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Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm2 and length = 20 cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is
0 = 4π × 10–7 Tm A–1)
  • a)
    2.4π × 10–5 H
  • b)
    4.8π × 10–4 H
  • c)
    4.8π × 10–5 H
  • d)
    2.4π × 10–4 H
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Two coaxial solenoids are made by winding thin insulated wire over a ...
Given, Area of cross-section of pipe, A = 10 cm2
Length of pipe, l = 20 cm
= 2.4π × 10–4 H
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Most Upvoted Answer
Two coaxial solenoids are made by winding thin insulated wire over a ...
To find the mutual inductance between two coaxial solenoids, we can use the formula:

M = µ₀ * N₁ * N₂ * A / l

where:
M is the mutual inductance
µ₀ is the permeability of free space (4π × 10⁻⁷ Tm/A)
N₁ is the number of turns in the first solenoid (300 turns)
N₂ is the number of turns in the second solenoid (400 turns)
A is the cross-sectional area of the solenoids (10 cm² = 10 × 10⁻⁴ m²)
l is the length of the solenoids (20 cm = 0.2 m)

Substituting the given values into the formula:

M = (4π × 10⁻⁷ Tm/A) * (300 turns) * (400 turns) * (10 × 10⁻⁴ m²) / (0.2 m)

Simplifying the expression:

M = (4π × 10⁻⁷ Tm/A) * (300) * (400) * (10⁻⁴ m²) / (0.2 m)

M = (4π × 10⁻⁷ Tm/A) * (120,000) * (10⁻⁴ m²) / (0.2 m)

M = (4π × 10⁻⁷ Tm/A) * (12) * (10⁵) * (10⁻⁴ m) / (1)

M = 4π × 10⁻⁷ Tm/A * 1.2 H

Simplifying further:

M = 4.8π × 10⁻⁷ H

So, the mutual inductance between the two coaxial solenoids is 4.8π × 10⁻⁷ H, which is equivalent to option D.
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Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm2 and length = 20 cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is(µ0 = 4π × 10–7 Tm A–1)a)2.4π × 10–5 Hb)4.8π × 10–4 Hc)4.8π × 10–5 Hd)2.4π × 10–4 HCorrect answer is option 'D'. Can you explain this answer?
Question Description
Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm2 and length = 20 cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is(µ0 = 4π × 10–7 Tm A–1)a)2.4π × 10–5 Hb)4.8π × 10–4 Hc)4.8π × 10–5 Hd)2.4π × 10–4 HCorrect 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 Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm2 and length = 20 cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is(µ0 = 4π × 10–7 Tm A–1)a)2.4π × 10–5 Hb)4.8π × 10–4 Hc)4.8π × 10–5 Hd)2.4π × 10–4 HCorrect 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 Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm2 and length = 20 cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is(µ0 = 4π × 10–7 Tm A–1)a)2.4π × 10–5 Hb)4.8π × 10–4 Hc)4.8π × 10–5 Hd)2.4π × 10–4 HCorrect answer is option 'D'. Can you explain this answer?.
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