑖𝑎 𝑑𝑒 𝑢𝑛 𝑐𝑎𝑏𝑙𝑒 𝑐𝑜𝑎𝑥𝑖𝑎𝑙 𝑑𝑒 𝑟𝑎𝑑𝑖𝑜 𝑖𝑛𝑡𝑒𝑟𝑛𝑜 𝑎 𝑦 𝑟𝑎𝑑𝑖𝑜 𝑒𝑥𝑡𝑒𝑟𝑛𝑜 𝑏.

SOLUCIÓN 𝑳 = 𝑳𝒊𝒏 + 𝑳𝒆𝒙𝒕 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑛𝑑𝑜 𝑙𝑎𝑠 𝑒𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠: 1 2 𝐵2 𝑈 = 𝐿𝐼 1 ; 𝑈 = න 𝑑𝑣 2 2μ 𝐷𝑒𝑠𝑝𝑒𝑗𝑎𝑛𝑑𝑜 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 1 : 𝐿 =

𝑳𝒊𝒏

(2)

2𝑈 𝑦 𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 é𝑠𝑡𝑎 𝑙𝑎 2 : 𝐼2

2 𝐵2 1 𝜇2 𝐼 2 𝜌2 𝜇2 𝐼 2 = 2 න 𝑑𝑣 = 2 ම 2 4 𝜌𝑑𝜌𝑑𝜑𝑑𝑧 = 2 ම 𝜌3 𝑑𝜌𝑑𝜑𝑑𝑧 2 4 𝐼 2μ 𝐼 μ 4𝜋 𝑎 𝐼 𝜇4𝜋 𝑎

𝑳𝒆𝒙𝒕

𝑙 2𝜋 𝑎 𝜇 𝜇2𝜋𝑙 𝑎4 𝝁𝒍 3 = 2 4 න 𝑑𝑧 න 𝑑𝜑 න 𝜌 𝑑𝜌 = 2 4 = 𝑯 4𝜋 𝑎 0 4𝜋 𝑎 4 𝟖𝝅 0 0 2 2 2 𝐵 1 𝜇 𝐼2 𝜇2 𝐼 2 1 = 2 න 𝑑𝑣 = 2 ම 2 2 𝜌𝑑𝜌𝑑𝜑𝑑𝑧 = 2 ම 𝑑𝜌𝑑𝜑𝑑𝑧 𝐼 2μ 𝐼 μ 4𝜋 𝜌 𝐼 𝜇4𝜋 2 𝜌 𝑙 2𝜋 𝑏 𝜇 𝑑𝜌 𝜇2𝜋𝑙 𝑏 𝝁𝒍 𝒃 = 2 න 𝑑𝑧 න 𝑑𝜑 න = 2 ln 𝑎 = 𝟐𝝅 𝐥𝐧 𝒂 𝑯 4𝜋 0 𝜌 4𝜋 0 𝑎

𝑳 = 𝑳𝒊𝒏 + 𝑳𝒆𝒙𝒕 =

𝜇𝑙 𝜇𝑙 𝑏 𝝁𝒍 𝟏 𝒃 + ln = + 𝐥𝐧 8𝜋 2𝜋 𝑎 𝟐𝝅 𝟒 𝒂

ENERGÍA MAGNÉTICA 𝐸𝑠 𝑙𝑎 𝑒𝑛𝑒𝑟𝑔í𝑎 𝑞𝑢𝑒 𝑠𝑒 𝑎𝑙𝑚𝑎𝑐𝑒𝑛𝑎 𝑒𝑛 𝑒𝑙 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑐𝑢𝑎𝑛𝑑𝑜 𝑙𝑎 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑑𝑎𝑑 𝑑𝑒 𝑐𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒 𝑎𝑢𝑚𝑒𝑛𝑡𝑎, 𝑦 𝑠𝑒 𝑙𝑖𝑏𝑒𝑟𝑎 𝑐𝑢𝑎𝑛𝑑𝑜 𝑙𝑎 𝑐𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒 𝑑𝑖𝑠𝑚𝑖𝑛𝑢𝑦𝑒.

ENERGÍA MAGNÉTICA 𝑃 = 𝜀𝑖

1 ;

𝑑𝑤 𝑃= 𝑑𝑡

2 ; 𝑃𝑑𝑡 = 𝑑𝑊 (3)

𝑑𝑖 𝑑𝑖 𝑃 = 𝜀𝑖 = 𝐿 𝑖 = 𝐿𝑖 𝑑𝑡 𝑑𝑡 𝐿𝑎 𝑒𝑛𝑒𝑟𝑔í𝑎 𝑑𝑈 𝑠𝑢𝑚𝑖𝑛𝑖𝑠𝑡𝑟𝑎𝑑𝑎 𝑎𝑙 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑑𝑢𝑟𝑎𝑛𝑡𝑒 𝑢𝑛 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 𝑑𝑒 𝑡𝑖𝑒𝑚𝑝𝑜 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒𝑠𝑖𝑚𝑎𝑙: 𝑑𝑈 = 𝑃𝑑𝑡 𝑑𝑈 = 𝐿𝑖𝑑𝑖 𝐼

න 𝑑𝑈 = න 𝐿𝑖𝑑𝑖 0

𝐼2 𝑈=𝐿 2 1 2 𝑈 = 𝐿𝐼 2

DENSIDAD DE ENERGÍA MAGNÉTICA 𝑈 𝐿𝑎 𝑑𝑒𝑛𝑠𝑖𝑑𝑎𝑑 𝑑𝑒 𝑒𝑛𝑒𝑟𝑔í𝑎 𝑚𝑎𝑔𝑛é𝑡𝑖𝑐𝑎 𝑒𝑠 𝑢 = 𝑉

𝐿𝑎 𝑑𝑒𝑛𝑠𝑖𝑑𝑎𝑑 𝑑𝑒 𝑒𝑛𝑒𝑟𝑔í𝑎 𝑚𝑎𝑔𝑛é𝑡𝑖𝑐𝑎 𝑠𝑒 𝑑𝑒𝑓𝑖𝑛𝑒 𝑐𝑜𝑚𝑜 𝑙𝑎 𝑒𝑛𝑒𝑟𝑔í𝑎 𝑚𝑎𝑔𝑛é𝑡𝑖𝑐𝑎 𝑝𝑜𝑟 𝑢𝑛𝑖𝑑𝑎𝑑 𝑑𝑒 𝑣𝑜𝑙𝑢𝑚𝑒𝑛.

2 1 2 1 𝜇 𝑁 𝐴 𝐼2 2 𝐼2 𝐿𝐼 𝜇𝑁 2 𝑙 𝑢=2 = = 𝐴𝑙 𝐴𝑙 2𝑙2 𝑁 𝐵=𝜇 𝐼 𝑙 2 𝑁 𝐵2 = 𝜇2 2 𝐼2 𝑙

𝐵2 𝑁2 2 =𝜇 2 𝐼 𝜇 𝑙 𝐵2 𝑢= 2𝜇

EJEMPLO 𝑈𝑛 𝑠𝑜𝑙𝑒𝑛𝑜𝑖𝑑𝑒 𝑚𝑢𝑦 𝑙𝑎𝑟𝑔𝑜 𝑐𝑜𝑛 𝑠𝑒𝑐𝑐𝑖ó𝑛 𝑡𝑟𝑎𝑛𝑠𝑣𝑒𝑟𝑠𝑎𝑙 𝑑𝑒 2𝑥2 𝑐𝑚 𝑡𝑖𝑒𝑛𝑒 𝑢𝑛 𝑛ú𝑐𝑙𝑒𝑜 𝑑𝑒 ℎ𝑖𝑒𝑟𝑟𝑜 𝜇𝑟 : 1000 𝑦 4000 𝑣𝑢𝑒𝑙𝑡𝑎𝑠Τ𝑚𝑒𝑡𝑟𝑜. 𝑆𝑖 𝑝𝑜𝑟𝑡𝑎 𝑢𝑛𝑎 𝑐𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒 𝑑𝑒 500𝑚𝐴, 𝑒𝑛𝑐𝑢𝑒𝑛𝑡𝑟𝑒 𝑎)𝑆𝑢 𝑎𝑢𝑡𝑜𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑖𝑎 𝑝𝑜𝑟 𝑚𝑒𝑡𝑟𝑜. 𝑏)𝐿𝑎 𝑒𝑛𝑒𝑟𝑔í𝑎 𝑝𝑜𝑟 𝑚𝑒𝑡𝑟𝑜 𝑎𝑙𝑚𝑎𝑐𝑒𝑛𝑎𝑑𝑎 𝑒𝑛 𝑠𝑢 𝑐𝑎𝑚𝑝𝑜.

SOLUCIÓN 𝐿 𝑁2 𝒂) = 𝜇 𝐴 𝑙 𝐿

μ = 𝜇0 𝜇𝑟 = 4𝜋 × 10−7 1000 = 4𝜋 × 10−4 𝐴 = (0.02𝑚) 0.02𝑚 = 4 × 10−4 𝑚

𝐿 = 4𝜋 × 10−4 4000 𝑙

2

−4

4 × 10

𝑈 1 2 1 𝒃) = 𝐿𝐼 = 8.042 500 × 10−3 𝑙 2 2

𝑯 = 𝟖. 𝟎𝟒𝟐 𝒎

2

𝑱 = 𝟏. 𝟎𝟎𝟓 𝒎