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Define mutual inductance between a pair of coils. Deriveexpression for the mutuel inductanceof two long coaial solenoids of same length wound one over the other. |
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Answer» Solution : MutualInductanceof two coils or circutits is defind as magnetic flux linked with the SECONDARY COIL due to the FLOW of unit current in the primary coil.  Let,I = length of each solenoid . `r_(1) , r_(2)` = radii of two solenoids A ` = pi r _(1)^(2)` = area of cross-section of inner solenoid `S_(1) ` `N_(1) , N_(2)`number of turns in the two solenoids, First , we pass a time varyingcurrent `I_(1)`through `S_(2)` . The magnetic flux is `mu_(0) n_(2) I_(2)`, where `n_(2)= N_(2)//I` = the number of turns per unit length of `S_(2)`. Total magnetic flux linked with the inner solenoid `S_(1)` is `Phi_(1) mu_(0) n_(2) I_(2) AN_(1)` `therefore` Mutual inductance of coil 1 w.r.t. coil 2 is `M_(12) (Phi_(1))/(I_(2)) =mu_(0) n_(2) I_(2) AN_(1)= (mu_(0) N_(1) N_(2) A)/(I)` Now, consider the flux linked with outer solenoid `S_(2)` due to the current `I_(1)`in the inner solenoid `S_(1) ` . The field `B_(1)`due to `I_(1)` inside `S_(1) = B_(1) mu_(0)n_(1) I_(1)` Total flux linked with the outer solenoid is ` Phi_(2) =B_(1) AN_(2) =mu_(0) n_(1) I_(1) AN_(2)= (mu_(0) n_(1) N_(2)AI_(1))/(I)` `therefore ` Mutual inductanceof coil 2 w. r . t coil 1 is `M_(21) = (phi_(2))/(I_(2)) = (mu_(0) N_(1) N_(2) A)/(I)` `thereforeM = (phi_(2))/(I_(2)) = (mu_(0) N_(1) N_(2) A)/(I)= mu_(0) n_(1)n_(2) AI = mu_(0) n_(1) n_(2) pir^(2) I`. If a medium of relative PERMEABILITY ` mu_(r)`is present within the solenoids , then ` M = mu_(r) mu_(0) n_(1) n_(2) pi r^(2) I` . |
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