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A long solenoid of length l having N turns carries a current I.Deduce the expression for the magnetic field in the interior of the solenoid?

A solenoid is a device used to generate a homogeneous magnetic field. A solenoid consists of thin conducting wire wound in a tight helical coil of many turns.

Let N=number of turns of solenoidl=length of solenoidn=number of turns per unit length of solenoidI=current flowing through the soleoid

The ideal solenoid has translational and rotational symmetry. As the magnetic field lines must form closed loops, the magnetic field can not be directed along a radial direction (otherwise field lines would be created or destroyed on the central axis of the solenoid). Therefore, the magnetic field lines in a solenoid must be parallel to the solenoid axis. The following figure shows an ideal solenoid.


Let us apply Ampère's circuital law to the rectangular loop abcd. We must first find the line integral of the magnetic field around abcd. Along bd and ac the magnetic field is essentially perpendicular to the loop, so there is no contribution to the lineintegral from these sections of the loop. Along cd the magnetic field is approximatelyuniform, of magnitude B, say, and is directed parallel to the loop. Along ab the magneticfield strength is essentially negligible, so this section of the loop makes no contribution to the line integral. Thus line integral of magnetic field around abcd will be∫abB.dl+∫bcB.dl+∫cdB.dl+∫daB.dl=BLFrom the ampere's circuital law, this line integral is equals to the μ0 times the current flowingthrough abcd. The total current which flows through the loop is nLI. We may writeBL=μ0nLIthat gives B=u0nI

 

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