Table of Contents
- 1. The private magnetic field of a closed circuit
- 2. What is self-inductance? The formula of self-inductance
- 3. Self-induced electromotive force (EMF) and the formula for calculating the coefficient of self-inductance
- 4. Applications of self-inductance in real life
- 5. Types of exercises on self-inductance and how to solve them
- 6. Multiple-choice exercises on self-inductance
1. The private magnetic field of a closed circuit
In a closed circuit (C) where there is an excessive current (i), the current running through the circuit creates a magnetic field. This magnetic field in turn generates a private magnetic flux Φ passing through the closed circuit (C), known as the private magnetic field of the circuit.
The private magnetic flux of a closed circuit with a current running through it is calculated as follows:
Φ = Li
Here, L is a coefficient that depends only on the structure and size of the closed circuit (C), known as the coefficient of self-inductance.
2. What is self-inductance? The formula of self-inductance
2.1. Defining self-inductance
Self-inductance is a phenomenon in which the electromagnetic induction in an electric circuit is caused by the changing current in the circuit itself.
2.2. Examples of self-inductance
- Experiment 1: When switch K1 is closed, K2 is closed, and K3 is open, bulb 2 immediately lights up, while bulb 1 lights up a bit slower than bulb 2.
Explanation: When switch K is closed, the current through the wire suddenly increases within an extremely short period of time (the current intensity increases from 0 to I), causing the magnetic field through this wire to increase. This increase in the magnetic flux through the wire leads to an induced current according to Lenz’s law, which opposes the increase in magnetic flux. This reduces the current intensity running through bulb 1, causing it to light up later than bulb 2.
- Experiment 2: When switch K1 is closed, K3 is closed, and K2 is open, bulb 3, which was previously off, suddenly lights up and then turns off.
Explanation: When the current through switch K is interrupted, the current decreases rapidly within a very short time period (from I to 0), causing the magnetic field through the coil L to decrease. This decrease in the magnetic flux through the coil L induces a current through the coil in the opposite direction, reducing the current intensity running through bulb 3. After a certain period of time, when there is no more variation in the magnetic flux, the induced current disappears, causing bulb 3 to turn off quickly.
2.3. What is the coefficient of self-inductance?
L is called the coefficient of self-inductance of closed circuit C, and this coefficient depends only on the structure and size of the closed circuit C, also known as the self-inductance of C.
3. Self-induced electromotive force (EMF) and the formula for calculating the coefficient of self-inductance
3.1. Defining self-induced electromotive force
Self-induced electromotive force is the electromotive force that appears in a circuit when there is a self-inductive phenomenon occurring.
The formula for calculating self-induced electromotive force is:
e_tc = -L * (Δi/Δt)
Where:
- e_tc: self-induced electromotive force (unit: V)
- L: coefficient of self-inductance of the coil (unit: H)
- Δi: change in current intensity (unit: A)
- Δt: time for the change in current intensity (unit: s)
- Δi/Δt: rate of change in current intensity (unit: A/s)
According to the formula above, the self-induced electromotive force is directly proportional to the rate of change in current intensity (i) in the circuit over time.
The coefficient of self-inductance for a wire is calculated as follows:
L = (4π 10^-7 n^2 V) / (l S)
Where:
- L: coefficient of self-inductance
- n: number of wire turns
- V: volume of the wire (m^3)
- l: length of the wire (m)
- S: cross-sectional area of the wire (m^2)
The unit of self-inductance is Henry (H), calculated as 1H = 1Wb/1A.
4. Applications of self-inductance in real life
Self-inductance has a wide range of applications in daily life, such as in AC circuits.
An inductor is an important component in AC circuits, oscillating circuits, and transformers, etc.
5. Types of exercises on self-inductance and how to solve them
Below are some exercises on the phenomenon of self-inductance to help you understand the basics and methods of solving problems in this area:
Exercise 1: Given a cylindrical wire with a length of l = 0.5 m and 1000 turns, and each wire loop has a diameter of 20 cm. Calculate the self-inductance of the wire.
Solution:
Exercise 2: Given a wire with a length of 40 cm, a total of 800 wire loops, and a cross-sectional area of 10 cm^2. Calculate the self-inductance of the wire.
Solution:
Exercise 3: A wire has a length of 40 cm, a total of 800 wire loops, and a cross-sectional area of 10 cm^2. The wire is connected to a power source that gradually increases the current intensity from 0 A to 4 A.
a) What is the self-inductance of the wire?
b) If the self-induced electromotive force of the wire has a magnitude of 1.2 V, determine the time for which the current intensity changes.
Solution:
Exercise 4: A long wire is wound with a density of 2000 turns per meter. The wire has a volume of 500 cm^3. The wire is connected to an electric circuit. After closing the switch, the current changes according to the graph below. The time when the switch is closed corresponds to t = 0. Calculate the self-induced electromotive force in the wire:
a) From t = 0 to t = 0.05 s.
b) After t = 0.05 s.
Solution:
Exercise 5: A coil has a self-inductance L = 50 mH and is connected in series with a resistor R = 20 Ω to a new power source with an electromotive force of 90 V and negligible internal resistance. Determine the rate of change in the current intensity I at:
a) The initial time when I = 0.
b) The time when I = 2 A.
Solution:
6. Multiple-choice exercises on self-inductance
Exercise 1: A wire with a current intensity of 3 A accumulates a magnetic field energy of 10 mJ. If a current of 9 A runs through the wire, the accumulated energy is:
A. 30 mJ
B. 60 mJ
C. 90 mJ
D. 120 mJ
Exercise 2: A wire with a length of 10 cm and a cross-sectional area of 10 cm^2 has a current intensity of 10 A running through it. The accumulated energy inside the wire is 0.1 J. Calculate the self-inductance of the wire:
A. 0.5 mH
B. 1 mH
C. 2 mH
D. 5 mH
Exercise 3: A wire with a self-inductance coefficient of 0.02 H has an energy of 0.25 J. Determine the current intensity running through the wire:
A. 5 A
B. 10 A
C. 15 A
D. 25 A
Exercise 4: A wire with a self-inductance coefficient of L = 0.01 H has a current intensity of i = 5 A running through it. Calculate the magnetic field energy inside the wire:
A. 0.025 J
B. 0.05 J
C. 0.125 J
D. 0.25 J
Exercise 5: If the wire loops are compressed on the wire, halving its length, how does the self-inductance coefficient of the wire change?
A. Decreases by two times
B. Increases by two times
C. Decreases by four times
D. Increases by four times
Exercise 6: A wire with a known length is wound onto a wire with double its length, but with the same cross-sectional area. What is the self-inductance coefficient of the wire?
A. 0.05 mH
B. 0.1 mH
C. 0.4 mH
D. 0.8 mH
Exercise 7: The coefficient of self-inductance of a wire depends on:
A. The number of wire loops
B. The volume of the wire
C. The magnetic flux through the wire when a current runs through
D. The magnetic field generated when a current runs through the wire
Exercise 8: A wire with a self-inductance coefficient of 0.4 H has a current that decreases uniformly over time from 1.2 A to 0.4 A within a time period of 0.2 s. Calculate the self-induced electromotive force in the wire:
A. 0.8 V
B. 1.6 V
C. 2.4 V
D. 3.2 V
Exercise 9: A wire has a current intensity that increases uniformly over time from 0.2 A to 1.8 A within a time period of 0.02 s. The wire has a self-inductance coefficient of 2 H. Calculate the self-induced electromotive force in the wire:
A. 10 V
B. 40 V
C. 80 V
D. 160 V
Exercise 10: A coil with a self-inductance coefficient of L = 1.2 H has a current that uniformly decreases from 2.4 A to 1.2 A within a time period of 0.5 minutes. Calculate the self-induced electromotive force that appears in the coil during this time period:
A. 38 mV
B. 84 mV
C. 48 mV
D. 83 mV
These exercises will help you solidify your understanding of self-inductance. Don’t forget to visit Kienthucykhoa.com to learn more about this fascinating topic and prepare for the upcoming national high school graduation exam.
