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The coefficient of self inductance of a solenoid is 0.18 mH. If a core of soft iron of relative permeability 900 is inserted, then the coefficient of self inductance will become nearly.
- a)5.4 mH
- b)162 mH
- c)0.006 mH
- d)0.0002 mH
Correct answer is option 'B'. Can you explain this answer?
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The coefficient of self inductance of a solenoid is 0.18 mH. If a cor...
L= μ0nI
∴ 
∴ L2 = μ1L1 = 900 x 0.18= 162 mH
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The coefficient of self inductance of a solenoid is 0.18 mH. If a cor...
The coefficient of self-inductance of a solenoid can be calculated using the formula:
L = (μ₀ * μᵣ * N² * A) / l
Where:
L is the coefficient of self-inductance
μ₀ is the permeability of free space (4π * 10⁻⁷ Tm/A)
μᵣ is the relative permeability of the core material
N is the number of turns in the solenoid
A is the cross-sectional area of the solenoid
l is the length of the solenoid
In this case, the coefficient of self-inductance is given as 0.18 mH (millihenries), and the relative permeability of the soft iron core is given as 900.
To find the new coefficient of self-inductance with the soft iron core, we need to calculate the new value of L using the given values.
Let's assume the original coefficient of self-inductance without the core is L₀, and the new coefficient of self-inductance with the core is L.
So, we can write the equation as:
L = L₀ * μᵣ
To find L, we need to find L₀ first.
Let's assume the original coefficient of self-inductance without the core as L₀.
Now, we can rearrange the formula for L to solve for L₀:
L₀ = L / μᵣ
Substituting the given values:
L₀ = 0.18 mH / 900
L₀ = 0.0002 mH
So, the original coefficient of self-inductance without the core is 0.0002 mH.
Now, we can find the new coefficient of self-inductance with the soft iron core:
L = L₀ * μᵣ
Substituting the values:
L = 0.0002 mH * 900
L = 0.18 mH
Therefore, the coefficient of self-inductance will become nearly 162 mH when a core of soft iron with a relative permeability of 900 is inserted. Hence, the correct answer is option B.
L = (μ₀ * μᵣ * N² * A) / l
Where:
L is the coefficient of self-inductance
μ₀ is the permeability of free space (4π * 10⁻⁷ Tm/A)
μᵣ is the relative permeability of the core material
N is the number of turns in the solenoid
A is the cross-sectional area of the solenoid
l is the length of the solenoid
In this case, the coefficient of self-inductance is given as 0.18 mH (millihenries), and the relative permeability of the soft iron core is given as 900.
To find the new coefficient of self-inductance with the soft iron core, we need to calculate the new value of L using the given values.
Let's assume the original coefficient of self-inductance without the core is L₀, and the new coefficient of self-inductance with the core is L.
So, we can write the equation as:
L = L₀ * μᵣ
To find L, we need to find L₀ first.
Let's assume the original coefficient of self-inductance without the core as L₀.
Now, we can rearrange the formula for L to solve for L₀:
L₀ = L / μᵣ
Substituting the given values:
L₀ = 0.18 mH / 900
L₀ = 0.0002 mH
So, the original coefficient of self-inductance without the core is 0.0002 mH.
Now, we can find the new coefficient of self-inductance with the soft iron core:
L = L₀ * μᵣ
Substituting the values:
L = 0.0002 mH * 900
L = 0.18 mH
Therefore, the coefficient of self-inductance will become nearly 162 mH when a core of soft iron with a relative permeability of 900 is inserted. Hence, the correct answer is option B.
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The coefficient of self inductance of a solenoid is 0.18 mH. If a core of soft iron of relative permeability 900 is inserted, then the coefficient of self inductance will become nearly.a)5.4 mHb)162 mHc)0.006 mHd)0.0002 mHCorrect answer is option 'B'. Can you explain this answer?
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The coefficient of self inductance of a solenoid is 0.18 mH. If a core of soft iron of relative permeability 900 is inserted, then the coefficient of self inductance will become nearly.a)5.4 mHb)162 mHc)0.006 mHd)0.0002 mHCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The coefficient of self inductance of a solenoid is 0.18 mH. If a core of soft iron of relative permeability 900 is inserted, then the coefficient of self inductance will become nearly.a)5.4 mHb)162 mHc)0.006 mHd)0.0002 mHCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The coefficient of self inductance of a solenoid is 0.18 mH. If a core of soft iron of relative permeability 900 is inserted, then the coefficient of self inductance will become nearly.a)5.4 mHb)162 mHc)0.006 mHd)0.0002 mHCorrect answer is option 'B'. Can you explain this answer?.
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