InterviewSolution
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InterviewSolution
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Calculate the self-inductance for very long solenoid. |
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Answer» SOLUTION :LET us calculate the SELF-inductance of a long solenoid of cross-sectional area A and length l, having n turns per unit length. The magnetic field DUE to a current I flowing in the solenoid is `B=mu_0nI`(NEGLECTING edge effects, as before) The total flux linked with the solenoid is `Nphi_B=NBA` `Nphi_B=(nl)(mu_0nI)(A) ""(because n=N/l)` `=mu_0n^2 AlI` where nl is the total number of turns. Thus, the self - inductance is, `L=(Nphi_B)/I` `L=mu_0n^2Al`...(1) If we fill the inside of the solenoid with a material of relative permeability u, (for example soft iron, which has a high value of relative permeability), then, `L=mu_r mu_0 n^2 Al`...(2) `[because mu_1=mu/mu_0]` The self-inductance of the coil depends on its geometry and on the permeability of the medium. |
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