Chapter 29
\text{Chapter 29}
Chapter 29
\text{Chapter 29}
Chapter 29
\text{Chapter 29}
Chapter 29
\text{Chapter 29}
Chapter 29
\text{Chapter 29}
Chapter 29
\text{Chapter 29}
Chapter 29
\text{Chapter 29}
Chapter 29
\text{Chapter 29}
A current is set up in a wire loop that is formed as shown in FIGURE 10, where
R
I
=
2.0
c
m
\text{A current is set up in a wire loop that is formed as shown in FIGURE 10, where $R_I=$ $2.0 \mathrm{~cm}$ }
A current is set up in a wire loop that is formed as shown in FIGURE 10, where
R
I
=
2.0
cm
\text{A current is set up in a wire loop that is formed as shown in FIGURE 10, where $R_I=$ $2.0 \mathrm{~cm}$ }
and
R
2
=
4.0
c
m
. The loop carries a current of
5.0
A
, as shown in the figure.
\text{and $R_2=4.0 \mathrm{~cm}$. The loop carries a current of $5.0 \mathrm{~A}$, as shown in the figure. }
and
R
2
=
4.0
cm
. The loop carries a current of 5.0
A
, as shown in the figure.
\text{and $R_2=4.0 \mathrm{~cm}$. The loop carries a current of $5.0 \mathrm{~A}$, as shown in the figure. }
What is the magnetic field at the center of the loop (C)?
\text{What is the magnetic field at the center of the loop (C)?}
What is the magnetic field at the center of the loop (C)?
\text{What is the magnetic field at the center of the loop (C)?}
A)
3.9
×
1
0
−
5
T
out of the page
\text{ A) $3.9 \times 10^{-5} \mathrm{~T}$ out of the page }
A) 3.9
×
1
0
−
5
T
out of the page
\text{ A) $3.9 \times 10^{-5} \mathrm{~T}$ out of the page }
B)
3.9
×
1
0
−
5
T
into the page
\text{ B) $3.9 \times 10^{-5} \mathrm{~T}$ into the page }
B) 3.9
×
1
0
−
5
T
into the page
\text{ B) $3.9 \times 10^{-5} \mathrm{~T}$ into the page }
C)
1.2
×
1
0
−
4
T
out of the page
\text{ C) $1.2 \times 10^{-4} \mathrm{~T}$ out of the page }
C) 1.2
×
1
0
−
4
T
out of the page
\text{ C) $1.2 \times 10^{-4} \mathrm{~T}$ out of the page }
D)
1.2
×
1
0
−
4
T
into the page
\text{ D) $1.2 \times 10^{-4} \mathrm{~T}$ into the page }
D) 1.2
×
1
0
−
4
T
into the page
\text{ D) $1.2 \times 10^{-4} \mathrm{~T}$ into the page }
E)
7.9
×
1
0
−
5
T
into of the page
\text{ E) $7.9 \times 10^{-5} \mathrm{~T}$ into of the page }
E) 7.9
×
1
0
−
5
T
into of the page
\text{ E) $7.9 \times 10^{-5} \mathrm{~T}$ into of the page }
A current is set up in a wire loop that is formed as shown in FIGURE 10, where
R
I
=
2.0
c
m
\text{A current is set up in a wire loop that is formed as shown in FIGURE 10, where $R_I=$ $2.0 \mathrm{~cm}$ }
A current is set up in a wire loop that is formed as shown in FIGURE 10, where
R
I
=
2.0
cm
\text{A current is set up in a wire loop that is formed as shown in FIGURE 10, where $R_I=$ $2.0 \mathrm{~cm}$ }
and
R
2
=
4.0
c
m
. The loop carries a current of
5.0
A
, as shown in the figure.
\text{and $R_2=4.0 \mathrm{~cm}$. The loop carries a current of $5.0 \mathrm{~A}$, as shown in the figure. }
and
R
2
=
4.0
cm
. The loop carries a current of 5.0
A
, as shown in the figure.
\text{and $R_2=4.0 \mathrm{~cm}$. The loop carries a current of $5.0 \mathrm{~A}$, as shown in the figure. }
What is the magnetic field at the center of the loop (C)?
\text{What is the magnetic field at the center of the loop (C)?}
What is the magnetic field at the center of the loop (C)?
\text{What is the magnetic field at the center of the loop (C)?}
B
1
=
μ
0
i
ϕ
4
π
R
1
→
out of the page
B
2
=
μ
0
i
ϕ
4
π
R
2
→
into the page
B
1
>
B
2
because
R
2
>
R
1
∴
B
c
=
B
1
−
B
2
=
μ
0
i
ϕ
4
π
(
1
R
1
−
1
R
2
)
=
4
π
×
1
0
−
7
×
5
×
π
4
π
(
1
2
−
1
4
)
×
1
0
2
=
3.9
×
1
0
5
T
→
out of the page
\begin{aligned} & \mathrm{B}_1=\frac{\mu_0 i_\phi}{4 \pi \mathrm{R}_1} \rightarrow \text { out of the page } \\ & \mathrm{B}_2=\frac{\mu_0 i_\phi}{4 \pi \mathrm{R}_2} \rightarrow \text { into the page } \\ & \mathrm{B}_1>\mathrm{B}_2 \text { because } \mathrm{R}_2>\mathrm{R}_1 \\ & \therefore \mathrm{B}_{\mathrm{c}}=\mathrm{B}_1-\mathrm{B}_2=\frac{\mu_0 i_\phi}{4 \pi}\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ & =\frac{4 \pi \times 10^{-7} \times 5 \times \pi}{4 \pi}\left(\frac{1}{2}-\frac{1}{4}\right) \times 10^2=3.9 \times 10^5 \mathrm{~T} \rightarrow \text { out of the page } \end{aligned}
B
1
=
4
π
R
1
μ
0
i
ϕ
→
out of the page
B
2
=
4
π
R
2
μ
0
i
ϕ
→
into the page
B
1
>
B
2
because
R
2
>
R
1
∴
B
c
=
B
1
−
B
2
=
4
π
μ
0
i
ϕ
(
R
1
1
−
R
2
1
)
=
4
π
4
π
×
1
0
−
7
×
5
×
π
(
2
1
−
4
1
)
×
1
0
2
=
3.9
×
1
0
5
T
→
out of the page
\begin{aligned} & \mathrm{B}_1=\frac{\mu_0 i_\phi}{4 \pi \mathrm{R}_1} \rightarrow \text { out of the page } \\ & \mathrm{B}_2=\frac{\mu_0 i_\phi}{4 \pi \mathrm{R}_2} \rightarrow \text { into the page } \\ & \mathrm{B}_1>\mathrm{B}_2 \text { because } \mathrm{R}_2>\mathrm{R}_1 \\ & \therefore \mathrm{B}_{\mathrm{c}}=\mathrm{B}_1-\mathrm{B}_2=\frac{\mu_0 i_\phi}{4 \pi}\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ & =\frac{4 \pi \times 10^{-7} \times 5 \times \pi}{4 \pi}\left(\frac{1}{2}-\frac{1}{4}\right) \times 10^2=3.9 \times 10^5 \mathrm{~T} \rightarrow \text { out of the page } \end{aligned}
In FIGURE 9, two infinitely long wires carry currents
i
. Each follows a
9
0
∘
arc on the
\text{In FIGURE 9, two infinitely long wires carry currents $\boldsymbol{i}$. Each follows a $90^{\circ}$ arc on the }
In FIGURE 9, two infinitely long wires carry currents
i
. Each follows a 9
0
∘
arc on the
\text{In FIGURE 9, two infinitely long wires carry currents $\boldsymbol{i}$. Each follows a $90^{\circ}$ arc on the }
circumference of the same circle of radius
R
.
\text{circumference of the same circle of radius $\boldsymbol{R}$.}
circumference of the same circle of radius
R
.
\text{circumference of the same circle of radius $\boldsymbol{R}$.}
What is the magnitude of the net magnetic field at the center of the circle (point
C
)?
\text{What is the magnitude of the net magnetic field at the center of the circle (point $\mathrm{C}$ )?}
What is the magnitude of the net magnetic field at the center of the circle (point
C
)?
\text{What is the magnitude of the net magnetic field at the center of the circle (point $\mathrm{C}$ )?}
A)
μ
o
i
2
π
R
\text{ A) $\frac{\mu_o i}{2 \pi R}$ }
A)
2
π
R
μ
o
i
\text{ A) $\frac{\mu_o i}{2 \pi R}$ }
B)
μ
o
i
π
R
\text{ B) $\frac{\mu_o i}{\pi R}$ }
B)
π
R
μ
o
i
\text{ B) $\frac{\mu_o i}{\pi R}$ }
C)
μ
o
i
4
π
R
\text{ C) $\frac{\mu_o i}{4 \pi R}$ }
C)
4
π
R
μ
o
i
\text{ C) $\frac{\mu_o i}{4 \pi R}$ }
D)
μ
o
i
2
π
R
+
μ
o
i
16
R
\text{ D) $\frac{\mu_o i}{2 \pi R}+\frac{\mu_o i}{16 R}$ }
D)
2
π
R
μ
o
i
+
16
R
μ
o
i
\text{ D) $\frac{\mu_o i}{2 \pi R}+\frac{\mu_o i}{16 R}$ }
E)
μ
o
i
π
R
+
μ
o
i
16
R
\text{ E) $\frac{\mu_o i}{\pi R}+\frac{\mu_o i}{16 R}$ }
E)
π
R
μ
o
i
+
16
R
μ
o
i
\text{ E) $\frac{\mu_o i}{\pi R}+\frac{\mu_o i}{16 R}$ }
In FIGURE 9, two infinitely long wires carry currents
i
. Each follows a
9
0
∘
arc on the
\text{In FIGURE 9, two infinitely long wires carry currents $\boldsymbol{i}$. Each follows a $90^{\circ}$ arc on the }
In FIGURE 9, two infinitely long wires carry currents
i
. Each follows a 9
0
∘
arc on the
\text{In FIGURE 9, two infinitely long wires carry currents $\boldsymbol{i}$. Each follows a $90^{\circ}$ arc on the }
circumference of the same circle of radius
R
.
\text{circumference of the same circle of radius $\boldsymbol{R}$.}
circumference of the same circle of radius
R
.
\text{circumference of the same circle of radius $\boldsymbol{R}$.}
What is the magnitude of the net magnetic field at the center of the circle (point
C
)?
\text{What is the magnitude of the net magnetic field at the center of the circle (point $\mathrm{C}$ )?}
What is the magnitude of the net magnetic field at the center of the circle (point
C
)?
\text{What is the magnitude of the net magnetic field at the center of the circle (point $\mathrm{C}$ )?}
* The current in the straight inner wire does not produce any magnetic field
\text{ * The current in the straight inner wire does not produce any magnetic field }
* The current in the straight inner wire does not produce any magnetic field
\text{ * The current in the straight inner wire does not produce any magnetic field }
* The magnetic fields due to the circular sections cancel.
\text{ * The magnetic fields due to the circular sections cancel. }
* The magnetic fields due to the circular sections cancel.
\text{ * The magnetic fields due to the circular sections cancel. }
* We are left with the outer straight wires:
\text{ * We are left with the outer straight wires: }
* We are left with the outer straight wires:
\text{ * We are left with the outer straight wires: }
B
c
=
2
×
semi - infinite wires
=
2
×
μ
0
i
4
π
R
=
μ
0
i
2
π
R
\text{ $\mathrm{B}_{\mathrm{c}}=2 \times$ semi - infinite wires $=2 \times \frac{\mu_0 \mathrm{i}}{4 \pi R}=\frac{\mu_0 \mathrm{i}}{2 \pi R}$ }
B
c
=
2
×
semi - infinite wires
=
2
×
4
π
R
μ
0
i
=
2
π
R
μ
0
i
\text{ $\mathrm{B}_{\mathrm{c}}=2 \times$ semi - infinite wires $=2 \times \frac{\mu_0 \mathrm{i}}{4 \pi R}=\frac{\mu_0 \mathrm{i}}{2 \pi R}$ }
FIGURE 10 shows two long, thin wires, parallel to the
z
axis, carrying currents in the positive
z
\text{FIGURE 10 shows two long, thin wires, parallel to the $z$ axis, carrying currents in the positive $z$ }
FIGURE 10 shows two long, thin wires, parallel to the
z
axis, carrying currents in the positive
z
\text{FIGURE 10 shows two long, thin wires, parallel to the $z$ axis, carrying currents in the positive $z$ }
direction. The
50
A
wire is in the
x
−
z
plane and is
5
m
from the
z
axis.
\text{direction. The $50\mathrm{A}$ wire is in the $x-z$ plane and is $5 \mathrm{~m}$ from the $z$ axis. }
direction. The 50
A
wire is in the
x
−
z
plane and is 5
m
from the
z
axis.
\text{direction. The $50\mathrm{A}$ wire is in the $x-z$ plane and is $5 \mathrm{~m}$ from the $z$ axis. }
The 40-A wire is in the
y
−
z
plane and is
4
m
from the
z
axis.
\text{The 40-A wire is in the $y-z$ plane and is $4 \mathrm{~m}$ from the $z$ axis. }
The 40-A wire is in the
y
−
z
plane and is 4
m
from the
z
axis.
\text{The 40-A wire is in the $y-z$ plane and is $4 \mathrm{~m}$ from the $z$ axis. }
What is the net magnetic field at the origin
O
due to the two wires?
\text{What is the net magnetic field at the origin $\mathrm{O}$ due to the two wires?}
What is the net magnetic field at the origin
O
due to the two wires?
\text{What is the net magnetic field at the origin $\mathrm{O}$ due to the two wires?}
A)
(
2
i
^
−
2
j
^
)
μ
T
\text{ A) $(2 \hat{i}-2 \hat{j}) \mu \mathrm{T}$ }
A)
(
2
i
^
−
2
j
^
)
μ
T
\text{ A) $(2 \hat{i}-2 \hat{j}) \mu \mathrm{T}$ }
B)
(
2
i
^
+
2
j
^
)
μ
T
\text{ B) $(2 \hat{i}+2 \hat{j}) \mu T$ }
B)
(
2
i
^
+
2
j
^
)
μ
T
\text{ B) $(2 \hat{i}+2 \hat{j}) \mu T$ }
C)
(
2
i
^
+
3
j
^
)
μ
T
\text{ C) $(2 \hat{i}+3 \hat{j}) \mu T$ }
C)
(
2
i
^
+
3
j
^
)
μ
T
\text{ C) $(2 \hat{i}+3 \hat{j}) \mu T$ }
D)
(
2
i
^
−
3
j
^
)
μ
T
\text{ D) $(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \mu \mathrm{T}$ }
D)
(
2
i
^
−
3
j
^
)
μ
T
\text{ D) $(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \mu \mathrm{T}$ }
E)
(
3
i
^
+
2
j
^
)
μ
T
\text{ E) $(3 \hat{i}+2 \hat{j}) \mu \mathrm{T}$ }
E)
(
3
i
^
+
2
j
^
)
μ
T
\text{ E) $(3 \hat{i}+2 \hat{j}) \mu \mathrm{T}$ }
FIGURE 10 shows two long, thin wires, parallel to the
z
axis, carrying currents in the positive
z
\text{FIGURE 10 shows two long, thin wires, parallel to the $z$ axis, carrying currents in the positive $z$ }
FIGURE 10 shows two long, thin wires, parallel to the
z
axis, carrying currents in the positive
z
\text{FIGURE 10 shows two long, thin wires, parallel to the $z$ axis, carrying currents in the positive $z$ }
direction. The
50
A
wire is in the
x
−
z
plane and is
5
m
from the
z
axis.
\text{direction. The $50\mathrm{A}$ wire is in the $x-z$ plane and is $5 \mathrm{~m}$ from the $z$ axis. }
direction. The 50
A
wire is in the
x
−
z
plane and is 5
m
from the
z
axis.
\text{direction. The $50\mathrm{A}$ wire is in the $x-z$ plane and is $5 \mathrm{~m}$ from the $z$ axis. }
The 40-A wire is in the
y
−
z
plane and is
4
m
from the
z
axis.
\text{The 40-A wire is in the $y-z$ plane and is $4 \mathrm{~m}$ from the $z$ axis. }
The 40-A wire is in the
y
−
z
plane and is 4
m
from the
z
axis.
\text{The 40-A wire is in the $y-z$ plane and is $4 \mathrm{~m}$ from the $z$ axis. }
What is the net magnetic field at the origin
O
due to the two wires?
\text{What is the net magnetic field at the origin $\mathrm{O}$ due to the two wires?}
What is the net magnetic field at the origin
O
due to the two wires?
\text{What is the net magnetic field at the origin $\mathrm{O}$ due to the two wires?}
B
→
1
=
μ
0
i
1
2
π
d
1
(
−
j
^
)
=
4
π
×
1
0
−
7
×
50
2
π
×
5
(
−
j
^
)
=
−
2
×
1
0
−
6
j
^
(
T
)
=
−
2
j
^
(
μ
T
)
B
→
2
=
+
μ
0
i
2
2
π
d
2
i
^
=
4
π
×
1
0
−
7
×
40
2
π
×
4
(
i
^
)
=
+
2
i
^
(
μ
T
)
∴
B
→
net
=
(
2
i
^
−
2
j
^
)
μ
T
\begin{aligned} & \overrightarrow{\mathrm{B}}_1=\frac{\mu_0 \mathrm{i}_1}{2 \pi \mathrm{d}_1}(-\hat{\mathrm{j}}) \\ & =\frac{4 \pi \times 10^{-7} \times 50}{2 \pi \times 5}(-\hat{\mathrm{j}})=-2 \times 10^{-6} \hat{\mathrm{j}}(\mathrm{T})=-2 \hat{\mathrm{j}}(\mu \mathrm{T}) \\ & \overrightarrow{\mathrm{B}}_2=+\frac{\mu_0 \mathrm{i}_2}{2 \pi \mathrm{d}_2} \hat{\mathrm{i}} \\ & =\frac{4 \pi \times 10^{-7} \times 40}{2 \pi \times 4}(\hat{\mathrm{i}})=+2 \hat{\mathrm{i}}(\mu \mathrm{T}) \\ & \therefore \overrightarrow{\mathrm{B}}_{\text {net }}=(2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}) \mu \mathrm{T} \end{aligned}
B
1
=
2
π
d
1
μ
0
i
1
(
−
j
^
)
=
2
π
×
5
4
π
×
1
0
−
7
×
50
(
−
j
^
)
=
−
2
×
1
0
−
6
j
^
(
T
)
=
−
2
j
^
(
μ
T
)
B
2
=
+
2
π
d
2
μ
0
i
2
i
^
=
2
π
×
4
4
π
×
1
0
−
7
×
40
(
i
^
)
=
+
2
i
^
(
μ
T
)
∴
B
net
=
(
2
i
^
−
2
j
^
)
μ
T
\begin{aligned} & \overrightarrow{\mathrm{B}}_1=\frac{\mu_0 \mathrm{i}_1}{2 \pi \mathrm{d}_1}(-\hat{\mathrm{j}}) \\ & =\frac{4 \pi \times 10^{-7} \times 50}{2 \pi \times 5}(-\hat{\mathrm{j}})=-2 \times 10^{-6} \hat{\mathrm{j}}(\mathrm{T})=-2 \hat{\mathrm{j}}(\mu \mathrm{T}) \\ & \overrightarrow{\mathrm{B}}_2=+\frac{\mu_0 \mathrm{i}_2}{2 \pi \mathrm{d}_2} \hat{\mathrm{i}} \\ & =\frac{4 \pi \times 10^{-7} \times 40}{2 \pi \times 4}(\hat{\mathrm{i}})=+2 \hat{\mathrm{i}}(\mu \mathrm{T}) \\ & \therefore \overrightarrow{\mathrm{B}}_{\text {net }}=(2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}) \mu \mathrm{T} \end{aligned}
Use the right hand rule for the direction of
B
⃗
\text{Use the right hand rule for the direction of $\vec{B}$}
Use the right hand rule for the direction of
B
\text{Use the right hand rule for the direction of $\vec{B}$}
FIGURE 11 shows a cross section of three parallel wires each carrying a current of
5.0
A
out of
\text{FIGURE 11 shows a cross section of three parallel wires each carrying a current of $5.0 \mathrm{~A}$ out of }
FIGURE 11 shows a cross section of three parallel wires each carrying a current of 5.0
A
out of
\text{FIGURE 11 shows a cross section of three parallel wires each carrying a current of $5.0 \mathrm{~A}$ out of }
the paper. If the distance
d
=
6.0
m
m
, what is the magnitude of the net magnetic force on
\text{the paper. If the distance $\boldsymbol{d}=6.0 \mathrm{~mm}$, what is the magnitude of the net magnetic force on }
the paper. If the distance
d
=
6.0
mm
, what is the magnitude of the net magnetic force on
\text{the paper. If the distance $\boldsymbol{d}=6.0 \mathrm{~mm}$, what is the magnitude of the net magnetic force on }
a
2.0
m
length of wire 1 ?
\text{a $2.0\mathrm{m}$ length of wire 1 ?}
a 2.0
m
length of wire 1 ?
\text{a $2.0\mathrm{m}$ length of wire 1 ?}
A)
2.9
m
N
\text{ A) $2.9 \mathrm{mN}$ }
A) 2.9
mN
\text{ A) $2.9 \mathrm{mN}$ }
B)
3.3
m
N
\text{ B) $3.3 \mathrm{mN}$ }
B) 3.3
mN
\text{ B) $3.3 \mathrm{mN}$ }
C)
2.1
m
N
\text{ C) $2.1 \mathrm{mN}$ }
C) 2.1
mN
\text{ C) $2.1 \mathrm{mN}$ }
D)
3.9
m
N
\text{ D) $3.9 \mathrm{mN}$ }
D) 3.9
mN
\text{ D) $3.9 \mathrm{mN}$ }
E)
1.7
m
N
\text{ E) $1.7 \mathrm{mN}$ }
E) 1.7
mN
\text{ E) $1.7 \mathrm{mN}$ }
FIGURE 11 shows a cross section of three parallel wires each carrying a current of
5.0
A
out of
\text{FIGURE 11 shows a cross section of three parallel wires each carrying a current of $5.0 \mathrm{~A}$ out of }
FIGURE 11 shows a cross section of three parallel wires each carrying a current of 5.0
A
out of
\text{FIGURE 11 shows a cross section of three parallel wires each carrying a current of $5.0 \mathrm{~A}$ out of }
the paper. If the distance
d
=
6.0
m
m
, what is the magnitude of the net magnetic force on
\text{the paper. If the distance $\boldsymbol{d}=6.0 \mathrm{~mm}$, what is the magnitude of the net magnetic force on }
the paper. If the distance
d
=
6.0
mm
, what is the magnitude of the net magnetic force on
\text{the paper. If the distance $\boldsymbol{d}=6.0 \mathrm{~mm}$, what is the magnitude of the net magnetic force on }
a
2.0
m
length of wire 1 ?
\text{a $2.0\mathrm{m}$ length of wire 1 ?}
a 2.0
m
length of wire 1 ?
\text{a $2.0\mathrm{m}$ length of wire 1 ?}
F
=
μ
0
L
i
2
2
π
d
=
4
π
×
1
0
−
7
×
2
×
25
2
π
×
6
×
1
0
−
3
=
1.67
×
1
0
−
3
N
F
net
=
2
. F
⋅
sin
60
=
2.9
×
1
0
−
3
N
\begin{aligned} & \mathrm{F}=\frac{\mu_0 \mathrm{Li}^2}{2 \pi \mathrm{d}}=\frac{4 \pi \times 10^{-7} \times 2 \times 25}{2 \pi \times 6 \times 10^{-3}} \\ & =1.67 \times 10^{-3} \mathrm{~N} \\ & \mathrm{~F}_{\text {net }}=2 \text {. F } \cdot \sin 60=2.9 \times 10^{-3} \mathrm{~N} \\ & \end{aligned}
F
=
2
π
d
μ
0
Li
2
=
2
π
×
6
×
1
0
−
3
4
π
×
1
0
−
7
×
2
×
25
=
1.67
×
1
0
−
3
N
F
net
=
2
. F
⋅
sin
60
=
2.9
×
1
0
−
3
N
\begin{aligned} & \mathrm{F}=\frac{\mu_0 \mathrm{Li}^2}{2 \pi \mathrm{d}}=\frac{4 \pi \times 10^{-7} \times 2 \times 25}{2 \pi \times 6 \times 10^{-3}} \\ & =1.67 \times 10^{-3} \mathrm{~N} \\ & \mathrm{~F}_{\text {net }}=2 \text {. F } \cdot \sin 60=2.9 \times 10^{-3} \mathrm{~N} \\ & \end{aligned}
A long, solid, cylindrical wire carries a uniformly distributed current. If the radius of the wire
\text{A long, solid, cylindrical wire carries a uniformly distributed current. If the radius of the wire}
A long, solid, cylindrical wire carries a uniformly distributed current. If the radius of the wire
\text{A long, solid, cylindrical wire carries a uniformly distributed current. If the radius of the wire}
is
3.5
m
m
, and the magnitude of the current density is
1.5
A
/
c
m
2
, what is the magnitude of
\text{is $3.5 \mathrm{~mm}$, and the magnitude of the current density is $1.5 \mathrm{~A} / \mathrm{cm}^2$, what is the magnitude of}
is 3.5
mm
, and the magnitude of the current density is 1.5
A
/
cm
2
, what is the magnitude of
\text{is $3.5 \mathrm{~mm}$, and the magnitude of the current density is $1.5 \mathrm{~A} / \mathrm{cm}^2$, what is the magnitude of}
the magnetic field at a distance of
2.5
m
m
from the axis of the wire?
\text{the magnetic field at a distance of $2.5 \mathrm{~mm}$ from the axis of the wire?}
the magnetic field at a distance of 2.5
mm
from the axis of the wire?
\text{the magnetic field at a distance of $2.5 \mathrm{~mm}$ from the axis of the wire?}
A)
2.4
×
1
0
−
5
T
\text{ A) $2.4 \times 10^{-5} \mathrm{~T}$ }
A) 2.4
×
1
0
−
5
T
\text{ A) $2.4 \times 10^{-5} \mathrm{~T}$ }
B)
3.3
×
1
0
−
5
T
\text{ B) $3.3 \times 10^{-5} \mathrm{~T}$ }
B) 3.3
×
1
0
−
5
T
\text{ B) $3.3 \times 10^{-5} \mathrm{~T}$ }
C)
1.3
×
1
0
−
5
T
\text{ C) $1.3 \times 10^{-5} \mathrm{~T}$ }
C) 1.3
×
1
0
−
5
T
\text{ C) $1.3 \times 10^{-5} \mathrm{~T}$ }
D)
6.9
×
1
0
−
5
T
\text{ D) $6.9 \times 10^{-5} \mathrm{~T}$ }
D) 6.9
×
1
0
−
5
T
\text{ D) $6.9 \times 10^{-5} \mathrm{~T}$ }
E) zero
\text{ E) zero }
E) zero
\text{ E) zero }
A long, solid, cylindrical wire carries a uniformly distributed current. If the radius of the wire
\text{A long, solid, cylindrical wire carries a uniformly distributed current. If the radius of the wire}
A long, solid, cylindrical wire carries a uniformly distributed current. If the radius of the wire
\text{A long, solid, cylindrical wire carries a uniformly distributed current. If the radius of the wire}
is
3.5
m
m
, and the magnitude of the current density is
1.5
A
/
c
m
2
, what is the magnitude of
\text{is $3.5 \mathrm{~mm}$, and the magnitude of the current density is $1.5 \mathrm{~A} / \mathrm{cm}^2$, what is the magnitude of}
is 3.5
mm
, and the magnitude of the current density is 1.5
A
/
cm
2
, what is the magnitude of
\text{is $3.5 \mathrm{~mm}$, and the magnitude of the current density is $1.5 \mathrm{~A} / \mathrm{cm}^2$, what is the magnitude of}
the magnetic field at a distance of
2.5
m
m
from the axis of the wire?
\text{the magnetic field at a distance of $2.5 \mathrm{~mm}$ from the axis of the wire?}
the magnetic field at a distance of 2.5
mm
from the axis of the wire?
\text{the magnetic field at a distance of $2.5 \mathrm{~mm}$ from the axis of the wire?}
∮
B
⃗
⋅
d
s
⃗
=
=
μ
0
i
e
n
c
⇒
B
(
2
π
r
)
=
μ
0
.
J
⋅
A
=
μ
0
J
.
π
r
2
B
=
μ
0
J
r
2
=
4
π
×
1
0
−
7
×
1.5
×
1
0
4
×
2.5
×
1
0
−
3
2
=
2.4
×
1
0
−
5
T
\begin{aligned} & \oint \vec{B} \cdot d \vec{s}==\mu_0 i_{\mathrm{enc}} \Rightarrow B(2 \pi r)=\mu_0 . J \cdot A=\mu_0 \mathrm{~J} . \pi r^2 \\ & B=\frac{\mu_0 \mathrm{Jr}}{2}=\frac{4 \pi \times 10^{-7} \times 1.5 \times 10^4 \times 2.5 \times 10^{-3}}{2}=2.4 \times 10^{-5} \mathrm{~T} \end{aligned}
∮
B
⋅
d
s
==
μ
0
i
enc
⇒
B
(
2
π
r
)
=
μ
0
.
J
⋅
A
=
μ
0
J
.
π
r
2
B
=
2
μ
0
Jr
=
2
4
π
×
1
0
−
7
×
1.5
×
1
0
4
×
2.5
×
1
0
−
3
=
2.4
×
1
0
−
5
T
\begin{aligned} & \oint \vec{B} \cdot d \vec{s}==\mu_0 i_{\mathrm{enc}} \Rightarrow B(2 \pi r)=\mu_0 . J \cdot A=\mu_0 \mathrm{~J} . \pi r^2 \\ & B=\frac{\mu_0 \mathrm{Jr}}{2}=\frac{4 \pi \times 10^{-7} \times 1.5 \times 10^4 \times 2.5 \times 10^{-3}}{2}=2.4 \times 10^{-5} \mathrm{~T} \end{aligned}
Resume presentation
C h a p t e r 2 9 \text{Chapter 29} C h a p t e r 2 9 \text{Chapter 29} C h a p t e r 2 9 \text{Chapter 29} C h a p t e r 2 9 \text{Chapter 29}
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