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Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACSelf and mutual inductorEasy2 minPYQ_2014

PhysicsEasysingle choice

A solenoid $30 cm$ long is made by winding 2000 loops of wire on an iron rod whose cross-section is $1.5 cm^{2}$ .If the relative permeability of the iron is $6000.$ What is the self-inductance of the solenoid?

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Answer:

Solution:

As we know, Self-inductance of the solenoid $L = \frac{μ _{r} \cdot μ _{0} N ^{2} A}{I} = \frac{600 \times 4 π \times 1 0 ^{- 7} \times ( 2000 ) ^{2} \times ( 1.5 ) \times 1 0 ^{- 4}}{0.3} = 1.5 H$

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

⏱ 2mℹ️ Source: PYQ_2014

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