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A long solenoid has 500 turns. When a currentof 2 ampere is passed through it, the resultingmagnetic flux linked with each turn of thesolenoid is 4 ×10–3 Wb. The self- inductance ofthe solenoid is [2008]
- a)2.5 henry
- b)2.0 henry
- c)1.0 henry
- d)40 henry
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A long solenoid has 500 turns. When a currentof 2 ampere is passed thr...
Total number of turns in the solenoid,
N = 500
Current, I = 2A.
Magnetic flux linked with each turn
= 4 × 10–3 Wb
N = 500
Current, I = 2A.
Magnetic flux linked with each turn
= 4 × 10–3 Wb
Most Upvoted Answer
A long solenoid has 500 turns. When a currentof 2 ampere is passed thr...
To find the magnetic flux linked with each turn of the solenoid, we can use the formula:
Magnetic Flux (Φ) = Number of Turns (N) * Magnetic Flux Density (B) * Area (A)
Given that the solenoid has 500 turns and a current of 2 amperes is passed through it, we can calculate the magnetic flux density using Ampere's Law:
Magnetic Flux Density (B) = (μ₀ * N * I) / L
Where:
- μ₀ is the permeability of free space (4π * 10^-7 T m/A)
- N is the number of turns (500)
- I is the current (2 A)
- L is the length of the solenoid (unknown)
Substituting the given values, we have:
B = (4π * 10^-7 T m/A) * 500 turns * 2 A / L
Now, we can substitute the magnetic flux density into the formula for magnetic flux:
Φ = N * B * A
Given that the magnetic flux linked with each turn of the solenoid is 4 units, we can write:
4 = 500 turns * B * A
Substituting the value of B, we have:
4 = 500 turns * ((4π * 10^-7 T m/A) * 500 turns * 2 A / L) * A
Simplifying the equation, we find:
4 = 500^2 * (4π * 10^-7 T m/A) * 2 A * A / L
Now, we can solve for L:
L = 500^2 * (4π * 10^-7 T m/A) * 2 A * A / 4
L = 500^2 * (π * 10^-7 T m/A) * 1000 A m^2 / 4
L = 125,000 * π * 10^-7 T m/A * 1000 A m^2 / 4
L = 125,000 * π * 10^-7 T m * 250
L = 125,000 * π * 2.5 * 10^-4 T m
Finally, we can calculate the length of the solenoid:
L = 125,000 * π * 2.5 * 10^-4 T m
L ≈ 0.981 meters
Therefore, the length of the solenoid is approximately 0.981 meters.
Magnetic Flux (Φ) = Number of Turns (N) * Magnetic Flux Density (B) * Area (A)
Given that the solenoid has 500 turns and a current of 2 amperes is passed through it, we can calculate the magnetic flux density using Ampere's Law:
Magnetic Flux Density (B) = (μ₀ * N * I) / L
Where:
- μ₀ is the permeability of free space (4π * 10^-7 T m/A)
- N is the number of turns (500)
- I is the current (2 A)
- L is the length of the solenoid (unknown)
Substituting the given values, we have:
B = (4π * 10^-7 T m/A) * 500 turns * 2 A / L
Now, we can substitute the magnetic flux density into the formula for magnetic flux:
Φ = N * B * A
Given that the magnetic flux linked with each turn of the solenoid is 4 units, we can write:
4 = 500 turns * B * A
Substituting the value of B, we have:
4 = 500 turns * ((4π * 10^-7 T m/A) * 500 turns * 2 A / L) * A
Simplifying the equation, we find:
4 = 500^2 * (4π * 10^-7 T m/A) * 2 A * A / L
Now, we can solve for L:
L = 500^2 * (4π * 10^-7 T m/A) * 2 A * A / 4
L = 500^2 * (π * 10^-7 T m/A) * 1000 A m^2 / 4
L = 125,000 * π * 10^-7 T m/A * 1000 A m^2 / 4
L = 125,000 * π * 10^-7 T m * 250
L = 125,000 * π * 2.5 * 10^-4 T m
Finally, we can calculate the length of the solenoid:
L = 125,000 * π * 2.5 * 10^-4 T m
L ≈ 0.981 meters
Therefore, the length of the solenoid is approximately 0.981 meters.
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A long solenoid has 500 turns. When a currentof 2 ampere is passed through it, the resultingmagnetic flux linked with each turn of thesolenoid is 4 ×10–3 Wb. The self- inductance ofthe solenoid is [2008]a)2.5 henryb)2.0 henryc)1.0 henryd)40 henryCorrect answer is option 'C'. Can you explain this answer? for NEET 2025 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 currentof 2 ampere is passed through it, the resultingmagnetic flux linked with each turn of thesolenoid is 4 ×10–3 Wb. The self- inductance ofthe solenoid is [2008]a)2.5 henryb)2.0 henryc)1.0 henryd)40 henryCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A long solenoid has 500 turns. When a currentof 2 ampere is passed through it, the resultingmagnetic flux linked with each turn of thesolenoid is 4 ×10–3 Wb. The self- inductance ofthe solenoid is [2008]a)2.5 henryb)2.0 henryc)1.0 henryd)40 henryCorrect answer is option 'C'. Can you explain this answer?.
A long solenoid has 500 turns. When a currentof 2 ampere is passed through it, the resultingmagnetic flux linked with each turn of thesolenoid is 4 ×10–3 Wb. The self- inductance ofthe solenoid is [2008]a)2.5 henryb)2.0 henryc)1.0 henryd)40 henryCorrect answer is option 'C'. Can you explain this answer? for NEET 2025 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 currentof 2 ampere is passed through it, the resultingmagnetic flux linked with each turn of thesolenoid is 4 ×10–3 Wb. The self- inductance ofthe solenoid is [2008]a)2.5 henryb)2.0 henryc)1.0 henryd)40 henryCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A long solenoid has 500 turns. When a currentof 2 ampere is passed through it, the resultingmagnetic flux linked with each turn of thesolenoid is 4 ×10–3 Wb. The self- inductance ofthe solenoid is [2008]a)2.5 henryb)2.0 henryc)1.0 henryd)40 henryCorrect answer is option 'C'. Can you explain this answer?.
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