a) a long solenoid with air core has n turns per unit length and carries a current i. Using amperes circuital law, derive an expression for the magnetic field b at an interior point on its axis. Write an expression for magnetic intensity h in the interior of the solenoid.
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A solenoid is a long wire wound in a close-packed helix carrying a current I and the length of the solenoid is much greater then its diameter
Figure below shows a section of a stretched out solenoid in xy and yz plane

The solenoid magnetic field is the vector sum of the field produced by the individual turns that make up the solenoid
Magnetic field B is nearly uniform and parallel to the axis of the solenoid at interior points near its center and external field near the center is very small
Consider a dashed closed path abcd as shown in figure .Let l be the length of side ab of the loop which is parallel to the is of the solenoid
Let us also consider that sides bc and da of the loop are very-very long so that side cd is very much far away from the solenoid and magnetic field at this side is negligibly small and for simplicity we consider its equal to 0
At side a magnetic field B is approximately parallel and constant. So for this side
∫B.dl=Bl
Magnetic field B is perpendicular to sides bc and da ,hence these portions of the loop does not make any contributions to the line integral as B.dl=0 for the side bc and da
Side cd lies at external points solenoid where B.dl=0 as B=0 or negligibly small outside the solenoid
Hence sum around the entire closed path reduces to Bl
If N are number of turns per unit length in a solenoid then number of turns in length l is nl.The total current through the rectangle abcd is NIl and from ampere 's law
Bl=μ0NlI
or B=μ0NI
Figure below shows a section of a stretched out solenoid in xy and yz plane

The solenoid magnetic field is the vector sum of the field produced by the individual turns that make up the solenoid
Magnetic field B is nearly uniform and parallel to the axis of the solenoid at interior points near its center and external field near the center is very small
Consider a dashed closed path abcd as shown in figure .Let l be the length of side ab of the loop which is parallel to the is of the solenoid
Let us also consider that sides bc and da of the loop are very-very long so that side cd is very much far away from the solenoid and magnetic field at this side is negligibly small and for simplicity we consider its equal to 0
At side a magnetic field B is approximately parallel and constant. So for this side
∫B.dl=Bl
Magnetic field B is perpendicular to sides bc and da ,hence these portions of the loop does not make any contributions to the line integral as B.dl=0 for the side bc and da
Side cd lies at external points solenoid where B.dl=0 as B=0 or negligibly small outside the solenoid
Hence sum around the entire closed path reduces to Bl
If N are number of turns per unit length in a solenoid then number of turns in length l is nl.The total current through the rectangle abcd is NIl and from ampere 's law
Bl=μ0NlI
or B=μ0NI
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