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Derive the expression for the self inductance of a long solenoid of cross sectional area A and length l, having n turns per unit length. |
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Answer» Solution :Magnetic field at a point inside the SOLENOID is `B=(mu_(0) NI)/(t)` where N is the total NUMBER of turns of the solenoid and L is its length. B is constant throughout the length of the solenoid. Magnetic flux through each turn `=B xx ` area of each turn. `therefore phi_(1)= mu_(0)(N)/(l) I xx A` where, A is the area of each turn. `therefore` Total magnetic flux linked with the solenoid `= PHI= mu_(0) (N)/(l)IA xx N` But from thedefinition of self inductance `(L), phi= LI`. `therefore LI=mu_(0)(N)/(l) IAxxN implies L=(mu_(0)N^(2)A)/(l)=mu_(0)n^(21A)` Henry. 
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