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A solenoid 4cm in diameter and 20 cm in length has 250 turns and carries a current of 15 ampere. calculate the flux through the surface of a disc of 10 cm radius that is positioned perpendicular to the axis of the solenoid?
Verified Answer
A solenoid 4cm in diameter and 20 cm in length has 250 turns and carri...
R = radius of solenoid = 2 cm
L = 20 cm, turns = N = 250
turns per unit length = n = 1,250 turns/meter
current i = 15 A
Magnetic field through the solenoid = μ₀ n i = 4π * 10^-7 * 1,250 * 15 units
B = 2.356 * 10^-2 H
The magnetic field outside the turns of the solenoid is zero, ie., it exists only in the area of π r�.
Flux = B * πr^2 = 0.0185 Volt-sec
This question is part of UPSC exam. View all NEET courses
Most Upvoted Answer
A solenoid 4cm in diameter and 20 cm in length has 250 turns and carri...
Given Data:
- Diameter of solenoid (d) = 4 cm
- Length of solenoid (l) = 20 cm
- Number of turns in solenoid (N) = 250
- Current passing through solenoid (I) = 15 A
- Radius of the disc (r) = 10 cm
Step 1: Calculating Magnetic Field Inside the Solenoid
The magnetic field inside a solenoid can be calculated using the formula:
B = μ₀ * N * I / l
where:
- B is the magnetic field
- μ₀ is the permeability of free space (4π × 10^-7 T·m/A)
- N is the number of turns
- I is the current passing through the solenoid
- l is the length of the solenoid
Substituting the given values:
B = (4π × 10^-7 T·m/A) * 250 * 15 A / 0.2 m
= (4π × 10^-7 T·m/A) * 3750 A / 0.2 m
= 235.62 × 10^-3 T
Step 2: Calculating Flux through the Disc
The flux through a surface can be calculated using the formula:
Φ = B * A * cos(θ)
where:
- Φ is the flux
- B is the magnetic field
- A is the area of the surface
- θ is the angle between the magnetic field and the surface normal (which is 90 degrees for a disc perpendicular to the solenoid axis)
The area of the disc can be calculated using the formula:
A = π * r²
Substituting the given values:
A = π * (0.1 m)²
= π * 0.01 m²
= 0.0314 m²
Since the angle between the magnetic field and the surface normal is 90 degrees, cos(θ) = cos(90°) = 0.
Therefore, the flux through the disc is:
Φ = (235.62 × 10^-3 T) * 0.0314 m² * 0
= 0
Step 3: Conclusion
The flux through the surface of the disc of 10 cm radius, positioned perpendicular to the axis of the solenoid, is zero. This indicates that no magnetic field lines pass through the disc since the angle between the magnetic field and the surface normal is 90 degrees.
- Diameter of solenoid (d) = 4 cm
- Length of solenoid (l) = 20 cm
- Number of turns in solenoid (N) = 250
- Current passing through solenoid (I) = 15 A
- Radius of the disc (r) = 10 cm
Step 1: Calculating Magnetic Field Inside the Solenoid
The magnetic field inside a solenoid can be calculated using the formula:
B = μ₀ * N * I / l
where:
- B is the magnetic field
- μ₀ is the permeability of free space (4π × 10^-7 T·m/A)
- N is the number of turns
- I is the current passing through the solenoid
- l is the length of the solenoid
Substituting the given values:
B = (4π × 10^-7 T·m/A) * 250 * 15 A / 0.2 m
= (4π × 10^-7 T·m/A) * 3750 A / 0.2 m
= 235.62 × 10^-3 T
Step 2: Calculating Flux through the Disc
The flux through a surface can be calculated using the formula:
Φ = B * A * cos(θ)
where:
- Φ is the flux
- B is the magnetic field
- A is the area of the surface
- θ is the angle between the magnetic field and the surface normal (which is 90 degrees for a disc perpendicular to the solenoid axis)
The area of the disc can be calculated using the formula:
A = π * r²
Substituting the given values:
A = π * (0.1 m)²
= π * 0.01 m²
= 0.0314 m²
Since the angle between the magnetic field and the surface normal is 90 degrees, cos(θ) = cos(90°) = 0.
Therefore, the flux through the disc is:
Φ = (235.62 × 10^-3 T) * 0.0314 m² * 0
= 0
Step 3: Conclusion
The flux through the surface of the disc of 10 cm radius, positioned perpendicular to the axis of the solenoid, is zero. This indicates that no magnetic field lines pass through the disc since the angle between the magnetic field and the surface normal is 90 degrees.
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A solenoid 4cm in diameter and 20 cm in length has 250 turns and carries a current of 15 ampere. calculate the flux through the surface of a disc of 10 cm radius that is positioned perpendicular to the axis of the solenoid?
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A solenoid 4cm in diameter and 20 cm in length has 250 turns and carries a current of 15 ampere. calculate the flux through the surface of a disc of 10 cm radius that is positioned perpendicular to the axis of the solenoid? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A solenoid 4cm in diameter and 20 cm in length has 250 turns and carries a current of 15 ampere. calculate the flux through the surface of a disc of 10 cm radius that is positioned perpendicular to the axis of the solenoid? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solenoid 4cm in diameter and 20 cm in length has 250 turns and carries a current of 15 ampere. calculate the flux through the surface of a disc of 10 cm radius that is positioned perpendicular to the axis of the solenoid?.
A solenoid 4cm in diameter and 20 cm in length has 250 turns and carries a current of 15 ampere. calculate the flux through the surface of a disc of 10 cm radius that is positioned perpendicular to the axis of the solenoid? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A solenoid 4cm in diameter and 20 cm in length has 250 turns and carries a current of 15 ampere. calculate the flux through the surface of a disc of 10 cm radius that is positioned perpendicular to the axis of the solenoid? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solenoid 4cm in diameter and 20 cm in length has 250 turns and carries a current of 15 ampere. calculate the flux through the surface of a disc of 10 cm radius that is positioned perpendicular to the axis of the solenoid?.
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