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Drive expression of M.F. Is a solenoid carrying current I & number of turn per unit length are n |
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Answer» Ampere's circuital law STATES that line integral of magnetic field around any closed loop is equal to μ o times the electric current FLOWING through the cross-section area enclosed by that loop.Mathematically, ∮B.dl=μILet the current flowing in the solenoid having number of turns per unit length N be I.Magnitude of magnetic field inside the solenoid is B while at outside is zero.Now ∮ loop B.dl=∫B ab .L+∫B bc .L ′ +∫B cd .L+∫B da .L ′ The value of first term ∫B ab .L=BLThe second and fourth term are zero because ANGLE between magnetic field and the length loop is 90 o .The third term is also zero as the value of magnetic field outside the solenoid is zero.Total current flowing through the loop I total =(nL)IFrom Ampere's circuital law, we get BL=μ o (nLI) ⟹ B=μ o nI |
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