Phy213: General Physics III
Chapter 29 Worksheet
2/15/2016
1
Biot-Savart Law
1.
A straight wire horizontal wire, L=0.3 m, has a 1.5 A current running through it.
i = 1.5 A
ds
y

y = .10 m
r
x
.P
L/2
a.
Using the Biot-Savart Law, derive the equation for the magnetic field vector produced by the
wire at an arbitrary point P, below the wire and halfway between the ends.
dB=
oi ds  r
4 r 3
L
2
L
2
-L
2
-L
2
 i ds  r  i dx
B= dB= o  3 = o  2 sin  kˆ
4
r
4 r
Ans.
L
2
 iy
i
dx
x
B= o 
kˆ = o
4  y2+x2 
4 y  y2+x2 
3
2
-L
2
b.
2
i
kˆ -L = o
4
2
L
1
2
L

L 
y  y2 + 
4

2
1
2
kˆ
Determine the magnitude and direction of the magnetic field vector at point P=0.1 m below
the wire.
B=
Ans.
o
4
1.5A 0.3m

0.1m  0.1m

c.
L/2
2
0.3m
2
+
4




1
2
kˆ = 2.50x10-6 T kˆ
Determine the magnitude and direction of the magnetic field produced by the wire at a point
0.01 m to the left of point P.
Ans.
B=
oi
x
4 y  y2+x2 
1
2




5L


6
oiL
5
1
ˆ
kˆ -L =

1 +
1  k
2 2
2 2
24 y  
6
 2 L   
 5L  
2
 y +    
  y + 
 
6


 6   


 





o (1.5A)(0.3m) 
5
1
-6
ˆ
B=
+
T kˆ

1
1  k = 2.07x10
2 2
2 2
24 (0.1m) 

 1.5m  
 0.3m   
2
2
 (0.1m) + 
  (0.1m) + 
 
  
 6  
 6   
 

Phy213: General Physics III
Chapter 29 Worksheet
d.
2/15/2016
2
Determine the magnitude and direction of the magnetic field at point P for a wire of “infinite”
length.
B=
Ans.

oi
i
x
kˆ = o kˆ = 3.0x10-6 T kˆ
-
4 y  y2+x2 
2 y
1
2
2.
A quarter circle wire arc (90o), with radius of curvature R=0.1 m, has a
2.0 A current running through it.
a.
Using the Biot-Savart Law, determine the equation for magnetic field
vector produced by the arc at point P, located at the center of
curvature.
R
oi ds  r
oi R 2d ˆ
 i d ˆ
=k=- o
k
3
3
4 r
4 R
4 R

i
i
B= dB=- o  d kˆ = - o  kˆ
4 R 0
4 R
dB=
Ans.
b.
B= -
oi
 kˆ = -3.1x10-6 T kˆ
4 R
i = 2.0 A
Derive the magnetic field vector at the center of a circular loop, of
radius R.
B= dB=-
Ans.
d.
.P
Calculate the magnetic field vector produced by the arc at point P.
Ans.
c.
i = 2.0 A
oi
4 R
R
2
.
P
i
0 d kˆ = - 2Ro kˆ
Determine the magnitude and direction of the magnetic field at
the center of a circular loop, where R=0.1 m.
B= -
Ans.
oi ˆ
k = -1.26x10-7 T kˆ
2R
3.
Consider the current loop shown below, consisting of 2 concentric 90o arc segments of radii,
R1 and R2, connected by 2 segments that are parallel to the center of curvature for the arcs.
The current flowing through the loop is 1.0 A, counterclockwise.
a.
Using the Biot-Savart Law, determine the equations for the
magnetic field vectors produced by each segment of the
loop arc at point P, located at the center of curvature for
both arc segments.
Ans.
Each arc produces an oppositely directed B-field:
B1= b.
oi ˆ
oi ˆ
k and B2=
k
4 R1
4 R 2
Calculate the total magnetic field vector at point P, where
R1=0.1 m and R2= 0.15 m.
Ans.
Each arc produces an oppositely directed B-field:
i = 1.0 A
R2
R1
.P
Phy213: General Physics III
Chapter 29 Worksheet
B= B1+B2=
c.
2/15/2016
3
oi  1 1  ˆ
-7
ˆ
 - +  k = -5.25x10 T k
4  R1 R2 
Determine the net magnetic force vector exerted on the outer arc segment of the loop due
to the other segments.
Ampere’s Law
4. An infinitely long horizontal wire has a 2A current running through it.
i=2A
y
r = 1.0 m
.P
a.
Ans.
Using Ampere’s Law, determine the magnitude and direction of magnetic field
produced by the wire at a point 1 m below the wire?
Apply a circular “Amperean loop” centered on the wire:
 Bd
B =
b.
x
= B (2 r) = oienc
oienc
= -4.01x10-7 T  B = -4.01x10-7 T kˆ
2 r
An electron travels with a velocity: v = 2x103 ms ˆj at point P. Determine the magnetic
force vector exerted on the electron at point P.
Ans.
c.
Ans.
FB = ev  B = evB ˆi = 1.28x10-22N ˆi
A second wire oriented parallel to wire 1 above, with a current of 1 A (in the same
direction), is then placed 1.0 meter directly below the first wire. What is the magnetic
field vector exactly halfway between the wires?
The 2 B-fields are additive:
B = B1+B2 = d.
Ans.
e.
o
i1+i2 kˆ = -1.20x10-6 T kˆ
2 r
What is the magnetic force acting on a 1 m segment of the 2nd wire?
FB = i2L2  B1 = -i2L2B1ˆj = -4.01x10-7N ˆj
Derive the equation for the magnetic field vector between the wires in the plane of the
wires.
Phy213: General Physics III
Chapter 29 Worksheet
2/15/2016
4
5. Two infinitely long horizontal wires separated by 0.01 m have a current running through them
(in opposite directions).
i1 = 2 A
r = 0.01 m
i2 = 1 A
d = 1.0 m
a. Using Ampere’s Law, determine the magnitude and direction of magnetic field produced by
both wires at a point 1 m below the lowest wire.
Ans.
Since r<<d, apply a circular “Amperean loop” centered halfway between the wires:
 Bd
B =
= B (2 r) = oienc =o i1-i2 
o i1-i2 
= 2.01x10-7 T  B = -2.01x10-7 T kˆ
2 r
b. Determine the magnitude and direction of the magnetic field produced by both wires at a
point halfway between the two wires in (b).
Ans.
Apply a circular “Amperean loop” around each wire separately:
B1 (2 r) = oi1  B1 = 8.02x10-5 T
B2 (2 r) = oi2  B2 = 4.01x10-5 T


Bnet = B1+B2 = - B1  B2 kˆ = -1.203x10-4 T kˆ
c. What is the magnetic force vector exerted by wire 1 on wire 2?
Ans.
F
i 
N ˆ
FB = i2L 2  B1 = -i2L 2  o 1  ˆj  B = -6.37x10-6 m
j
2

r
L


2
d. If the separation between the wires were 1 m, determine the magnitude and direction of
magnetic field produced by both wires at a point 1 m below the lowest wire.
Ans.
Apply a circular “Amperean loop” around each wire separately:
oi1
4 d
i
= o2
2 d
B1 (4 d) = oi1  B1 =
B2 (2 d) = oi2  B2


i 
i
Bnet = B1+B2 = B1 - B2 kˆ =  o 1 - o 2  kˆ = 0kˆ
 4 d 2  d 
Phy213: General Physics III
Chapter 29 Worksheet
2/15/2016
5
e. The current in wire 2 is increased to 2A. What is the magnetic force vector exerted by
wire 2 on wire 1?
F
 i 
N ˆ
FB = i1L1  B2 = i1L1  o 2  ˆj  B = 1.27x10-5 m
j
L1
 2 r 
Ans.
6. Consider a current carrying coil (i=0.5 A, counter-clockwise as viewed looking down), where
the number of coils is 20 and the radius is 5.0 cm.
a. Calculate the magnitude of the magnetic moment
for the coil.
Ans.
z
o = NiA ˆj = 7.85x10-2 A  m2 ˆj
i
b. Derive the magnetic field B along the central axis
(i.e. in the z direction) of the coil.
Ans.
R
oi ds  r
oi r2d
oi r 2d  R  ˆ
ˆ
dB=
=
sin k =
  k
4 r 3
4 r 3
4 r 3  r 
B= dB=-
2
NoiR

4 z +R
2
2
d

NoiR
kˆ =

2 z2+R 2
0

y
kˆ
x
c. What is the magnetic field at z=20 cm above the coil?
Ans.
B=
NoiR

2
2 z +R
2

kˆ = 7.87x10-7 T kˆ
d. What is the magnetic field vector at z=40 cm above the coil?
Ans.
B=
NoiR

2
2 z +R
2

kˆ = 1.96x10-7 T kˆ
e. What is the magnetic field vector at z=80 cm above the coil?
Ans.
B=
NoiR

2
2 z +R
2

kˆ = 4.90x10-8 T kˆ
f. At what z distance does the magnetic field become effectively independent of the radius,
i.e. the magnetic field can be calculated to within 3% when the radius is ignored?
Ans.
BR=0
BR 0
z2 
NoiR
2z2
=
NoiR
2 z2+R 2

=


1
z2
1
2
z +R 2
=

z2+R 2
 0.97
z2
0.97 2
0.97
R  z2 
R = 5.9R  or 28.4 cm
1-0.97
1-0.97