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A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is
- a)6.6 x 10-9 weber
- b)9.1 x 10-11 weber
- c)6 x 10-11 weber
- d)3.3 x 10-11 weber
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A circular loop of radius 0.3 cm lies parallel to a much bigger circu...
Given:
Radius of the smaller loop, r = 0.3 cm = 0.003 m
Radius of the bigger loop, R = 20 cm = 0.2 m
Distance between their centers, d = 15 cm = 0.15 m
Current flowing through the smaller loop, I = 2.0 A
To find: Flux linked with the bigger loop.
Formula:
The magnetic flux through a loop is given by the formula:
Φ = μ₀ * N * A * B
where
Φ is the magnetic flux,
μ₀ is the permeability of free space (4π × 10⁻⁷ T m/A),
N is the number of turns in the loop,
A is the area of the loop, and
B is the magnetic field.
Calculating the magnetic field at the center of the smaller loop:
The magnetic field at the center of a circular loop is given by the formula:
B = (μ₀ * I * R²) / (2 * (R² + x²)^(3/2))
where
I is the current flowing through the loop,
R is the radius of the loop, and
x is the distance between the center of the loop and the point where the field is to be calculated.
Calculating the magnetic field at the center of the bigger loop:
The distance between the center of the bigger loop and the center of the smaller loop is given as d = 0.15 m.
Therefore, the distance between the center of the bigger loop and the point where the field is to be calculated is x = R - r = 0.2 - 0.003 = 0.197 m.
Using the formula for magnetic field, we can calculate:
B = (4π × 10⁻⁷ * 2.0 * (0.2)²) / (2 * ((0.2)² + (0.197)²)^(3/2))
Calculating the area of the bigger loop:
The area of a circular loop is given by the formula:
A = π * R²
Calculating the flux linked with the bigger loop:
Using the formula for magnetic flux, we can calculate:
Φ = (4π × 10⁻⁷) * (1) * (π * (0.2)²) * B
Simplifying the expression and calculating the value, we get:
Φ = (4π × 10⁻⁷) * (π * 0.04) * B
Therefore, the flux linked with the bigger loop is given by:
Φ = (4π × 10⁻⁷) * (π * 0.04) * B = 9.1 × 10⁻¹¹ Weber (approximately)
Hence, the correct answer is option 'B' (9.1 × 10⁻¹¹ Weber).
Radius of the smaller loop, r = 0.3 cm = 0.003 m
Radius of the bigger loop, R = 20 cm = 0.2 m
Distance between their centers, d = 15 cm = 0.15 m
Current flowing through the smaller loop, I = 2.0 A
To find: Flux linked with the bigger loop.
Formula:
The magnetic flux through a loop is given by the formula:
Φ = μ₀ * N * A * B
where
Φ is the magnetic flux,
μ₀ is the permeability of free space (4π × 10⁻⁷ T m/A),
N is the number of turns in the loop,
A is the area of the loop, and
B is the magnetic field.
Calculating the magnetic field at the center of the smaller loop:
The magnetic field at the center of a circular loop is given by the formula:
B = (μ₀ * I * R²) / (2 * (R² + x²)^(3/2))
where
I is the current flowing through the loop,
R is the radius of the loop, and
x is the distance between the center of the loop and the point where the field is to be calculated.
Calculating the magnetic field at the center of the bigger loop:
The distance between the center of the bigger loop and the center of the smaller loop is given as d = 0.15 m.
Therefore, the distance between the center of the bigger loop and the point where the field is to be calculated is x = R - r = 0.2 - 0.003 = 0.197 m.
Using the formula for magnetic field, we can calculate:
B = (4π × 10⁻⁷ * 2.0 * (0.2)²) / (2 * ((0.2)² + (0.197)²)^(3/2))
Calculating the area of the bigger loop:
The area of a circular loop is given by the formula:
A = π * R²
Calculating the flux linked with the bigger loop:
Using the formula for magnetic flux, we can calculate:
Φ = (4π × 10⁻⁷) * (1) * (π * (0.2)²) * B
Simplifying the expression and calculating the value, we get:
Φ = (4π × 10⁻⁷) * (π * 0.04) * B
Therefore, the flux linked with the bigger loop is given by:
Φ = (4π × 10⁻⁷) * (π * 0.04) * B = 9.1 × 10⁻¹¹ Weber (approximately)
Hence, the correct answer is option 'B' (9.1 × 10⁻¹¹ Weber).
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Community Answer
A circular loop of radius 0.3 cm lies parallel to a much bigger circu...
Field due to current loop 1 at an axial point,
Flux linked with smaller loop 2 due to B1,
The coefficient of mutual inductance between the loops,
Flux linked with bigger loop 1,
Substituting the given values, we get
ф1= 9.1 10-11 weber
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A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop isa)6.6 x 10-9 weberb)9.1 x 10-11 weberc)6 x 10-11 weberd)3.3 x 10-11 weberCorrect answer is option 'B'. Can you explain this answer?
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A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop isa)6.6 x 10-9 weberb)9.1 x 10-11 weberc)6 x 10-11 weberd)3.3 x 10-11 weberCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop isa)6.6 x 10-9 weberb)9.1 x 10-11 weberc)6 x 10-11 weberd)3.3 x 10-11 weberCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop isa)6.6 x 10-9 weberb)9.1 x 10-11 weberc)6 x 10-11 weberd)3.3 x 10-11 weberCorrect answer is option 'B'. Can you explain this answer?.
A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop isa)6.6 x 10-9 weberb)9.1 x 10-11 weberc)6 x 10-11 weberd)3.3 x 10-11 weberCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop isa)6.6 x 10-9 weberb)9.1 x 10-11 weberc)6 x 10-11 weberd)3.3 x 10-11 weberCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop isa)6.6 x 10-9 weberb)9.1 x 10-11 weberc)6 x 10-11 weberd)3.3 x 10-11 weberCorrect answer is option 'B'. Can you explain this answer?.
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ample number of questions to practice A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop isa)6.6 x 10-9 weberb)9.1 x 10-11 weberc)6 x 10-11 weberd)3.3 x 10-11 weberCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
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