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A long solenoid has 1000 turns. When a current of 4A flows through it, the magnetic flux linked with each turn of the solenoid is 4 x 10-3 Wb. The self-inductance of the solenoid is
- a)1H
- b)4H
- c)3H
- d)2H
Correct answer is option 'A'. Can you explain this answer?
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A long solenoid has 1000 turns. When a current of 4A flows through it,...
Given:
Number of turns, N = 1000
Current, I = 4 A
Magnetic flux per turn, Φ = 4 × 10-3 Wb
To find: Self-inductance, L
Formula: L = Φ / I
Calculation:
L = Φ / I
L = (4 × 10-3 Wb) / (4 A)
L = 1 × 10-3 H
L = 1 H (since 1 H = 1000 mH = 1000 × 10-3 H)
Therefore, the self-inductance of the solenoid is 1 H.
Explanation:
The self-inductance of a solenoid depends on its geometry, i.e., number of turns (N), length (l), and radius (r), and the permeability of the material inside the solenoid. However, in this question, we are given the magnetic flux per turn (Φ) and the current (I) flowing through the solenoid.
The self-inductance of a solenoid is defined as the ratio of the magnetic flux linked with each turn to the current flowing through the solenoid. Hence, using the formula L = Φ / I, we can calculate the self-inductance of the solenoid.
In this case, the magnetic flux per turn is given as 4 × 10-3 Wb and the current flowing through the solenoid is 4 A. Therefore, substituting these values in the formula, we get the self-inductance of the solenoid as 1 H.
Number of turns, N = 1000
Current, I = 4 A
Magnetic flux per turn, Φ = 4 × 10-3 Wb
To find: Self-inductance, L
Formula: L = Φ / I
Calculation:
L = Φ / I
L = (4 × 10-3 Wb) / (4 A)
L = 1 × 10-3 H
L = 1 H (since 1 H = 1000 mH = 1000 × 10-3 H)
Therefore, the self-inductance of the solenoid is 1 H.
Explanation:
The self-inductance of a solenoid depends on its geometry, i.e., number of turns (N), length (l), and radius (r), and the permeability of the material inside the solenoid. However, in this question, we are given the magnetic flux per turn (Φ) and the current (I) flowing through the solenoid.
The self-inductance of a solenoid is defined as the ratio of the magnetic flux linked with each turn to the current flowing through the solenoid. Hence, using the formula L = Φ / I, we can calculate the self-inductance of the solenoid.
In this case, the magnetic flux per turn is given as 4 × 10-3 Wb and the current flowing through the solenoid is 4 A. Therefore, substituting these values in the formula, we get the self-inductance of the solenoid as 1 H.
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