16. Define Ampere circuital law. Derive the expression for magnetic field inside the solenoid.
Answers
Explanation:
Along the paths Q→RandS→P,→B is perpendicular to →dl inside the solenoid while →B=0 outside. By Ampere's law, in free space, ∮→B⋅→dl=μ0Iencl, where μ0 is the permeability of free space. This is the required expression
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=μI
Let 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=BL
The 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)I
From Ampere's circuital law, we get BL=μ
o
(nLI)
⟹ B=μ
o
nI