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Show that the magnetic field at any poin...

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))`.

Text Solution

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Consider a solenoid having radius R consists of n number of turns per unit length .
Let P be the point at a distance . X form the origin of the solenoid .
The current carrying element dx at a distance . x from origin and the distance r from point p .
` r = sqrt(R^(2) + (x' - x)^(2))`
The magnetic field due to current carrying circular 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 cot phi `
` (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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