PH317 MAGNETISM
WINTER Quarter 2009-2010
Solution to Homework Assignment 2
Assigned: Monday 12/07/09
Due: Monday 12/14/09
A straight wire segment lies on the x-axis and extends from x=-a to
x=0. It carries a constant current, I, directed along the positive
x-axis. Use the Biot-Savart law for this calculation.
I=8.00 A
a = 12.0 cm
(a) Choose an arbitrary point (x,y) in the first quadrant and
derive a formula for the magnetic field: B(x,y) due to the current
in the wire segment.
The program I use to type up the solutions does NOT have script characters. I
will use R for the distance variable.
dB
=
moI
4p
dl�×R
R3
Let (x,y) be the coordinates of the field point. Let
(x�,y�) be the coordinates of a point on the
current line segment.
R
=
(x-x�)
^
x
+ (y-y�)
^
y
=
(x-x�)
^
x
+ y
^
y
R
=
�
(x-x�)2+y2
R3
=
[(x-x�)2+y2]3/2
dl�
=
dx�
^
x
Combining the parts, one obtains
B(x,y)
=
moI
^
z
4p
� �
0
-a
dx�
[(x-x� )2+y2]3/2
=
moI
^
z
4py2
� �
�
x+a
�
(x+a)2+y2
-
x
�
x2+y2
� �
�
(b) Using Maple, plot the magnitude B(x,y) along the line
y = (0.0100)x+(0.0200 m)
in the range 0 � x � 0.0300 m.
Substitute the line equation into the magnetic field equation and let Maple
do the plot. See the file hw317-2s.mws in the homework solution folder for the
graph.
In class, an expression for the magnetic field on the planar axis of
a semi-circular current-carrying loop was found to be
B
=
moIa2
4p
� �
p
0
dq
(a2+y2o+2ayosinq)3/2
^
z
+
moIayo
4p
� �
p
0
sinq dq
(a2+y2o+2ayo sinq)3/2
^
z
(a) Using the Biot-Savart law as was done in Physics III, show that
the magnetic field at the center of curvature of the semi-circular wire is
Bc=moI/4a
dB
=
moI
4p
dl�×R
R3
The individual terms are (in cylindrical):
R
=
-a
^
s
R
=
a
dl�
=
dl
^
q
=
a dq
^
q
Substituting,
B
=
moI
4p
� �
p
0
� �
a dq
^
f
� �
×(-a
^
s
)
a3
=
+
moI
^
z
4pa
� �
p
0
dq
=
+
moI
^
z
4 a
(b) Using the above integral expression, show that it reduces to the
same expression you found in part (a).
Just set yo=0. The second integral vanishes. The first integral becomes
B
=
moI
^
z
4pa
� �
p
0
dq
=
moI
^
z
4 a
(c) Plot the integrands for a=0.100 m, yo=0.100 m, and
I=12.0 A in the interval 0 � q � p.
See Maple file hw317-2s.mws. Both integrands are well-behaved. There should
be no trouble with the integration.
(d) Evaluate the two integrals for the numbers given in part (c) and
show that the total magnetic field is
B(yo=0.1 m)=1.06×10-5 T
The two integrals have values
B1
=
6.88676×10-6 T
B2
=
3.68972×10-6 T
The sum is
B=1.06×10-6 T
The magnetic field inside a long, straight cylindrical wire carrying
a constant current is
B
=
� �
mo I
2pa2
� �
s
^
j
Calculate the current density, J, inside the wire. Express the curl in
cylindrical coordinates to do this calculation.
The magnetic field is in the azimuthal direction.
B = B
^
f
The equation for the current density is
moJ = �×B
In cylindrical coordinates, the only surviving term in the curl expression is
�×B
=
1
s
�
�s
(sBf)
^
z
=
1
s
d
ds
(sBf)
^
z
=
1
2
� �
d
ds
� �
moI
2pa2
� �
s2
� �
^
z
=
moI
pa2
Therefore,
J =
I
pa2
^
z
A circular wire of radius a lies in the x-y plane and is centered
at the origin. The wire carries a constant current I in the
counter-clockwise direction. The magnetic field at point P on the z-axis
(as determined by the Biot-Savart Law) is
B(P)
=
moIa2
2(z2+a2)3/2
^
z
The existence of �Bz/�z is clearly seen from the
equation for B. The other two partial derivatives
�Bx/�x and �By/�y also exist even
though the field equation does not contain either Bx or By! Use the
divergence of B to find the other two derivatives. Comment on why
they must NOT be zero.
�·B
=
�Bx
�x
+
�By
�y
+
�Bz
�z
=
0
By symmetry, the first two terms are equal. Therefore,
�By
�y
=
-
1
2
�Bz
�z
=
-
1
2
� �
-3moIa2z
4(a2+z2) 5/2
� �
=
3moIa2z
8(a2+z2)5/2
COMMENT: The divergence of the magnetic field cannot be zero if the first
two terms are zero. The other way to get this result is to calculate the
field off-axis. Now, the field is a function of all three coordinates. The
individual partial derivatives can now be taken and then one can set
x=y=0.
Take the x-axis to the right, the y-axis up, and, the z-axis
out of the page. Let Kupper be the surface current density in the
upper surface and Klower be the surface current density in the
lower surface. The vector expressions for the surface current densities are:
Kupper=sv=sv
^
x
=K
^
x
Klower=-sv=-sv
^
x
=-K
^
x
The magnetic fields above the upper surface, in between the two surfaces and
below the lower surface are (using the results of Example 5.8, page 226):
Bupper-above
=
+(mo/2)K
^
z
Bupper-below
=
-(mo/2)K
^
z
Blower-above
=
-(mo/2)K
^
z
Blower-below
=
+(mo/2)K
^
z
(a) Above the upper surface, the first and third field expressions
cancel. Between the two surfaces, the second and third field expressions add,
giving
Bmiddle=-moK
^
z
Below the lower surface, the second and fourth field expressions cancel.
(b) Take a small area segment of size A on the upper surface. The
force on that segment is
F
=
� �
(Kupper×Blower-above)da
=
� �
(K
^
x
)×(-moK\l
^
z
/2)da
=
+(moK2A/2)
^
y
=
+(mos2v2A/2)
^
y
The magnetic force per unit area is
fm
=
F
A
=
+(mos2v2/2)
^
y
(c) The electric field at the area segment A is uniform and has a
magnitude E=s/2eo. This can be found from Gauss' law. The
charge on the area A is Q=sA. The electric force on A is
F=QE=(sA)(s/2eo)=s2A/2eo
The electric force per unit area is
fe=F/A=s2/2eo
Set the two force densities equal to each other.
fm
=
fe
mos2v2
2
=
s2
2eo
mov2
=
1
eo
v
=
1
�
moeo
=
c
File translated from
TEX
by
TTH,
version 3.85. On 15 Dec 2009, 17:06.