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A long solenoid has 500 turns . When a current of 2 ampere is passed through it , the resulting magnetic flux linked with each turn of the solenoid is 4 x 10-3 Wb. The self-inductance of the solenoid is
- a)4.0 henry
- b)2.5 henry
- c)2.0 henry
- d)1.0 henry
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A long solenoid has 500 turns . When a current of 2 ampere is passed t...
For a long solenoid,
B = μ0ni = μ0 N/l i
Flux = μ0 N/l iA
given flux per turn = 4 x 10-3, i = 2A
∴ Total flux = 4 x 10-3
B = μ0ni = μ0 N/l i
Flux = μ0 N/l iA
given flux per turn = 4 x 10-3, i = 2A
∴ Total flux = 4 x 10-3
Most Upvoted Answer
A long solenoid has 500 turns . When a current of 2 ampere is passed t...
Given:
n = 500 (number of turns)
I = 2 A (current)
Φ = 4 × 10⁻³ Wb (magnetic flux)
We know that self-inductance (L) of a solenoid can be calculated using the formula:
L = (μ₀n²A)/l
where,
μ₀ = permeability of free space = 4π × 10⁻⁷ T m A⁻¹
n = number of turns
A = area of cross-section of the solenoid
l = length of the solenoid
To find the self-inductance of the solenoid, we need to calculate the area of cross-section and length of the solenoid.
Area of cross-section of the solenoid:
We know that the magnetic flux linked with each turn of the solenoid is given by:
Φ = B × A
where,
B = magnetic field inside the solenoid
A = area of cross-section of the solenoid
From this equation, we can calculate the area of cross-section as:
A = Φ/B
We know that the magnetic field inside a solenoid is given by:
B = μ₀nI
where,
I = current passing through the solenoid
Substituting the given values, we get:
B = (4π × 10⁻⁷) × 500 × 2
B = 1 × 10⁻³ T
Substituting the values of Φ and B, we get:
A = Φ/B
A = (4 × 10⁻³)/(1 × 10⁻³)
A = 4 m²
Length of the solenoid:
We know that the magnetic field inside a solenoid is uniform and given by:
B = μ₀nI
where,
I = current passing through the solenoid
We also know that the magnetic field inside a solenoid is given by:
B = μ₀n²I/l
where,
l = length of the solenoid
Equating the above two equations, we get:
μ₀nI = μ₀n²I/l
l = n
Substituting the value of n, we get:
l = 500 m
Now, we can calculate the self-inductance of the solenoid using the formula:
L = (μ₀n²A)/l
Substituting the given values, we get:
L = (4π × 10⁻⁷ × 500² × 4)/(500)
L = 1 H
Therefore, the self-inductance of the solenoid is 1 H.
n = 500 (number of turns)
I = 2 A (current)
Φ = 4 × 10⁻³ Wb (magnetic flux)
We know that self-inductance (L) of a solenoid can be calculated using the formula:
L = (μ₀n²A)/l
where,
μ₀ = permeability of free space = 4π × 10⁻⁷ T m A⁻¹
n = number of turns
A = area of cross-section of the solenoid
l = length of the solenoid
To find the self-inductance of the solenoid, we need to calculate the area of cross-section and length of the solenoid.
Area of cross-section of the solenoid:
We know that the magnetic flux linked with each turn of the solenoid is given by:
Φ = B × A
where,
B = magnetic field inside the solenoid
A = area of cross-section of the solenoid
From this equation, we can calculate the area of cross-section as:
A = Φ/B
We know that the magnetic field inside a solenoid is given by:
B = μ₀nI
where,
I = current passing through the solenoid
Substituting the given values, we get:
B = (4π × 10⁻⁷) × 500 × 2
B = 1 × 10⁻³ T
Substituting the values of Φ and B, we get:
A = Φ/B
A = (4 × 10⁻³)/(1 × 10⁻³)
A = 4 m²
Length of the solenoid:
We know that the magnetic field inside a solenoid is uniform and given by:
B = μ₀nI
where,
I = current passing through the solenoid
We also know that the magnetic field inside a solenoid is given by:
B = μ₀n²I/l
where,
l = length of the solenoid
Equating the above two equations, we get:
μ₀nI = μ₀n²I/l
l = n
Substituting the value of n, we get:
l = 500 m
Now, we can calculate the self-inductance of the solenoid using the formula:
L = (μ₀n²A)/l
Substituting the given values, we get:
L = (4π × 10⁻⁷ × 500² × 4)/(500)
L = 1 H
Therefore, the self-inductance of the solenoid is 1 H.
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A long solenoid has 500 turns . When a current of 2 ampere is passed through it , the resulting magnetic flux linked with each turn of the solenoid is 4 x 10-3 Wb. The self-inductance of the solenoid isa)4.0 henryb)2.5 henryc)2.0 henryd)1.0 henryCorrect answer is option 'D'. Can you explain this answer?
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A long solenoid has 500 turns . When a current of 2 ampere is passed through it , the resulting magnetic flux linked with each turn of the solenoid is 4 x 10-3 Wb. The self-inductance of the solenoid isa)4.0 henryb)2.5 henryc)2.0 henryd)1.0 henryCorrect answer is option 'D'. 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 A long solenoid has 500 turns . When a current of 2 ampere is passed through it , the resulting magnetic flux linked with each turn of the solenoid is 4 x 10-3 Wb. The self-inductance of the solenoid isa)4.0 henryb)2.5 henryc)2.0 henryd)1.0 henryCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A long solenoid has 500 turns . When a current of 2 ampere is passed through it , the resulting magnetic flux linked with each turn of the solenoid is 4 x 10-3 Wb. The self-inductance of the solenoid isa)4.0 henryb)2.5 henryc)2.0 henryd)1.0 henryCorrect answer is option 'D'. Can you explain this answer?.
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