derive an expression for magnetic field at a point due to long straight cunductor of length L and current i.
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Answers
Answer:
Magnetic induction due to infinitely long straight conductor carrying current: XY, is an infinitely long straight conductor carrying a current I(Figure).
P is a point at a distance a from the conductor. AB is a small element of length dl. θ is the angle between the current element Idl and the line joining the element dl and the point P. According to Biot-Savart law, the magnetic induction at the point P due to the current element Idl is.
dB=
4π
μ
o
r
2
Idl⋅sinθ
......(1)
AC is drawn perpendicular to BP from A.
∠OPA=ϕ, ∠APB=dϕ
In ΔABC, sinθ=
AB
AC
=
dl
AC
∴AC=dlsinθ ..........(2)
From ΔAPC, AC=rdϕ .........(3)
From equations (2) and (3),
rdϕ=dlsinθ ............(4)
Substituting equation (4) in equation (1)
dB=
4π
μ
o
r
2
Irdϕ
=
4π
μ
o
r
Idϕ
.............(5)
In ΔOPA, cosϕ=
r
a
∴r=
cosϕ
a
..........(6)
Substituting equation (6) in equation (5)
dB=
4π
μ
o
a
1
cosϕdϕ
The total magnetic induction at P due to the conductor XY is
B=
−ϕ
1
∫
ϕ
2
dB=
−ϕ
1
∫
ϕ
2
4πa
μ
o
I
cosϕdϕ
B=
4πa
μ
o
I
[sinϕ
1
+sinϕ
2
]
For infinitely long conductor,
ϕ
1
=ϕ
2
=90
o
∴B=
2πa
μ
o
I
If the conductor is places in a medium