The magnetic flux near the axis and inside the air-core solenoid of length 60 cm carrying current I is 1.57 × 10 6 Wb . Its magnetic moment will be (cross-sectional area of a solenoid is very small as compared to its length, μ 0 = 4 π × 10 - 7 SI units)

Magnetic induction inside the solenoid B = μ 0 N I L Magnetic flux, ϕ = B A = μ 0 N I . A L Magnetic moment = N I A = ϕ L μ 0 = 1.57 × 10 - 6 × 0.6 4 × 3.14 × 10 - 7 = 0.75 A m 2

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Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACFaraday law/Lenz lawEasy2 minPYQ_2020

PhysicsEasysingle choice

The magnetic flux near the axis and inside the air-core solenoid of length $60 cm$ carrying current $I$ is $1.57 \times 10^{6} Wb .$ Its magnetic moment will be (cross-sectional area of a solenoid is very small as compared to its length, $μ_{0} = 4 π \times 10^{- 7}$ SI units)

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Answer:

Solution:

Magnetic induction inside the solenoid

$B = \frac{μ_{0} N I}{L}$

Magnetic flux, $ϕ = B A$

$= \frac{μ_{0} N I . A}{L}$

Magnetic moment $= N I A = \frac{ϕ L}{μ_{0}}$

$= \frac{1.57 \times 10^{- 6} \times 0.6}{4 \times 3.14 \times 10^{- 7}}$

$= 0.75 {A m}^{2}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:EMI/ACSubtopic:Faraday law/Lenz law

⏱ 2mℹ️ Source: PYQ_2020

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