FISIKA Kelas 12 - Induksi Elektromagnetik: Induktansi Diri | GIA Academy

Summary

This educational YouTube video by Gia Akademi explores the concept of self-induction in electrical circuits. It explains how the gradual brightening and dimming of lights when switched on and off is due to the changing magnetic flux, inducing an electromotive force (EMF) that opposes the current change. The video delves into the principles of self-induction, including the formula for induced EMF, \( \varepsilon = -L \frac{\Delta I}{\Delta t} \), where \( L \) is the inductance, \( \Delta I \) is the change in current, and \( \Delta t \) is the change in time. It also covers the calculation of inductance in solenoids and toroids, the energy stored in inductors, and mutual inductance between two coils. The video concludes with practical examples and problems to solidify the understanding of self-induction.

Takeaways

• 🔌 The concept of self-induction is applied in everyday life, such as the gradual dimming of a light when switched off due to changes in magnetic flux.
• 💡 When turning on a light, the switch closes, and electricity flows through the circuit, creating a change in magnetic flux which causes the light to gradually brighten rather than instantly.
• 🌀 The gradual fading of a light when switched off is due to the change in magnetic flux as the electric current in the circuit is interrupted.
• ⚖️ The self-induced electromotive force (emf) is calculated using the formula ε = -L * (ΔI/Δt), where L is the self-inductance, and ΔI/Δt is the rate of change of current over time.
• 👨‍🔬 Joseph Henry, an American scientist, formulated the law stating that the magnitude of self-induced emf is proportional to the rate of change of current with respect to time.
• 🧲 The self-inductance of a coil can be determined by the formula L = n * ΔΦ/ΔI, where n is the number of turns, and ΔΦ is the change in magnetic flux.
• 📏 The self-inductance at the center of a solenoid can be calculated using the formula L = μ₀ * n² * a / l, with μ₀ being the permeability of free space.
• 🔗 The energy stored in a coil or inductor is given by the formula W = 1/2 * L * I², where W is the energy in joules, L is the inductance in henries, and I is the current in amperes.
• 🔄 Mutual or reciprocal inductance occurs between two coils when a changing current in one coil induces a changing magnetic flux in the other, calculated by M = N₁ * Φ₂ / I₂ = N₂ * Φ₁ / I₁.
• 📡 The inductance of a toroid can be determined using the formula L = μ₀ * N² * a / 2π, where a is the mean radius, N is the number of turns, and μ₀ is the permeability of free space.

Q & A

What is the concept of self-induction explained in the video?

Self-induction is a phenomenon where a changing electric current in a circuit induces an electromotive force (EMF) in the same circuit due to the change in magnetic flux. This concept is demonstrated through the example of a light bulb, which doesn't turn on or off instantly when the switch is flipped.

How is the EMF of self-induction calculated?

The EMF of self-induction is calculated using the formula ε = -L * (ΔI/Δt), where ε is the induced EMF, L is the self-inductance, and (ΔI/Δt) is the rate of change of current with respect to time.

Who is Joseph Henry and what is his contribution to the understanding of self-induction?

Joseph Henry was an American scientist who formulated the law of self-induction, stating that the magnitude of the induced EMF is proportional to the rate of change of the magnetic flux, which can be expressed as ε = -L * (dI/dt).

What is the unit of self-inductance and what is its symbol?

The unit of self-inductance is the Henry, symbolized as 'H'.

What is the relationship between the change in magnetic flux and the induced EMF according to Faraday's law?

According to Faraday's law, a change in magnetic flux through a coil induces an EMF in the coil, which is proportional to the rate of change of the magnetic flux, expressed as ε = -n * (ΔΦ/Δt), where n is the number of turns in the coil and ΔΦ is the change in magnetic flux.

How is the self-inductance of a solenoid calculated?

The self-inductance of a solenoid is calculated using the formula L = μ₀ * n² * A / l, where μ₀ is the permeability of free space, n is the number of turns per unit length, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.

What is the energy stored in an inductor and how is it calculated?

The energy stored in an inductor is given by the formula W = 1/2 * L * I², where W is the energy in joules, L is the inductance in henries, and I is the current in amperes.

What is mutual induction and how does it differ from self-induction?

Mutual induction occurs when a changing current in one coil (primary) induces a current in a nearby coil (secondary) due to the changing magnetic field. It differs from self-induction, which occurs within the same coil. Mutual induction is calculated using the formula M = N₁ * Φ₂ / I₂ = N₂ * Φ₁ / I₁, where M is the mutual inductance, N₁ and N₂ are the number of turns in the primary and secondary coils, and Φ₁ and Φ₂ are the magnetic fluxes in the primary and secondary coils, respectively.

How is the self-inductance of a toroid calculated?

The self-inductance of a toroid is calculated using the formula L = μ₀ * n² * A / (2π), where μ₀ is the permeability of free space, n is the number of turns, A is the cross-sectional area, and the division by 2π accounts for the toroid's geometry.

What is the significance of the problem-solving examples provided in the video?

The problem-solving examples in the video demonstrate the practical application of the concepts of self-induction, mutual induction, and energy storage in inductors. They help in understanding how to calculate the induced EMF, self-inductance, and stored energy in different types of inductors.

Outlines

00:00

💡 Introduction to Self-Inductance

The script begins with an introduction to the concept of self-inductance using the everyday example of a light switch. It explains that when a light is turned on or off, the light doesn't immediately change state due to the principle of electromagnetic induction. The video aims to explore self-inductance in detail, including how it occurs and its applications in daily life. The script delves into the scientific explanation of self-inductance, mentioning the role of changing magnetic flux and how it induces an electromotive force (EMF) that opposes the change in current, as described by Faraday's law of induction and further elaborated by Joseph Henry.

05:01

🧲 Calculation of Self-Inductance in Solenoids and Toroids

This paragraph focuses on the calculation of self-inductance in solenoids and toroids. It discusses the formula for self-inductance at the center of a solenoid, which depends on the number of turns per unit length, the cross-sectional area, and the permeability of free space. The script also covers the self-inductance of a toroid, explaining how it is determined by similar factors. The mathematical relationships are simplified to provide formulas for calculating self-inductance in these coil configurations.

10:05

🔋 Energy Storage in Inductors and Mutual Inductance

The script moves on to discuss how inductors can store energy in a magnetic field, with the energy stored in an inductor being proportional to the square of the current flowing through it, as per the formula W = 1/2 * L * I^2. It then introduces the concept of mutual inductance, which occurs when two coils are in proximity and a changing current in one coil induces a voltage in the other. The mutual inductance is calculated using the number of turns in each coil and the magnetic flux linkage. The script provides examples of how mutual inductance is calculated for both solenoids and toroids.

15:07

📐 Practical Problems on Self-Inductance and Mutual Inductance

The final paragraph presents a series of practical problems to apply the concepts of self-inductance and mutual inductance. It includes problems such as calculating the induced EMF for a given coil with a time-varying current, determining the self-inductance of a coil based on the change in current and the resulting magnetic flux, and calculating the energy stored in a toroid. The script also covers the calculation of self-inductance for a telephone cable and mutual inductance between two solenoids. Each problem is solved step by step, demonstrating the application of the formulas introduced earlier in the script.

Keywords

💡Induction

Induction in the context of the video refers to the phenomenon where a change in the magnetic field within a circuit induces an electromotive force (EMF). This is a fundamental concept in electromagnetism and is central to the operation of many electrical devices. In the script, induction is used to explain why lights do not turn on or off instantly when a switch is flipped, illustrating the concept with the example of a light bulb and its circuit.

💡Self-Induction

Self-induction is a specific type of electromagnetic induction that occurs within a single coil or circuit. It is the process by which a change in current in a circuit induces a voltage in the same circuit. The video script discusses self-induction in the context of explaining the gradual brightening and dimming of a light when the switch is turned on or off, respectively.

💡Electromotive Force (EMF)

EMF, or electromotive force, is the work needed to move a unit charge from a point of zero electric potential to a point of higher electric potential. In the video, EMF is induced due to changes in the magnetic field and is represented by the equation ε = -L * (ΔI/Δt), where ε is the induced EMF, L is the inductance, and ΔI/Δt is the rate of change of current.

💡Inductance

Inductance is a property of an electrical conductor by which a change in current flowing through it induces an electromotive force in the conductor. The video script introduces inductance as a measure of an inductor's ability to induce EMF and discusses how it is calculated using the formula L = n * ΔΦ/ΔI, where n is the number of turns, ΔΦ is the change in magnetic flux, and ΔI is the change in current.

💡Magnetic Flux

Magnetic flux is the measure of the magnetic field passing through a given area. In the video, the concept of magnetic flux is used to explain how changes in the magnetic field can induce voltages in a circuit, as seen in the formula ΔΦ = B * A, where B is the magnetic field and A is the area through which the field lines pass.

💡Joseph Henry

Joseph Henry was an American scientist and a significant figure in the development of electromagnetism. The video script mentions Henry in relation to his law, which states that the induced EMF is proportional to the rate of change of the magnetic flux. This law is fundamental to understanding self-induction and is used to derive the formula for induced EMF.

💡Faraday's Law

Faraday's Law of electromagnetic induction states that a change in the magnetic field within a closed loop induces an electromotive force in the loop. The video script references Faraday's Law to explain the generation of induced EMF in a coil when the magnetic field changes, which is a key concept in understanding how inductors and transformers work.

💡Mutual Induction

Mutual induction occurs when the change in current in one coil induces a voltage in a nearby coil. This is discussed in the video script in the context of two coils (primary and secondary) that are close to each other. The induced voltage in one coil due to changes in the other is a critical principle behind transformers and other inductive coupling devices.

💡Energy Storage

The video script touches on the concept of energy storage in an inductor, where energy is stored in the magnetic field created by the current flowing through the inductor. The formula W = 1/2 * L * I^2 is used to calculate the energy stored, where W is the energy, L is the inductance, and I is the current. This is an important aspect of inductors' utility in electrical circuits.

💡Solenoid

A solenoid is a type of electromagnet, usually consisting of a coil of wire that forms a helix. The video script explains how the inductance of a solenoid can be calculated using the formula L = μ₀ * n² * A / l, where μ₀ is the permeability of free space, n is the number of turns per unit length, A is the cross-sectional area, and l is the length of the solenoid. Solenoids are used to generate controlled magnetic fields.

💡Toroid

A toroid is a doughnut-shaped inductor that has a higher inductance-to-size ratio compared to a solenoid. The video script discusses how the inductance of a toroid can be determined using the formula L = μ₀ * n² * A / (2π), where μ₀ is the permeability of free space, n is the number of turns, A is the cross-sectional area, and π is the mathematical constant pi. Toroids are used in various applications due to their compact design and high inductance.

Highlights

Induction is a concept applied in everyday life, such as the gradual dimming of lights when switched off.

Self-induction occurs when a change in electric current in a circuit induces a voltage in the same circuit.

The induced electromotive force (EMF) is proportional to the rate of change of the magnetic flux, as described by Faraday's law of induction.

Joseph Henry's law states that the magnitude of self-induced EMF is proportional to the rate of change of electric current over time.

The formula for self-induced EMF is given by ε = -L * (ΔI/Δt), where L is the self-inductance, and ΔI and Δt represent changes in current and time, respectively.

Self-inductance is measured in henries (H) and is calculated based on the change in magnetic flux and current in a coil.

The self-inductance of a solenoid can be determined using the formula L = μ₀ * n² * A / l, where μ₀ is the permeability of free space, n is the number of turns, A is the cross-sectional area, and l is the length of the solenoid.

The energy stored in an inductor can be calculated using the formula W = 1/2 * L * I², where W is the energy in joules, L is the inductance, and I is the current.

Mutual induction or transformer action occurs when a changing current in one coil induces a current in a nearby coil.

The mutual inductance between two coils is given by the formula M = N₁ * Φ₂ / I₂ = N₂ * Φ₁ / I₁, where M is the mutual inductance, N₁ and N₂ are the number of turns in each coil, and Φ₁ and Φ₂ are the magnetic fluxes.

The self-inductance of a toroid can be calculated using the formula L = μ₀ * n² * A / (2π), where A is the cross-sectional area and n is the number of turns.

An example problem demonstrates how to calculate the self-induced EMF given a time-varying current and inductance.

Another example shows how to determine the self-inductance of a coil given a change in current and the resulting change in magnetic flux.

The self-inductance of a telephone cable can be calculated by considering its spiral shape and the number of turns.

The energy stored in a toroid can be determined by calculating its self-inductance and the square of the current flowing through it.

Mutual inductance between two solenoids is calculated using the formula M = μ₀ * N₁ * N₂ * A / l, taking into account the geometry and number of turns of each solenoid.

The video concludes with a summary of the concepts of self-induction and mutual induction, encouraging viewers to continue learning about these topics.

Transcripts

[Musik]

Halo teman-teman selamat datang kembali

di channel YouTube Gia Akademi Semoga

teman-teman selalu sehat dan terus

semangat teman-teman biasa nya ketika

akan tidur kita mematikan lampu kamar

Pernahkah teman-teman memperhatikan

bahwa setelah kita menekan saklar lampu

tidak langsung padam melainkan redup

dahulu baru kemudian padam begitu juga

saat kita menghidupkan lampu setelah

saklar kita tekan lampu tidak langsung

menyala terang peristiwa ini merupakan

salah satu penerapan konsep induktansi

diri dalam kehidupan sehari-hari tahukah

teman-teman Apa itu induktansi diri

Bagaimanakah proses terjadi nya

induktansi diri kita akan membahas

materi ini secara lengkap di video kali

ini jadi di video ini kita akan belajar

tentang induktansi diri simak terus

videonya

ya teman-teman saat menghidupkan lampu

saklar ditutup lampu akan menyala dan

pada kumparan akan mengalir arus listrik

yang mengakibatkan timbulnya perubahan

fluks magnetik dari nol menuju nilai

tertentu Itulah sebabnya lampu tidak

langsung menyala terang saat kita

hidupkan sedangkan saat mematikan lampu

saklar dibuka lampu masih menyala sesaat

karena arus listrik di dalam rangkaian

akan terputus akibatnya pada kumparan

akan terjadi perubahan fluks magnetik

dari nilai tertentu kembali menjadi nol

ini juga yang menyebabkan lampu tidak

langsung padam ketika kita matikan

arus listrik yang berubah-ubah pada

rangkaian listrik menyebabkan fluks

magnetik pada kumparan juga mengalami

perubahan perubahan fluks magnetik Ini

menimbulkan GGL induksi diri yang

dinyatakan dengan persamaan εeps = - l *

Delta I / dengan Delta t dengan eps S

GGL induksi diri satuannya volt l

induktansi diri satuannya Henry Delta I

perubahan arus listrik satuannya amper

dan Delta t perubahan waktu satuannya

sek persamaan tersebut dirumuskan oleh

seorang ilmuwan Amerika bernama Joseph

Henry ia menyatakan bahwa besarnya GGL

induksi diri yang timbul akibat

perubahan fluks magnetik sebanding

dengan laju perubahan arus terhadap

waktu

Jika perubahan arus listrik terjadi

dalam waktu singkat atau Delta t

mendekati 0 GGL induksi diri ditentukan

dengan persamaan eps = - l *

di/dt dengan eps S GGL induksi diri

satuannya volt l induktansi diri

satuannya Henry di/dt turunan pertama

fungsi arus listrik terhadap waktu

satuannya ampere

perss pada materi sebelumnya kita sudah

mengetahui bahwa menurut farade adanya

perubahan garis gaya magnetik pada

kumparan akan menimbulkan GGL induksi

yang ditentukan dengan persamaan eps in

= -n * Delta Pi Dib Delta t menurut

Joseph Henry besarnya laju perubahan

arus listrik pada kumparan juga

bergantung pada perubahan garis gaya

magnetik sehingga dapat dinyatakan bahwa

ep s = εep in - l * Delta I / Delta t =

-n * Delta pi / dengan Delta t kita

Sederhanakan sehingga kita peroleh

persamaan induktansi diri l = n * Delta

V / Delta I

dengan l induktansi diri satuannya Henry

n jumlah lilitan Delta Pi perubahan

fluks magnetik satuannya Weber Delta I

perubahan arus listrik satuannya amper

dari persamaan tersebut dapat

disimpulkan bahwa induktansi diri suatu

kumparan sama dengan jumlah fluks

magnetik yang melingkupi kumparan tiap

satuan kuat arus listrik yang mengalir

pada kumparan

kita juga dapat menentukan besar

induktansi diri pada pusat solenoida

yaitu l = n * Pi / I karena Pi = b * a

maka l = n * b * a / i induksi magnetik

pada pusat solenoida adalah b = mi0 * n

* i / l maka Mak l = n * a / i * μ0 * n

* i / dengan l kita Sederhanakan

sehingga besar induktansi diri pada

pusat solenoida dapat ditentukan dengan

persamaan l = μ0 * n² * a /

l dengan l induktansi diri solenoida

satuannya Henry mu 0 permeabilitas ruang

hampa nilainya 4pi *

10^-7

weber/ampm n jumlah lilitan a luas

bidang kumparan satuannya met² dan l

panjang solenoida satuannya

meter dengan cara yang sama kita juga

dapat menentukan besar induktansi diri

pada toroida karena induksi magnetik

pada toroida adalah b = μ0 * n * i / 2P

maka l = n * a / i * μ0 * n * i / 2P

kita Sederhanakan sehingga besar

induktansi diri pada teroroida dapat

ditentukan dengan persamaan l = μ0 * n²

* a / 2π dengan l induktansi diri

toroida satuannya Henry miol

permeabilitas ruang hampa n jumlah

lilitan a luas bidang kumparan satuannya

meter² dan a jari-jari toroida satuannya

meter pemasangan kumparan pada rangkaian

listrik dapat menimbulkan GGL induksi

Oleh karena itu kumparan dikenal juga

dengan induktor energi yang tersimpan

dalam kumparan atau induktor dapat

ditentukan dengan persamaan w = 1/2 * l

* i² dengan W energi yang tersimpan

dalam induktor satuannya Joule l

induktansi diri satuannya Henry I kuat

arus listrik satuannya

amper selanjutnya kita akan membahas

mengenai induktansi silang atau dikenal

juga dengan induktansi timbal balik pada

gambar terlihat dua kumparan yaitu

kumparan primer P dan kumparan sekunder

S yang berdekatan satu sama lain apabila

hambatan reostat R diubah-ubah maka akan

mengalir arus yang berubah-ubah pada

kumparan primer sehingga akan timbul GGL

induksi pada kumparan sekunder GGL

induksi pada kumparan sekunder ini

menimbulkan arus induksi yang

berubah-ubah sehingga kembali

menimbulkan GGL induksi pada kumparan

primer GGL yang timbul pada kumparan

primer maupun kumparan sekunder ini

disebut induktansi silang atau

induktansi timbal balik yang dapat

ditentukan dengan persamaan m = N1 * pi2

/ I2 = N2 * pi1 /

I1 dengan m induktansi timbal balik

satuannya Henri N1 jumlah lilitan pada

kumparan 1 N2 jumlah lilitan pada

kumparan 2 pi1 fluks magnetik pada

kumparan 1 pi2 fluks magnetik pada

kumparan 2 keduanya dalam satuan Weber

I1 arus listrik pada kumparan 1 I2 arus

listrik pada kumparan dua keduanya dalam

satuan

amper dengan mensubstitusikan persamaan

fluks magnetik Pi = b * a dan induksi

magnetik solenoida b = μ0 * n * i / l

maka diperoleh rumus induktansi silang

pada solenoida sebagai berikut m = 0 *

N1 * N2 * a / l sedangkan besar

induktansi silang pada toroida

ditentukan dengan persamaan m = μ0 * N1

* N2 * a Dib dengan 2

pi dengan m induktansi timbal balik

satuannya Henri N1 jumlah lilitan pada

kumparan 1 N2 jumlah lilitan pada

kumparan 2 mi0 permeabilitas ruang hampa

a luas bidang kumparan satuannya meter²

F panjang solenoida dan a jari-jari

toroida keduanya dalam satuan meter

sampai di sini teman-teman paham

ya agar teman-teman semakin paham Mari

kita selesaikan contoh soal

berikut soal pertama Diketahui sebuah

kumparan dengan induktansi l = 0,8 Henry

dialiri arus listrik yang berubah-ubah

terhadap waktu dengan persamaan I = 2t^

- 4T + 3 I AMP dan t se saat GGL induksi

diri ep S terjadi sebesar

-1,6 volt kita diminta menentukan waktu

T untuk menjawab soal ini kita gunakan

persamaan

s = - l *

di/dt

-1,6 = -0 0,8 *

d/dt dari 2t² - 4T + 3 -1,6

/

-0,8 = 4T - 4 2 = 4T - 4 4T = 6 t = 6 /

4 = 1,5 se jadi jawabannya

C soal kedua Diketahui sebuah kumparan

terdiri atas n = 200 lilitan dalam waktu

Delta t = 0,1 se mengalami perubahan

kuat arus dari I1 = 6 AMP menjadi I2 = 8

AMP dan menimbulkan fluks magnetik

sebesar Delta Pi =

10^-3 Weber kita diminta menentukan

besar induk ansi diri l yang diberikan

kumparan untuk menjawab soal ini kita

gunakan persamaan l = n * deltaap /

dengan Delta I = 200 *

10^-3 di/ dengan 2 =

10^-1 sama juga dengan 0,1 Henry

Jawabannya

d soal selanjutnya diketahui kabel

telepon berbentuk spiral berdiameter d =

1 cm jumlah lilitan kabel n = 100

lilitan apabila panjang kabel saat tidak

direnggangkan l = 50 cm dan

permeabilitas ruang hampa 0 4P *

10^--7 weber/ AMPM kita diminta

menentukan induktansi diri l pada kabel

telepon tersebut untuk menjawab soal ini

terlebih dahulu kita tentukan luas kabel

telepon a =

1/4πd^ = 1/4 * pi * 1² =

0,25π cm² =

2,5p *

10^-5 m² selanjutnya kita Tentukan

induktansi diri kabel telepon berbentuk

spiral dengan persamaan l = 0 * n² * A /

dengan l kita masukkan nilainya kita

lakukan perhitungan kita peroleh nilai

induktansi diri dari kabel telepon

sebesar 2p^ *

10^-7 hry jadi jawabannya

B soal keempat diketahui sebuah toroida

terdiri dari n = 1000 lilitan dengan

jari-jari a = 20 cm toroida kemudian

dialiri arus listrik I = 10 AMP kita

diminta menentukan besar energi yang

tersimpan dalam toroida tersebut yaitu W

untuk menjawab soal ini terlebih dahulu

kita tentukan luas penampang toroida a =

pi a² = Pi * 2 *

10^-1 di = 4P *

10^-2 m² selanjutnya kita tentukan besar

induktansi diri toroida dengan persamaan

l = μ0 * n² * a / 2π kita masukkan

nilainya kita lakukan perhitungan kita

peroleh nilai l = 8p * 10^ -2 Henri

terakhir kita tentukan nilai energi yang

tersimpan dalam toroida dengan persamaan

w = 1/2 * l * i² = 1/2 * 8p * 10^ -2 *

10² = 4π Jou jawabannya

C soal terakhir diketahui solenoida

sepanjang l = 2

cm terdiri atas N1 = 500 lilitan di luar

solenoida dililitkan kawat sebanyak N2 =

200 lilitan Apabila luas penampang

solenoida a = 20

cm^ kita diminta menentukan induktansi

timbal balik kedua solenoida yaitu m

untuk menjawab soal ini kita gunakan

persamaan m = 0 * n 1 * N2 * a / l kita

masukkan nilainya kita lakukan

perhitungan kita peroleh nilai m = 40p *

10^-5 Henry ama juga dengan 4pi *

10^-4 Henry Jawabannya

a Oke teman-teman demikianlah pembahasan

kita tentang induktansi diri jangan lupa

tonton terus video-video terbaru di

channel kita ya sampai jumpa di video

[Musik]

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