751
26-2 CU R R E NT DE NSITY
R/2
R
(a)
R
(e)
R/2
(d)
(b)
dr
(c)
We want the current in the area
between these two radii.
Our job is to sum the current in
all rings from this smallest one ...
... to this largest one.
Its area is the product of the
circumference and the width.
The current within the ring is
the product of the current density
and the ring’s area.
If the current is nonuniform, we start with a
ring that is so thin that we can approximate
the current density as being uniform within it.
Figure 26-6 (a) Cross section of a wire of
radius R. If the current density is uniform, the current is just the product of
the current density and the area. (b)–(e)
If the current is nonuniform, we must
first find the current through a thin ring
and then sum (via integration) the currents in all such rings in the given area.
A
(b) Suppose, instead, that the current density through a
cross section varies with radial distance r as J ϭ ar
2
, in which
a ϭ 3.0 ϫ 10
11
A/m
4
and r is in meters. What now is the
current through the same outer portion of the wire?
KEY IDEA
Because the current density is not uniform across a cross
section of the wire, we must resort to Eq. 26-4
and integrate the current density over the portion of the
wire from r ϭ R/2 to r ϭ R.
Calculations: The current density vector
(along the
wire’s length) and the differential area vector
(perpendicular to a cross section of the wire) have the same
direction. Thus,
J
: ؒ dA
: ϭ J dA cos 0 ϭ J dA.
dA
:
J
:
(i ϭ ͵ J
: ؒ dA
: )
We need to replace the differential area dA with something we can actually integrate between the limits r ϭ R/2
and r ϭ R. The simplest replacement (because J is given as a
function of r) is the area 2pr dr of a thin ring of circumference 2pr and width dr (Fig. 26-6b). We can then integrate
with r as the variable of integration. Equation 26-4 then
gives us
(Answer)
ϭ
15
32
p (3.0 ϫ 10
11
A/m
4
)(0.0020 m)
4 ϭ 7.1 A.
ϭ 2pa ΄
r
4
4 ΅
R
R/2
ϭ
pa
2 ΄ R
4 Ϫ
R
4
16 ΅ ϭ
15
32
paR
4
ϭ ͵
R
R/2
ar
2
2pr dr ϭ 2pa͵
R
R/2
r
3
dr
i ϭ ͵J
: ؒ dA
: ϭ ͵ J dA
Additional examples, video, and practice available at WileyPLUS
26-2 CU R R E NT DE NSITY
R/2
R
(a)
R
(e)
R/2
(d)
(b)
dr
(c)
We want the current in the area
between these two radii.
Our job is to sum the current in
all rings from this smallest one ...
... to this largest one.
Its area is the product of the
circumference and the width.
The current within the ring is
the product of the current density
and the ring’s area.
If the current is nonuniform, we start with a
ring that is so thin that we can approximate
the current density as being uniform within it.
Figure 26-6 (a) Cross section of a wire of
radius R. If the current density is uniform, the current is just the product of
the current density and the area. (b)–(e)
If the current is nonuniform, we must
first find the current through a thin ring
and then sum (via integration) the currents in all such rings in the given area.
A
(b) Suppose, instead, that the current density through a
cross section varies with radial distance r as J ϭ ar
2
, in which
a ϭ 3.0 ϫ 10
11
A/m
4
and r is in meters. What now is the
current through the same outer portion of the wire?
KEY IDEA
Because the current density is not uniform across a cross
section of the wire, we must resort to Eq. 26-4
and integrate the current density over the portion of the
wire from r ϭ R/2 to r ϭ R.
Calculations: The current density vector
(along the
wire’s length) and the differential area vector
(perpendicular to a cross section of the wire) have the same
direction. Thus,
J
: ؒ dA
: ϭ J dA cos 0 ϭ J dA.
dA
:
J
:
(i ϭ ͵ J
: ؒ dA
: )
We need to replace the differential area dA with something we can actually integrate between the limits r ϭ R/2
and r ϭ R. The simplest replacement (because J is given as a
function of r) is the area 2pr dr of a thin ring of circumference 2pr and width dr (Fig. 26-6b). We can then integrate
with r as the variable of integration. Equation 26-4 then
gives us
(Answer)
ϭ
15
32
p (3.0 ϫ 10
11
A/m
4
)(0.0020 m)
4 ϭ 7.1 A.
ϭ 2pa ΄
r
4
4 ΅
R
R/2
ϭ
pa
2 ΄ R
4 Ϫ
R
4
16 ΅ ϭ
15
32
paR
4
ϭ ͵
R
R/2
ar
2
2pr dr ϭ 2pa͵
R
R/2
r
3
dr
i ϭ ͵J
: ؒ dA
: ϭ ͵ J dA
Additional examples, video, and practice available at WileyPLUS