Answer :
Answer:To calculate the magnetic field generated at a distance
x from the center of a finite straight wire with a steady current
I and length
2
2a, we can use the Biot-Savart law. Let's consider the wire to be along the
z-axis, with the center of the wire at the origin, and
−
≤
≤
−a≤z≤a.
Explanation:Biot-Savart Law
The Biot-Savart law states that the magnetic field
B at a point due to a current element
Idl is given by:
=
0
4
×
∣
∣
3
dB=
4π
μ
0
I
∣r∣
3
dl×r
where
r is the vector from the current element to the point where the magnetic field is being calculated.
For a straight wire, consider a current element
Idz at position
z on the wire. The position vector from this element to the point where we are calculating the magnetic field (at distance
x from the wire) is:
=
^
−
^
r=x
x
^
−z
z
^
The differential length element is:
=
^
dl=dz
z
^
The cross product
×
dl×r is:
×
=
^
×
(
^
−
^
)
=
^
dl×r=dz
z
^
×(x
x
^
−z
z
^
)=xdz
y
^
The magnitude of
r is:
∣
∣
=
2
+
2
∣r∣=
x
2
+z
2
Thus, the differential magnetic field is:
=
0
4
^
(
2
+
2
)
3
/
2
dB=
4π
μ
0
I
(x
2
+z
2
)
3/2
xdz
y
^
Integrate this from
=
−
z=−a to
=
z=a to find the total magnetic field at distance
x from the wire:
=
0
4
∫
−
(
2
+
2
)
3
/
2
B=
4π
μ
0
Ix
∫
−a
a
(x
2
+z
2
)
3/2
dz
Solving the Integral
Let's solve the integral:
∫
−
(
2
+
2
)
3
/
2
∫
−a
a
(x
2
+z
2
)
3/2
dz
This can be done using the substitution
=
tan
(
)
z=xtan(θ):
=
sec
2
(
)
dz=xsec
2
(θ)dθ
2
+
2
=
2
sec
2
(
)
x
2
+z
2
=x
2
sec
2
(θ)
(
2
+
2
)
3
/
2
=
3
sec
3
(
)
(x
2
+z
2
)
3/2
=x
3
sec
3
(θ)
The integral limits change accordingly. When
=
−
z=−a,
=
−
tan
−
1
(
/
)
θ=−tan
−1
(a/x). When
=
z=a,
=
tan
−
1
(
/
)
θ=tan
−1
(a/x). So,
∫
−
(
2
+
2
)
3
/
2
=
∫
−
tan
−
1
(
/
)
tan
−
1
(
/
)
sec
2
(
)
3
sec
3
(
)
=
1
2
∫
−
tan
−
1
(
/
)
tan
−
1
(
/
)
cos
(
)
∫
−a
a
(x
2
+z
2
)
3/2
dz
=∫
−tan
−1
(a/x)
tan
−1
(a/x)
x
3
sec
3
(θ)
xsec
2
(θ)dθ
=
x
2
1
∫
−tan
−1
(a/x)
tan
−1
(a/x)
cos(θ)dθ
=
1
2
[
sin
(
)
]
−
tan
−
1
(
/
)
tan
−
1
(
/
)
=
1
2
(
sin
(
tan
−
1
(
)
)
−
sin
(
−
tan
−
1
(
)
)
)
=
x
2
1
[sin(θ)]
−tan
−1
(a/x)
tan
−1
(a/x)
=
x
2
1
(sin(tan
−1
(
x
a
))−sin(−tan
−1
(
x
a
)))
Since
sin
(
tan
−
1
(
/
)
)
=
2
+
2
sin(tan
−1
(a/x))=
x
2
+a
2
a
, we get:
∫
−
(
2
+
2
)
3
/
2
=
2
2
2
+
2
∫
−a
a
(x
2
+z
2
)
3/2
dz
=
x
2
x
2
+a
2
2a
Magnetic Field
Thus, the magnetic field at a distance
x from the center of the wire is:
=
0
4
⋅
2
2
2
+
2
=
0
2
(
2
+
2
)
1
/
2
B=
4π
μ
0
Ix
⋅
x
2
x
2
+a
2
2a
=
2πx(x
2
+a
2
)
1/2
μ
0
Ia
Limit as Length Goes to Infinity
Take the limit as
→
∞
a→∞:
=
lim
→
∞
0
2
(
2
+
2
)
1
/
2
B=lim
a→∞
2πx(x
2
+a
2
)
1/2
μ
0
Ia
For large
a:
(
2
+
2
)
1
/
2
≈
(x
2
+a
2
)
1/2
≈a
So,
=
lim
→
∞
0
2
=
0
2
B=lim
a→∞
2πxa
μ
0
Ia
=
2πx
μ
0
I
Comparison with Ampere's Law
Using Ampere's Law for an infinite straight wire, the magnetic field at a distance
x from the wire is:
=
0
2
B=
2πx
μ
0
I
This result matches the one obtained by taking the limit of the finite wire's magnetic field as its length goes to infinity. Thus, the calculation using the Biot-Savart law and the result from Ampere's law are consistent.
