An infinite current sheet lies in the z=0 plane. It carries a surface
current density of
K=50.9 A/m
^
y
A wire segment of length L is stretched between two points specified by
P1(-3.00 m,-4.00 m,+5.00 m)
P2(+5.00 m,+12.0 m,+13.0 m)
A constant current of 25.0 A flows from P1 to P2.
(a) Calculate the magnetic field (magnitude and direction) due to the
current sheet for z > 0.
SOLUTION
The magnetic field above the plane is given in Example 5.8 on page 226.
B
=
(mo/2)K
^
x
=
1
2
(4p×10-7 T·m/A) (50.9 A/m)
^
x
=
+3.20×10-5
^
x
T
The long way is to use Eqn. 5.39 on page 219.
B(r)
=
mo
4p
ó õ
K( r¢)×R
R3
The calculation is messy, but doable and gives the same result. It is a lot
easier to use Ampere's law in the line integral form to get the result.
(b) Calculate the force which the current sheet exerts on the wire.
SOLUTION
Note that the magnetic field is constant everywhere above the plane. The
problem reduces to a straight current-carrying wire in a uniform magnetic
field. The force becomes
F = I L×B
The vector length is
L
=
[(5)-(-3)]
^
x
+ [(12)-(-4)]
^
y
+ [13-5]
^
z
=
8
^
x
+16
^
y
+8
^
z
Substituting the length and magnetic field vectors into the force equation
gives
F
=
æ è
6.40
^
y
-12.8
^
z
ö ø
10-3 N
The magnetic fields inside and outside the charged spinning sphere of
example 5.11 are
Bin
=
2
3
moM
^
z
=
2
3
moM
æ è
cosq
^
r
-sinq
^
q
ö ø
(r < R)
Bout
=
moMR3
3r3
æ è
2cosq
^
r
+ sinq
^
q
ö ø
(r > R)
(a) Verify that the divergence of the magnetic field both inside and
outside the sphere is zero.
SOLUTION
Since the magnetic field varies only with r and q and Bf=0,
the divergence equation, in spherical coordinates, reduces to
Ñ·B
=
1
r2
¶
¶r
(r2Br) +
1
rsinq
¶
¶q
(sinq Bq)
Isolating the magnetic field components,
Br(in)
=
2
3
moMcosq
Bq(in)
=
-
2
3
moMsinq
Br(out)
=
2moMR3cosq
3r3
Bq(out)
=
-
moMR3sinq
3r3
Inside the sphere,
Ñ·Bin
=
4moMcosq
3r
-
4moMcosq
3r
=
0 r £ R
Outside the sphere,
Ñ·Bout
=
-
2moMR3cosq
3r4
+
2moMR3cosq
3r4
=
0 r ³ R
Each of the above two calculations is valid for all respective r including
r=R.
(b) Using Ampere's law in the form
Ñ×B = moJ
Determine the equation for the current density on the surface. Use the
spherical coordinate version for the curl.
SOLUTION
Again, since the magnetic field depends only on r and q, and
Bf=0, the curl equation reduces to
Ñ×B
=
1
r
é ë
¶
¶r
(rBq)-
¶Br
¶q
ù û
^
f
The curl of the field should be zero both inside and outside the sphere
because there is no current anywhere but on the surface. As a check,
Inside the sphere
Ñ×Bin
=
é ë
-
2moMsinq
3r
+
2moMsinq
3r
ù û
^
f
=
0 for r < R
Outside the sphere
Ñ×Bout
=
é ë
-
2moMR3sinq
3r4
+
2moMR3 sinq
3r4
ù û
^
f
=
0 for r > R
For the surface, set r=R. The results are
Ñ×Bin|r=R
=
1
R
é ë
¶
¶r
æ è
-
2moMR sinq
3
ö ø
-
¶
¶q
æ è
2moMcosq
3
ö ø
ù û
^
f
=
2moMsinq
3R
^
f
Ñ×Bout|r=R
=
1
R
é ë
¶
¶r
æ è
moMR sinq
3
ö ø
-
¶
¶q
æ è
2moMcosq
3
ö ø
ù û
^
f
=
2moMsinq
3R
^
f
The current density is then
J
=
2Msinq
3R
^
f
(c) Where does the current density have a maximum value? a minimum
value?
SOLUTION
The current density is a maximum in the equatorial plane, (q = p/2),
and zero at the poles (q = 0, p).
A parabolic line segment (y=kx2) carries a constant current, I.
The ends of the wire are defined by the points (-a,a) and (a,a). The goal is to find the magnetic field at the point (0,c) on the y-axis using the
Biot-Savart law in the form
dB
=
moI
4p
dl¢×R
R3
The values of the constants are
a=0.200 m
c=0.400 m
I=15.0 A
(a) Find the value of k.
SOLUTION
k=1/a=5/m
(b) Draw a figure identifying the relevant variables in the above
equation. Make sure you get the directions right.
(c) Determine the equations for dl¢, R, and,
R ( the first two in terms of the unit vectors).
SOLUTION
dl
=
dx¢
^
x
+dy¢
^
y
=
dx¢
^
x
+2kx¢dx¢
^
y
=
(
^
x
+2kx¢
^
y
)dx¢
R
=
(0-x¢)
^
x
+(c-y¢ )
^
y
=
-x¢
^
x
+(c-y¢)
^
y
R
=
Ö
x¢2+(c-y¢)2
=
Ö
x¢2+(c-kx¢2)2
=
Ö
k2x¢4-(2kc-1)x¢2+c2
=
Ö
25x¢4-3x¢2+0.160
(d) Set up the final equation for obtaining B.
SOLUTION
Substituting the results of part (c) into the dB equation and
combining terms,
B
=
2
æ è
moI
4p
ö ø
ó õ
a
0
(c+kx ¢2)dx¢
[k2x¢4- (2kc-1)x¢2+c2]3/2
^
z
(e) Use Maple to solve for the numerical value of B. To make sure Maple
does the integration, surround the entire int command with evalf. The result
is
B = 6.97×10-6 T
^
z
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version 3.85. On 04 Jan 2010, 15:20.