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Define self-inductance. Write its SI unit. Derive an expression for self-inductance of a long, aircored solenoid of length l, cross-sectional area A (radius r), and having N number of turns. |
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Answer» Solution :Self-inductance of a coil is numerically equal to the amount of magnetic flux linked with it when unit current flows through the coil. Alternately, self-inductance of a coil is numerically equal to the emf induced in the coil when rate of change of current through the coil is unity. SI unit of self-inductance is henry (H). Expression for self-inductance: Let a current Iis passed through a solenoid coil of N turns, LENGTH I and A the area enclosed by each turn of coil. The magnetic field B at any point inside the solenoid is: `B = (mu_(0)NI)/l` `THEREFORE` Magnetic flux through each turn of the solenoid `phi_(B) = B.A = (mu_(0)NI)/l. A` `therefore` TOTAL magnetic flux linked with the coil `Nphi_(B)= ((mu_(0)NI)/l.A).N = (mu_(0)N^(2)IA)/l` By definition `Nphi_(B) = L I` ` therefore (mu_(0)N^(2)IA)/l = LI` `implies` Self-inductance of solenoid `L = (mu_(0)N^(2)A)/l` If r be the radius of solenoid coil, then `A = pir^(2)` and we can write `L = (mu_(0)N^(2))/l.pir^(2)` If instead of air the solenoid coil has been wound on a ferromagnetic material (say soft iron) of relative PERMEABILITY `mu_(r), then we have ` L = (mu_(0)mu_(r).N^(2)A)/l = (mu_(0)mu_(r).N^(2)pir^(2))/l` |
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