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Using Ampere's circuital law, obtain an expression for the magnetic field along the axis of a current carrying solenoid of length l and havingN number of turns. |
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Answer» Solution :Magnetic field due to solenoid : Consider a rectangular amperianloop ABCD near the middle ofsolenoid as shown in flg. Where PQ=l. Let the mangetic field along the path ab be B and in zeroalong CD. As the paths bc and DA are perpendicular to the axis of solenoid, the magnetic field component along these path is zero. Therefore, the path bc and da will not contribute to the line integral of magnetic field B. Total number of turns in length l=Nl The line integral of magnetic field induction B over the closed path abcd is `underset(abcd)ointvec(B). vec(dl)=underset(a)overset(b)int vec(B).vec(dl)+underset(b)overset(c)int vec(B).vec(dl)+underset(c)overset(d) int vec(B).vec(dl)+underset(d) overset(a)int vec(B).vec(dl)` `:' underset(a)overset(b) int vec(B).vec(dl)=underset(a)overset(b)int B dl cos 0^(@) = Bl andunderset(b)overset(c) int vec(B).vec(dl) = underset(b)overset(c) B dl cos 90^(@) = 0 = underset(d) overset(a) vec(B).vec(dl)` Also, `underset(c) overset(d) vec(B).vec(dl) = 0 ` (`:'` Outside the solenoid, B=0) `:. underset(abcd)oint vec(B).vec(dl) = Bl + 0+ 0+0=Bl` ...(i) Using AMPERE's circuital law `underset(abcd) oint vec(B).vec(dl)=mu_(0) xx` total current in rectangle abcd `=mu_(0)xx`no. of turns in rectangle `xx` current `= mu_(0) xx Nl xx I = mu_(0) NlI "" ...(ii)` From (1) and (2), we have `Bl=mu_(0) Nl.I` `:. B=mu_(0)NI`. 
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