line integral magnetic induction
Need help with this line integral problem. I've been stuck on this problem for a while any help will be much appreciated.
$$ Problem $$
Experiments show that a steady current in a long wire produces a magnetic field B that is tangent to any circle that lies in
the plane perpendicular to the wire and whose center is the axis
of the wire. Ampère’s Law relates the electric current to its magnetic effects and states that
$$int_c B ,dr=mu I $$
where $$ I $$ is the net current that passes through any surface
bounded by a closed curve C , and $$ mu$$ is a constant called the
permeability of free space. By taking C to be a circle with
radius r , show that the magnitude B of the magnetic
field at a distance r from the center of the wire is $$ B = mu I/2 pi r $$
calculus integration multivariable-calculus line-integrals
asked Nov 28 '15 at 8:17
James
84
add a comment |
Need help with this line integral problem. I've been stuck on this problem for a while any help will be much appreciated.
$$ Problem $$
Experiments show that a steady current in a long wire produces a magnetic field B that is tangent to any circle that lies in
the plane perpendicular to the wire and whose center is the axis
of the wire. Ampère’s Law relates the electric current to its magnetic effects and states that
$$int_c B ,dr=mu I $$
where $$ I $$ is the net current that passes through any surface
bounded by a closed curve C , and $$ mu$$ is a constant called the
permeability of free space. By taking C to be a circle with
radius r , show that the magnitude B of the magnetic
field at a distance r from the center of the wire is $$ B = mu I/2 pi r $$
calculus integration multivariable-calculus line-integrals
asked Nov 28 '15 at 8:17
James
84
add a comment |
Need help with this line integral problem. I've been stuck on this problem for a while any help will be much appreciated.
$$ Problem $$
Experiments show that a steady current in a long wire produces a magnetic field B that is tangent to any circle that lies in
the plane perpendicular to the wire and whose center is the axis
of the wire. Ampère’s Law relates the electric current to its magnetic effects and states that
$$int_c B ,dr=mu I $$
where $$ I $$ is the net current that passes through any surface
bounded by a closed curve C , and $$ mu$$ is a constant called the
permeability of free space. By taking C to be a circle with
radius r , show that the magnitude B of the magnetic
field at a distance r from the center of the wire is $$ B = mu I/2 pi r $$
calculus integration multivariable-calculus line-integrals
asked Nov 28 '15 at 8:17
James
84
Need help with this line integral problem. I've been stuck on this problem for a while any help will be much appreciated.
$$ Problem $$
Experiments show that a steady current in a long wire produces a magnetic field B that is tangent to any circle that lies in
the plane perpendicular to the wire and whose center is the axis
of the wire. Ampère’s Law relates the electric current to its magnetic effects and states that
$$int_c B ,dr=mu I $$
where $$ I $$ is the net current that passes through any surface
bounded by a closed curve C , and $$ mu$$ is a constant called the
permeability of free space. By taking C to be a circle with
radius r , show that the magnitude B of the magnetic
field at a distance r from the center of the wire is $$ B = mu I/2 pi r $$
calculus integration multivariable-calculus line-integrals
calculus integration multivariable-calculus line-integrals
asked Nov 28 '15 at 8:17
James
84
asked Nov 28 '15 at 8:17
James
84
edited Nov 28 '15 at 8:22
asked Nov 28 '15 at 8:17
James
84
asked Nov 28 '15 at 8:17
James
84
asked Nov 28 '15 at 8:17
James
84
84
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Ampere's Law:
$$oint vec B cdot , d vec l = mu_0 I$$
We assume that the magnetic field $vec B$ is constant, therefore we take it out of the integral:
$$ B oint d l = mu_0 I$$
$$ B cdot (text{circumference of circle}) = mu_0 I$$
$$ B cdot (2 pi r) = mu_0 I$$
$$ B = frac{mu_0 I}{2 pi r}$$
answered Nov 28 '15 at 8:36
Aldon
1,4121219
I think it is important to detail what $vec{B}$ constant implies. It would be more complete, in order to understand how $vec{B}cdot dvec{l}$ becomes $B dl$.
– Kuifje
Nov 28 '15 at 18:47
No, what you've written is really very wrong.
– Ted Shifrin
Nov 18 '18 at 0:44
add a comment |
Ampere's law states that
$$
int_C vec{B}cdot dvec{r} = mu I.
$$
If $C$ is a circle of radius $R$, it can be parametrized as follows:
$$
vec{r}(t)=Rcos(t)vec{i}+Rsin(t)vec{j},quad 0le t le 2pi,
$$
so Ampere's law can be rewritten as
$$
int_0^{2pi} vec{B}(vec{r}(t))cdot vec{r}'(t)dt = mu I.
$$
But if the magnetic field $vec{B}$ is tangent to any circle that lies in the plane perpendicular to the wire and whose center is the axis of the wire, then its direction is given by $vec{r}'(t)$, more precisely by $frac{vec{r}'(t)}{|vec{r}'(t)|}$ if we want a vector with norm 1. And since $vec{B}$ has the same intensity all around the wire, it does not depend on variable $t$. It follows that
$$
vec{B}(vec{r}(t))=B(R) frac{vec{r}'(t)}{|vec{r}'(t)|},
$$
and that
$$
int_0^{2pi} vec{B}(vec{r}(t))cdot vec{r}'(t)dt = int_0^{2pi} B(R)frac{vec{r}'(t)}{|vec{r}'(t)|}cdot vec{r}'(t)dt
=int_0^{2pi} B(R)frac{|vec{r}'(t)|^2}{|vec{r}'(t)|}dt\
= B(R) int_0^{2pi} |vec{r}'(t) | dt
= B(R) int_0^{2pi} R ;dt =2pi R B(R)
$$
Solving for $B(R)$ yields
$$
B(R )=frac{mu I }{2pi R}
$$
answered Nov 28 '15 at 18:46
Kuifje
7,1082725
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Ampere's Law:
$$oint vec B cdot , d vec l = mu_0 I$$
We assume that the magnetic field $vec B$ is constant, therefore we take it out of the integral:
$$ B oint d l = mu_0 I$$
$$ B cdot (text{circumference of circle}) = mu_0 I$$
$$ B cdot (2 pi r) = mu_0 I$$
$$ B = frac{mu_0 I}{2 pi r}$$
answered Nov 28 '15 at 8:36
Aldon
1,4121219
I think it is important to detail what $vec{B}$ constant implies. It would be more complete, in order to understand how $vec{B}cdot dvec{l}$ becomes $B dl$.
– Kuifje
Nov 28 '15 at 18:47
No, what you've written is really very wrong.
– Ted Shifrin
Nov 18 '18 at 0:44
add a comment |
Ampere's Law:
$$oint vec B cdot , d vec l = mu_0 I$$
We assume that the magnetic field $vec B$ is constant, therefore we take it out of the integral:
$$ B oint d l = mu_0 I$$
$$ B cdot (text{circumference of circle}) = mu_0 I$$
$$ B cdot (2 pi r) = mu_0 I$$
$$ B = frac{mu_0 I}{2 pi r}$$
answered Nov 28 '15 at 8:36
Aldon
1,4121219
I think it is important to detail what $vec{B}$ constant implies. It would be more complete, in order to understand how $vec{B}cdot dvec{l}$ becomes $B dl$.
– Kuifje
Nov 28 '15 at 18:47
No, what you've written is really very wrong.
– Ted Shifrin
Nov 18 '18 at 0:44
add a comment |
Ampere's Law:
$$oint vec B cdot , d vec l = mu_0 I$$
We assume that the magnetic field $vec B$ is constant, therefore we take it out of the integral:
$$ B oint d l = mu_0 I$$
$$ B cdot (text{circumference of circle}) = mu_0 I$$
$$ B cdot (2 pi r) = mu_0 I$$
$$ B = frac{mu_0 I}{2 pi r}$$
answered Nov 28 '15 at 8:36
Aldon
1,4121219
Ampere's Law:
$$oint vec B cdot , d vec l = mu_0 I$$
We assume that the magnetic field $vec B$ is constant, therefore we take it out of the integral:
$$ B oint d l = mu_0 I$$
$$ B cdot (text{circumference of circle}) = mu_0 I$$
$$ B cdot (2 pi r) = mu_0 I$$
$$ B = frac{mu_0 I}{2 pi r}$$
answered Nov 28 '15 at 8:36
Aldon
1,4121219
answered Nov 28 '15 at 8:36
Aldon
1,4121219
answered Nov 28 '15 at 8:36
Aldon
1,4121219
answered Nov 28 '15 at 8:36
Aldon
1,4121219
1,4121219
I think it is important to detail what $vec{B}$ constant implies. It would be more complete, in order to understand how $vec{B}cdot dvec{l}$ becomes $B dl$.
– Kuifje
Nov 28 '15 at 18:47
No, what you've written is really very wrong.
– Ted Shifrin
Nov 18 '18 at 0:44
add a comment |
I think it is important to detail what $vec{B}$ constant implies. It would be more complete, in order to understand how $vec{B}cdot dvec{l}$ becomes $B dl$.
– Kuifje
Nov 28 '15 at 18:47
No, what you've written is really very wrong.
– Ted Shifrin
Nov 18 '18 at 0:44
I think it is important to detail what $vec{B}$ constant implies. It would be more complete, in order to understand how $vec{B}cdot dvec{l}$ becomes $B dl$.
– Kuifje
Nov 28 '15 at 18:47
I think it is important to detail what $vec{B}$ constant implies. It would be more complete, in order to understand how $vec{B}cdot dvec{l}$ becomes $B dl$.
– Kuifje
Nov 28 '15 at 18:47
No, what you've written is really very wrong.
– Ted Shifrin
Nov 18 '18 at 0:44
No, what you've written is really very wrong.
– Ted Shifrin
Nov 18 '18 at 0:44
add a comment |
Ampere's law states that
$$
int_C vec{B}cdot dvec{r} = mu I.
$$
If $C$ is a circle of radius $R$, it can be parametrized as follows:
$$
vec{r}(t)=Rcos(t)vec{i}+Rsin(t)vec{j},quad 0le t le 2pi,
$$
so Ampere's law can be rewritten as
$$
int_0^{2pi} vec{B}(vec{r}(t))cdot vec{r}'(t)dt = mu I.
$$
But if the magnetic field $vec{B}$ is tangent to any circle that lies in the plane perpendicular to the wire and whose center is the axis of the wire, then its direction is given by $vec{r}'(t)$, more precisely by $frac{vec{r}'(t)}{|vec{r}'(t)|}$ if we want a vector with norm 1. And since $vec{B}$ has the same intensity all around the wire, it does not depend on variable $t$. It follows that
$$
vec{B}(vec{r}(t))=B(R) frac{vec{r}'(t)}{|vec{r}'(t)|},
$$
and that
$$
int_0^{2pi} vec{B}(vec{r}(t))cdot vec{r}'(t)dt = int_0^{2pi} B(R)frac{vec{r}'(t)}{|vec{r}'(t)|}cdot vec{r}'(t)dt
=int_0^{2pi} B(R)frac{|vec{r}'(t)|^2}{|vec{r}'(t)|}dt\
= B(R) int_0^{2pi} |vec{r}'(t) | dt
= B(R) int_0^{2pi} R ;dt =2pi R B(R)
$$
Solving for $B(R)$ yields
$$
B(R )=frac{mu I }{2pi R}
$$
answered Nov 28 '15 at 18:46
Kuifje
7,1082725
add a comment |
Ampere's law states that
$$
int_C vec{B}cdot dvec{r} = mu I.
$$
If $C$ is a circle of radius $R$, it can be parametrized as follows:
$$
vec{r}(t)=Rcos(t)vec{i}+Rsin(t)vec{j},quad 0le t le 2pi,
$$
so Ampere's law can be rewritten as
$$
int_0^{2pi} vec{B}(vec{r}(t))cdot vec{r}'(t)dt = mu I.
$$
But if the magnetic field $vec{B}$ is tangent to any circle that lies in the plane perpendicular to the wire and whose center is the axis of the wire, then its direction is given by $vec{r}'(t)$, more precisely by $frac{vec{r}'(t)}{|vec{r}'(t)|}$ if we want a vector with norm 1. And since $vec{B}$ has the same intensity all around the wire, it does not depend on variable $t$. It follows that
$$
vec{B}(vec{r}(t))=B(R) frac{vec{r}'(t)}{|vec{r}'(t)|},
$$
and that
$$
int_0^{2pi} vec{B}(vec{r}(t))cdot vec{r}'(t)dt = int_0^{2pi} B(R)frac{vec{r}'(t)}{|vec{r}'(t)|}cdot vec{r}'(t)dt
=int_0^{2pi} B(R)frac{|vec{r}'(t)|^2}{|vec{r}'(t)|}dt\
= B(R) int_0^{2pi} |vec{r}'(t) | dt
= B(R) int_0^{2pi} R ;dt =2pi R B(R)
$$
Solving for $B(R)$ yields
$$
B(R )=frac{mu I }{2pi R}
$$
answered Nov 28 '15 at 18:46
Kuifje
7,1082725
add a comment |
Ampere's law states that
$$
int_C vec{B}cdot dvec{r} = mu I.
$$
If $C$ is a circle of radius $R$, it can be parametrized as follows:
$$
vec{r}(t)=Rcos(t)vec{i}+Rsin(t)vec{j},quad 0le t le 2pi,
$$
so Ampere's law can be rewritten as
$$
int_0^{2pi} vec{B}(vec{r}(t))cdot vec{r}'(t)dt = mu I.
$$
But if the magnetic field $vec{B}$ is tangent to any circle that lies in the plane perpendicular to the wire and whose center is the axis of the wire, then its direction is given by $vec{r}'(t)$, more precisely by $frac{vec{r}'(t)}{|vec{r}'(t)|}$ if we want a vector with norm 1. And since $vec{B}$ has the same intensity all around the wire, it does not depend on variable $t$. It follows that
$$
vec{B}(vec{r}(t))=B(R) frac{vec{r}'(t)}{|vec{r}'(t)|},
$$
and that
$$
int_0^{2pi} vec{B}(vec{r}(t))cdot vec{r}'(t)dt = int_0^{2pi} B(R)frac{vec{r}'(t)}{|vec{r}'(t)|}cdot vec{r}'(t)dt
=int_0^{2pi} B(R)frac{|vec{r}'(t)|^2}{|vec{r}'(t)|}dt\
= B(R) int_0^{2pi} |vec{r}'(t) | dt
= B(R) int_0^{2pi} R ;dt =2pi R B(R)
$$
Solving for $B(R)$ yields
$$
B(R )=frac{mu I }{2pi R}
$$
answered Nov 28 '15 at 18:46
Kuifje
7,1082725
Ampere's law states that
$$
int_C vec{B}cdot dvec{r} = mu I.
$$
If $C$ is a circle of radius $R$, it can be parametrized as follows:
$$
vec{r}(t)=Rcos(t)vec{i}+Rsin(t)vec{j},quad 0le t le 2pi,
$$
so Ampere's law can be rewritten as
$$
int_0^{2pi} vec{B}(vec{r}(t))cdot vec{r}'(t)dt = mu I.
$$
But if the magnetic field $vec{B}$ is tangent to any circle that lies in the plane perpendicular to the wire and whose center is the axis of the wire, then its direction is given by $vec{r}'(t)$, more precisely by $frac{vec{r}'(t)}{|vec{r}'(t)|}$ if we want a vector with norm 1. And since $vec{B}$ has the same intensity all around the wire, it does not depend on variable $t$. It follows that
$$
vec{B}(vec{r}(t))=B(R) frac{vec{r}'(t)}{|vec{r}'(t)|},
$$
and that
$$
int_0^{2pi} vec{B}(vec{r}(t))cdot vec{r}'(t)dt = int_0^{2pi} B(R)frac{vec{r}'(t)}{|vec{r}'(t)|}cdot vec{r}'(t)dt
=int_0^{2pi} B(R)frac{|vec{r}'(t)|^2}{|vec{r}'(t)|}dt\
= B(R) int_0^{2pi} |vec{r}'(t) | dt
= B(R) int_0^{2pi} R ;dt =2pi R B(R)
$$
Solving for $B(R)$ yields
$$
B(R )=frac{mu I }{2pi R}
$$
answered Nov 28 '15 at 18:46
Kuifje
7,1082725
answered Nov 28 '15 at 18:46
Kuifje
7,1082725
answered Nov 28 '15 at 18:46
Kuifje
7,1082725
answered Nov 28 '15 at 18:46
Kuifje
7,1082725
7,1082725
add a comment |
add a comment |
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