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A circular loop of radius 0.3 cm lies parallel to amuch 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)9.1 × 10–11 weber
- b)6 × 10–11 weber
- c)3.3 × 10–11 weber
- d)6.6 × 10–9 weber
Correct answer is option 'A'. Can you explain this answer?
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A circular loop of radius 0.3 cm lies parallel to amuch bigger circula...
As we know, Magnetic flux, φ = B.A
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A circular loop of radius 0.3 cm lies parallel to amuch bigger circula...
The magnetic field produced by the small loop at the position of the bigger loop can be calculated using the formula for the magnetic field at the center of a circular loop:
B = (μ₀ * I * R²) / (2 * (R² + r²)^(3/2))
Where B is the magnetic field, μ₀ is the permeability of free space, I is the current, R is the radius of the loop producing the field, and r is the distance between the center of the producing loop and the point where the field is measured.
Plugging in the values, we get:
B = (4π * 10^(-7) * 2.0 * (0.3 * 10^(-2))²) / (2 * ( (0.3 * 10^(-2))² + (20 * 10^(-2))²)^(3/2))
B ≈ 0.0627 μT (T = Tesla)
The flux linked with the bigger loop can then be calculated using the formula:
Φ = B * A
Where Φ is the flux, B is the magnetic field, and A is the area of the loop.
The area of the bigger loop can be calculated using the formula for the area of a circle:
A = π * R²
Plugging in the values, we get:
Φ = (0.0627 * 10^(-6)) * π * (20 * 10^(-2))²
Φ ≈ 0.0248 Wb (Wb = Weber)
Therefore, the flux linked with the bigger loop is approximately 0.0248 Weber.
B = (μ₀ * I * R²) / (2 * (R² + r²)^(3/2))
Where B is the magnetic field, μ₀ is the permeability of free space, I is the current, R is the radius of the loop producing the field, and r is the distance between the center of the producing loop and the point where the field is measured.
Plugging in the values, we get:
B = (4π * 10^(-7) * 2.0 * (0.3 * 10^(-2))²) / (2 * ( (0.3 * 10^(-2))² + (20 * 10^(-2))²)^(3/2))
B ≈ 0.0627 μT (T = Tesla)
The flux linked with the bigger loop can then be calculated using the formula:
Φ = B * A
Where Φ is the flux, B is the magnetic field, and A is the area of the loop.
The area of the bigger loop can be calculated using the formula for the area of a circle:
A = π * R²
Plugging in the values, we get:
Φ = (0.0627 * 10^(-6)) * π * (20 * 10^(-2))²
Φ ≈ 0.0248 Wb (Wb = Weber)
Therefore, the flux linked with the bigger loop is approximately 0.0248 Weber.
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A circular loop of radius 0.3 cm lies parallel to amuch 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)9.1 × 10–11 weberb)6 × 10–11 weberc)3.3 × 10–11 weberd)6.6 × 10–9 weberCorrect answer is option 'A'. Can you explain this answer?
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A circular loop of radius 0.3 cm lies parallel to amuch 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)9.1 × 10–11 weberb)6 × 10–11 weberc)3.3 × 10–11 weberd)6.6 × 10–9 weberCorrect answer is option 'A'. 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 amuch 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)9.1 × 10–11 weberb)6 × 10–11 weberc)3.3 × 10–11 weberd)6.6 × 10–9 weberCorrect answer is option 'A'. 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 amuch 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)9.1 × 10–11 weberb)6 × 10–11 weberc)3.3 × 10–11 weberd)6.6 × 10–9 weberCorrect answer is option 'A'. Can you explain this answer?.
A circular loop of radius 0.3 cm lies parallel to amuch 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)9.1 × 10–11 weberb)6 × 10–11 weberc)3.3 × 10–11 weberd)6.6 × 10–9 weberCorrect answer is option 'A'. 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 amuch 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)9.1 × 10–11 weberb)6 × 10–11 weberc)3.3 × 10–11 weberd)6.6 × 10–9 weberCorrect answer is option 'A'. 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 amuch 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)9.1 × 10–11 weberb)6 × 10–11 weberc)3.3 × 10–11 weberd)6.6 × 10–9 weberCorrect answer is option 'A'. Can you explain this answer?.
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