Physics Lesson 16.9.1 - The Meaning of Inductance

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Welcome to our Physics lesson on The Meaning of Inductance, this is the first lesson of our suite of physics lessons covering the topic of Inductance and Self-Induction, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

The Meaning of Inductance

So far, we have seen many similarities between electric and magnetic phenomena and we have used the analogy between the corresponding quantities to have a better understanding on the new quantities (usually magnetic ones, because electricity was explained earlier). For instance, we have described the magnetic field through the help of electric field, magnetic flux using the analogy with electric flux and so on.

Let's use the same method to explain a new magnetic-related concept. In tutorial 14.7 "Capacitance and Capacitors", we have seen that capacitors are circuit components used to store electric charges in their plates. In this way, we can produce a desired electric field between the plates of a capacitor because they are charged by opposite signs. We considered the basic arrangement of capacitors as the parallel plate capacitor, having the symbol (-| |-).

Similarly, we call an "inductor" a device used to produce a desired magnetic field. The symbol of inductor is ( Physics and Math Symbols: Inductor Symbol ). A solenoid is the most typical example of conductor.

Let's consider as solenoid connected to an electric circuit that contains a battery and a rheostat (variable resistor) as shown in the figure. This is a kind of electromagnet because the current passing around the coil produces a magnetic field similar to that of a bar magnet.

Physics Tutorials: This image provides visual information for the physics tutorial Inductance and Self-Induction

The solenoid has N turns and its length is l. The rheostat is initially in the position 1 and therefore, its resistance is R₀ and the current produced in the circuit is I₀. As a result, the initial magnetic field produced by this electromagnet is B₀.

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, as shown in the figure below.

The initial current flowing through the solenoid is

I₀ = ε/R₀

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 = μ₀ ∙ 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

Φ₀ = B₀ ∙ A
= μ₀ ∙ N ∙ I₀/l ∙ A

where A is the area of solenoid loops, and the final flux through the solenoid is

Φ = B ∙ A
= μ₀ ∙ 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-Φ₀/Δt
= -N ∙ μ₀ ∙ N ∙ I/L ∙ A-μ₀ ∙ N ∙ I₀/l ∙ A/Δt
= -N ∙ (μ₀ ∙ N ∙ A/l) ∙ (I-I₀₀/Δt)

Thus, we obtain for the self-induced emf in the coil:

ε' = -μ₀ ∙ N² ∙ A/I ∙ ∆I/Δt

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 = μ₀ ∙ N² ∙ A/I

we obtain for the self-induced emf in the coil in terms of inductance:

ε' = -L ∙ ∆I/Δt

As stated at the beginning of this tutorial, coils are known as "inductors" just because of their property of self-inductance.

Example 1

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.

Calculate the self-inductance of the solenoid

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 1

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' = ?

First, we calculate the number of turns in the solenoid. We have
N = N ∙ l
= 500 turns/m ∙ 0.20 m
= 100 turns
= 10² turns
The self-inductance of the solenoid is
L = μ₀ ∙ N² ∙ A/I
= (4 ∙ 3.14 ∙ 10^-7 N/A² ) ∙ (10²)² ∙ (4 × 10^-4 m² )/(0.20 m)
= 2.512 × 10^-5 H

You have reached the end of Physics lesson 16.9.1 The Meaning of Inductance. There are 3 lessons in this physics tutorial covering Inductance and Self-Induction, you can access all the lessons from this tutorial below.

More Inductance and Self-Induction Lessons and Learning Resources

Magnetism Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
16.9	Inductance and Self-Induction
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
16.9.1	The Meaning of Inductance
16.9.2	Another Approach on Inductance
16.9.3	Applications of Inductance in Technology

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