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|16.9||Inductance and Self-Induction|
In this Physics tutorial, you will learn:
What methods do you know that can be used induce emf in a coil?
Do you think the number of turs in the coil has any effect in the induced emf?
Think about changing the value of resistor connected to the coil. What happens to the emf induced in the coil?
In this tutorial, we will discuss about a phenomenon called "self-induction". We encounter it in many electronic devices and it is used to protect them from damage caused by sudden current changes.
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 (). 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.
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
When the sliding contact of rheostat is moved in the new position, the current flowing through the solenoid becomes
We have explained in the tutorial 16.2 that the magnetic field produced by a solenoid is
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
where A is the area of solenoid loops, and the final flux through the solenoid is
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
Thus, we obtain for the self-induced emf in the coil:
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).
we obtain for the self-induced emf in the coil in terms of inductance:
As stated at the beginning of this tutorial, coils are known as "inductors" just because of their property of self-inductance.
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.
(Take the figure shown in theory section as a reference)
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' = ?
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:
where N is the number of turns in the inductor. Indeed, since
We obtain by combining the two above formulae (considering also the fact that the magnetic field of a solenoid is B = μ0 ∙ N ∙ I/L)
Therefore, the two formulae of inductance given above are equivalent.
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?
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 = ?
First, we calculate the current flowing through the circuit through the Ohm's law. We have
Now, let's calculate the inductance in the solenoid. We have
Now, let's use the other formula of inductance to find the magnetic field through the solenoid. Thus, giving that
There is a wide range of inductance application in modern electronic devices. Some of them include:
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