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When the length of the solenoid is doubled without any change in the number of turns and the area of the coil.Then its self-inductance will
- a)Nine times
- b)Half times
- c)Doubled
- d)Unchanged
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
When the length of the solenoid is doubled without any change in the n...
Concept:
- Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
- As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
- This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
- Self-inductance of a solenoid is given by –
Where μo = Absolute permeability, N = Number of turns, l = length of the solenoid, and A = Area of the solenoid.
Explanation
Given - l2 = 2l1
- Self-inductance of a solenoid is given by:
- According to the question, the length of the solenoid is doubled without any change in the number of turns and the area of the coil
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When the length of the solenoid is doubled without any change in the n...
Explanation:
In order to understand why the self-inductance of a solenoid halves when its length is doubled without any change in the number of turns and the area of the coil, we need to consider the formula for the self-inductance of a solenoid.
The self-inductance of a solenoid is given by the formula:
L = (μ₀N²A)/l
Where:
L = self-inductance of the solenoid
μ₀ = permeability of free space (constant)
N = number of turns in the solenoid
A = area of the cross-section of the solenoid
l = length of the solenoid
Analysis:
When the length of the solenoid is doubled, the value of 'l' in the formula is doubled.
If we substitute the new values into the formula, we get:
L' = (μ₀N²A)/(2l)
Now, let's compare the new self-inductance (L') with the original self-inductance (L):
L' = (μ₀N²A)/(2l)
L = (μ₀N²A)/l
Comparison:
To compare L' and L, we can divide L' by L:
L'/L = [(μ₀N²A)/(2l)] / [(μ₀N²A)/l]
L'/L = l / (2l)
L'/L = 1/2
This shows that the new self-inductance (L') is half of the original self-inductance (L).
Conclusion:
Therefore, when the length of a solenoid is doubled without any change in the number of turns and the area of the coil, the self-inductance of the solenoid halves.
In order to understand why the self-inductance of a solenoid halves when its length is doubled without any change in the number of turns and the area of the coil, we need to consider the formula for the self-inductance of a solenoid.
The self-inductance of a solenoid is given by the formula:
L = (μ₀N²A)/l
Where:
L = self-inductance of the solenoid
μ₀ = permeability of free space (constant)
N = number of turns in the solenoid
A = area of the cross-section of the solenoid
l = length of the solenoid
Analysis:
When the length of the solenoid is doubled, the value of 'l' in the formula is doubled.
If we substitute the new values into the formula, we get:
L' = (μ₀N²A)/(2l)
Now, let's compare the new self-inductance (L') with the original self-inductance (L):
L' = (μ₀N²A)/(2l)
L = (μ₀N²A)/l
Comparison:
To compare L' and L, we can divide L' by L:
L'/L = [(μ₀N²A)/(2l)] / [(μ₀N²A)/l]
L'/L = l / (2l)
L'/L = 1/2
This shows that the new self-inductance (L') is half of the original self-inductance (L).
Conclusion:
Therefore, when the length of a solenoid is doubled without any change in the number of turns and the area of the coil, the self-inductance of the solenoid halves.
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When the length of the solenoid is doubled without any change in the number of turns and the area of the coil.Then its self-inductance willa)Nine timesb)Half timesc)Doubledd)UnchangedCorrect answer is option 'B'. Can you explain this answer?
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