Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Get immediate and reliable solutions to your questions from a community of experienced experts on our Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
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.
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
