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Deduce an expression for the mutual inductance of two long coaxial solenoids but having different radii and different number of turns. Or Obtain the expression for the mutual inductance of two long coaxial solenoids S_(1) and S_(2) wound one over the other, each of length I and radii r_(1) and r_(2), and number of turns per unit length n_(1) and n_(2) respeetively, when a current I is set up in the outer solenoid S_(2). |
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Answer» Solution :Consider two long coaxial solenoids each of length l. Let radius of inner solenoid `S_(1), be r_(1)`, for cross SECTION area be `A_(1)` and `n_(1)` be the number of turns per unit length in it. The corresponding quantities for outer solenoid `S_(1) be r_(1), A^(2) and n_(2)`. Total number of turns in two solenoids are `N_(1) = n_(1)l` and `N_(2) = n_(2)l` RESPECTIVELY. Let a current `I_(2)`, is set up through outer solenoid `S_(2)` so that magnetic field developed along its axial line `B = mu_(0)n_(0)I_(2)` `therefore` Magnetic flux linked per unit turn of solenoid coil `S_(1), phi_(1) = BA_(1) = mu_(0)n_(2)I_(2)A_(1)` and total magnetic flux linked with solenoid `S_(1),N_(1) phi_(1) = N_(1)mu_(0)n_(2)A_(1)I_(2) = mun_(1)n_(2)l.pir_(1)^(2)I_(2)` As per definition of mutual inductance `Nphi_(1) = M_(2)I_(2)` `therefore` mu_(0)n_(1)n_(2)l.pir_(1)^(2)I^(2)` `implies M_(12) = mu_(0)n_(1)n_(2)l.pir_(1)^(2) = mu_(0)n_(1)n_(2)lA_(1) = (mu_(0)N_(1)N_(2)A_(1))/l` For a pair of two solenoid coils `M_(12) = M_(21) = M`, HENCE mutual inductance of pair of solenoid coil is `M = mu_(0)n_(1)n_(2)l.pir_(1)^(2) = (mu_(0)N_(1)N_(2)A_(1))/l` If a soft iron core of relative permeability `mu_(r)`, is also INSERTED `S_(1)`, turns inside the solenoid coil S, the mutual inductance becomes `M = mu_(0)n_(1)n_(2)l pir_(1)^(2) = (mu_(0)N_(1)N_(2)A_(1))/l` 
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