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Show that the magnetic field at any point on the axis of the solenoid having n turns per unit length is B =(1)/(2) mu_(0) n I (cos theta_(1) - cos theta_(2)). |
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Answer» SOLUTION :Consider a solenoid havingradius R consistsof n NUMBER of TURNS PER unit length . Let P be the pointat a distance. Xformthe origin of thesolenoid. The current carrying element dx ata distance. xfromorigin andthe distance r from point p . ` r = sqrt(R^(2) + (x'- x)^(2))` The magnetic field due to current carryingcircular coil along its axis is `dB = mu_(0)/2 . (IR^(2))/r^(2) xx N ` where ` N= ndx ` then ` dB = mu_(0)/2.(nIR^(2))/r^(3) . dx ` `sin theta= R/ r ` ` r = R cosec phi ` .....(1) `tan phi = R/ (x' - x) ` ` x - x= R cotphi ` ` (dx)/(d phi) = R cosec^(2) phi ` ` dx = R cosec^(2) phi d phi ` .....(2) from above equation , `dB = mu_(0)/ 2 . (nIR^(2).R cosec ^(2) phi .d phi )/(R^(3) cosec ^(3) phi ) ` `dB = mu_(0)/2 . nI sin phi d phi ` Total magnetic field can be obtained by INTEGRATING `B= (mu_(0) nI)/2 underset(phi_(1))overset(phi_(2)) int sin phi d phi ` ` B = ( mu_(0)nI)/2 [ - cos phi ]_(phi_(1))^(phi_(2))` `B = (mu_(0)nI)/2 ( cos phi_(1) - cos phi_(2))` |
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