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In this Physics tutorial, you will learn:

- What is inductance?
- What is an inductor?
- What is the self-induced emf?
- How do we measure the self-inductance?
- What are the factors affecting the self-inductance?
- What are some applications of inductance in technology?

I_{0} = ε/R_{0}

When the sliding contact of rheostat is moved in the new position, the current flowing through the solenoid becomes I = ε/R

We have explained in the tutorial 16.2 that the magnetic field produced by a solenoid is B = (μ_{0} ∙ N ∙ I)/l

where μ0 is the vacuum permeability, N is the number of turns in the solenoid, I is the current flowing through the circuit and l is the solenoid's length. The initial magnetic flux through the solenoid therefore is Φ_{0} = B_{0} ∙ A = (μ_{0} ∙ N ∙ I_{0})/l ∙ A

where A is the area of solenoid loops, and the final flux through the solenoid is Φ = B ∙ A = (μ_{0} ∙ N ∙ I)/l ∙ A

Therefore, the induced emf produced by the solenoid (known as self-induced emf because it is not generated by the flux change due to any motion in respect to an external magnetic field), is ε^' = -N ∙ ∆Φ/Δt = -N ∙ (Φ-Φ_{0})/Δt = -N ∙ ((μ_{0} ∙ N ∙ I)/l ∙ A-(μ_{0} ∙ N ∙ I_{0})/l ∙ A)/Δt = -N ∙ ((μ_{0} ∙ N ∙ A)/l) ∙ ((I-I_{0})/∆t)

Thus, we obtain for the self-induced emf in the coil: ε^' = -((μ_{0} ∙ N^{2} ∙ A)/l) ∙ ∆I/∆t

We denote by L the expression inside the brackets. This quantity is known as self-inductance (or simply inductance). It depends only on the physical features of the solenoid (number of turns, area, length) and not on the electric properties of the circuit. The unit of self-inductance is known as Henry (H). Thus, having L = (μ_{0} ∙ N^{2} ∙ A)/l

we obtain for the self-induced emf in the coil in terms of inductance: ε^' = -L ∙ ∆I/∆t

As stated at the beginning of this tutorial, coils are known as "inductors" just because of their property of self-inductance. Example: A 20 cm long solenoid having a cross sectional area of 4 cm2 contains 500 turns per metre. The solenoid is connected to a 12 V battery through a rheostat. a) Calculate the self-inductance of the solenoid b) Find the self-induced emf of solenoid if the resistance in the rheostat decreases from 24 Ω to 6 Ω in 0.4 seconds. (Take the figure shown in theory section as a reference) Solution Clues: l = 20 cm = 0.20 m A = 4 cm2 = 0.0004 m2 = 4 × 10-4 m2 n = 500 turns/metre = 5 × 102 turns/metre ε = 12 V (μ0 = 4π × 10-7 N/A2) R0 = 24 Ω R = 6 Ω Δt = 0.4 s a) L = ? b) ε' = ? a) First, we calculate the number of turns in the solenoid. We have N = n ∙ l = 500 turns/m ∙ 0.20 m = 100 turns = 10^{2} turns

The self-inductance of the solenoid is L = (μ_{0} ∙ N^{2} ∙ A)/l = ((4 ∙ 3.14 ∙ 10^{-7} N/A^{2} ) ∙ 〖(10^{2})〗^{2} ∙ (4 × 10^{-4} m^{2} ))/((0.20 m) ) = 2.512 × 10^{-5} H

Another Approach on Inductance If a current I is flowing through the turns of a solenoid (called henceforth an "inductor"), it produces a magnetic flux Φm through the central region of the inductor. As a result, we obtain for the inductance of inductor in terms of magnetic flux: L = (N ∙ Φ_m)/I

where N is the number of turns in the inductor. Indeed, since L = (μ_{0} ∙ N^{2} ∙ A)/l

and Φ_m = N ∙ B ∙ A

We obtain by combining the two above formulae (considering also the fact that the magnetic field of a solenoid is B = (μL = (N ∙ Φ_m)/I = (N ∙ B ∙ A)/I = (N ∙ (μ_{0} ∙ N ∙ I)/l ∙ A)/I = (μ_{0} ∙ N^{2} ∙ A)/l

Therefore, the two formulae of inductance given above are equivalent. Example: When a 10 cm long solenoid containing 200 turns and having the cross-sectional area of each turn equal to 4 cm2 is connected to a 24 V power source. The resistance of the circuit is 48 Ω. What is the magnetic field the solenoid generates in such conditions? Clues: l = 10 cm = 10-1 m N = 200 turns = 2 × 102 turns A = 4 cm2 = 4 × 10-4 m2 ε = 24V R = 48 Ω (μ0 = 4π × 10-7 N/A2) B = ? Solution First, we calculate the current flowing through the circuit through the Ohm's law. We have I = ε/R = (24 V)/(48 Ω) = 0.5 A

Now, let's calculate the inductance in the solenoid. We have L = (μ_{0} ∙ N^{2} ∙ A)/l = ((4 ∙ 3.14 × 10^{-7} N/A^{2} ) ∙ (2 × 10^{2} )^{2} ∙ (4 × 10^{-4} m^{2} ))/((10^{-1} m) ) = 2 × 10^{-4} H

Now, let's use the other formula of inductance to find the magnetic field through the solenoid. Thus, giving that L = (N ∙ Φ_m)/I = (N ∙ B ∙ A)/I

we obtain B = (I ∙ L)/(N ∙ A) = ((0.5 A) ∙ (2 × 10^{-4} H))/((2 × 10^{2} ) ∙ (4 × 10^{-4} m^{2} ) ) = 1.25 × 10^{-3} T = 1.25 mT

Applications of Inductance in Technology There is a wide range of inductance application in modern electronic devices. Some of them include: a) Filters. Inductors are used together with capacitors and resistors (as we will see in the upcoming tutorials) to create filters that prevent undesired frequencies from mixing with the signal and allowing only certain bands of signal frequency (the desired ones) to pass through. b) Sensors. Inductors are very good in detecting magnetic fields or the presence of magnetic materials from a certain distance. Therefore, they are used in cars, traffic lights, measuring devices etc. However, inductive sensors are limited in two major ways. Either the object to be sensed must be magnetic and induce a current in the sensor, or the sensor must be connected to a power source to detect the presence of materials that interact with a magnetic field. These parameters limit the applications of inductive sensors and influence the designs that use them. c) Transformers. They are devices that are used to change the value of potential difference in a power transmission system in order to save energy or to make the values fit the electric appliances. Transformers are devices that are built by combining inductors that have a shared magnetic path. d) Electric Motors. We have seen that electric motors are operated by making a coil turn between two magnets. This action bring a rotation of the circuit that powers the coil as well. This is not practical and consumes a lot of energy. To prevent this, an inductor is installed in the place where the circuit and the coil meet. The inductor does not require any direct contact between the parts it connects, so the coil can rotate while the supplying circuit remains stationary. e) Tape recorder. Although an outdated device, tape recorder is one of the most important inventions of the last century. It converts sound into electric signal through a microphone and an amplifier. This current then passes through the coil of an electromagnet. After a few processes, the tape is magnetised, causing a varying flux through the playback coil. As a result, the original voice is reproduced in the original frequencies giving an identical pattern when replayed. Summary Inductors are devices used to produce a desired magnetic field. The symbol of inductor is ( ). A solenoid is the most typical example of conductor. Inductors are analogue to capacitors, which are circuit components used to store electric charges in their plates producing in this way a desired electric field between their plates as they are charged by opposite signs. We can change the magnetic field produced in the coil by changing the value of resistance of the circuit. This can be achieved by moving the sliding contact of rheostat in another position. As a result, we will obtain the new values: R, I and B for the corresponding quantities from R0, I0 and B0 they were initially. The induced emf produced by the solenoid (known as self-induced emf because it is not generated by the flux change due to any motion in respect to an external magnetic field), is ε^' = -N ∙ ∆Φ/Δt

Since the magnetic field of a solenoid is B = (μ_{0} ∙ N ∙ I)/l

and the change in magnetic flux through the solenoid is ∆Φ = A ∙ ∆B

then, the self-induced emf in the inductor is ε^' = -((μ_{0} ∙ N^{2} ∙ A)/l) ∙ ∆I/∆t

The expression inside the brackets is called inductance, L. It is measured in Henry [H]. Hence, the self-induced emf in terms of inductance is ε^' = -L ∙ ∆I/∆t

If a current I is flowing through the turns of a solenoid (called henceforth an "inductor"), it produces a magnetic flux Φm through the central region of the inductor. As a result, we obtain for the inductance of inductor in terms of magnetic flux: L = (N ∙ Φ_m)/I

where N is the number of turns in the inductor. There is a wide range of inductance application in modern electronic devices. Some of them include filters, sensors, transformers, electric motors, tape recorders, etc. **1)** What is the magnitude of the self-induced emf in a 200 mH coil if the current changes steadily from 5A to 20A in 0.02s?

- 15 V
- 150 V
- 50 V
- 200 V

**Correct Answer: B**

**2)** How many turns must a solenoid have to make a 0.5 H conductor if the solenoid is 15 cm long and its diameter is 4 cm?

- 7
- 14
- 21
- 28

**Correct Answer: B**

**3)** A 12 V emf is induced in a 2 H coil by a current that rises uniformly from zero to I in 0.1 s. What is the value of the current I?

- 0.6 A
- 1.2 A
- 6 A
- 60 A

**Correct Answer: A**

We hope you found this Physics tutorial "Inductance and Self-Induction" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of Magnetism with our Physics tutorial on Induction and Energy Transfers. Induced Electric Fields .

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