Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACSelf and mutual inductorEasy2 minPYQ_2020

PhysicsEasysingle choice

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area $A = 10 {cm}^{2}$ and length = $20 cm$ . If one of the solenoids has $300$ turns and the other $400$ turns, their mutual inductance is $(μ_{0} = 4 π \times 10^{- 7} T m A^{-1})$

Options:

Answer:

Solution:

Mutual inductance

$M = \frac{flux in (2)}{current in (1)}$

$= \frac{ϕ_{2}}{i_{1}}$

$= \frac{B_{1} . A_{2} N_{2}}{i_{1}}$

$= \frac{μ_{0} N_{1} i_{1}}{l} \times \frac{A_{2} N_{2}}{i_{1}}$

$= \frac{μ_{0} N_{1} . N_{2} . A}{l}$

$A_{1} = A_{2} = A$

$A = π r_{1}^{2} = 10 {cm}^{2}$ , $l = 20 cm$ , $N_{1} = 300$ , $N_{2} = 400$ .

$M = \frac{μ_{0} N_{1} N_{2} A}{l} = \frac{4 π \times 10^{- 7} \times 300 \times 400 \times 10 \times 10^{- 4}}{0.20}$

$= 2.4 π \times 10^{- 4} H$ .

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Self and mutual inductor

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...