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Welcome to our Physics lesson on Another Approach on Inductance, this is the second 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.
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
and
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?
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 = ?
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
we obtain
You have reached the end of Physics lesson 16.9.2 Another Approach on Inductance. There are 3 lessons in this physics tutorial covering Inductance and Self-Induction, you can access all the lessons from this tutorial below.
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