Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
http://pami.uwaterloo.ca/~akrem/
University of Waterloo,
Electrical and Computer Engineering Dep.
Magnetic Fields Due to Currents
Objectives:
1.
2.
3.
4.
To study Biot-Savart law.
To study Ampere’s law.
To learn how to compute the magnetic field for a current-carrying wire.
To learn how to compute the force between two parallel current-carrying wires.
Biot–Savart Law
Biot and Savart found that the magnitude of the magnetic field contribution (dB) is directly
proportional to the amount of current (i) and the length of the small segment of wire (ds).They
also found that the magnitude of the magnetic field at a point (P) decreases as the inverse
square of the distance between the segment of the wire and point P.
dB
ds
θ
p
r
i
|| 4 x r
4
Example:#1
In the following figure, a straight wire of length L carrying current i.
Circuits (MTE 120) (Spring 2010)
at a distance R from the segment along the perpendicular
(a) Use Biot-Savart law to find !
bisector.
if the wire is very long (infinite length).
(b) Find !
Solution:
For a small segment (dy) of the wire:
" … … … 1
4 It is straightforward to show that:
2
Circuits (MTE 120) (Spring 2010)
%" & ' (
sin180 Substitute r and in Equation (1):
"
'
%" & ' (
4 %" & ' (
'
%" & ' (
' "
4 " & ' /
Due to the symmetry:
20
2 0
1/
1/
' "
' 1/
"
0
… . . 2
2
4 " & ' /
" & ' /
Use the following trigonometric substitution:
" ' tan5
Then
" '
1
cos
u
5
In Eq. 2
1
'
0
2
9:
9:
'
1
5
cos u
' tan 5 & ' 1
' 9:1/ ' cos u 5
0
1
2
9:
' /
cos u
1
9:
;sin5<=:L/
0 cos5 5 =:
2 ' 9:
2 '
3
Circuits (MTE 120) (Spring 2010)
Since y ' tan5 sinu Then,
=
(
%=( ?R(
L/
y
;sin 5<=:L/
A(
D
=:
2 '
2 ' %y & R
2'
2 '
/2
E & ' 2
√ & 4 ' 2 ' √ & 4 ' GH IJH KH '
b) For a wire of infinite length (long wire: L>>R)
2'
L H" J
M H
Using the right-hand rule the direction of B is into the page
4
Circuits (MTE 120) (Spring 2010)
Example #2:
at a point P (R,b)
Find the magnetic field Solution:
For a small segment (dy) of the wire:
" … … … 3
4 5
Circuits (MTE 120) (Spring 2010)
It is straightforward to show that:
%O " & ' (
'
%O " & ' (
Then Equation (3) becomes:
"
'
%O " & ' (
4 %O " & ' (
' "
4 O " & ' /
There is NO symmetry:
1
0 0
1
' "
' 1
"
0
/
/
4
4 O " & ' O " & ' Substitute u= O ", then dy=-du
6
Circuits (MTE 120) (Spring 2010)
if y=L u=b-L
if y=0 u=b
' PQ1
5
0
4
5 & ' /
P
' P
5
0
4 PQ1 5 & ' /
P
'
5
R (
S
4 ' √5 & ' PQ1
'
O
O
A
D
(
(
4 ' √O & ' %O & ' O
O
A(
(
D
4 ' √O & ' %O & ' Remark:
If b=L/2, then we get the same result of the previous example. (Do it by yourself)
Example #3
) at C due to current-length element in a circular arc; as show in the
Find the magnetic field figure below:
Solution:
7
Circuits (MTE 120) (Spring 2010)
x r
4
We notice that is always perpendicular to r 90 sin(90 1.
Then,
4 Use the following identity to change the integration from to U
' U
Then,
' U
4
0
' U ' U
0 U
4
4 ' U
4
As we can see that r is always equal to R then:
U
4 '
Note: U: WXY HMHH
8
Circuits (MTE 120) (Spring 2010)
Example #4
at P.
For the following figure, find Solution:
The wire can be divided into three segments:
B1=0, because the angle θ between ds and r is equal to 0; (sin(0)=0)
B3=0, because the angle θ between ds and r is equal to 180; (sin(180)=0)
B2 can be obtained using the previous example:
U /2 4 '
4
'
8'
Z[\ ] & & 8'
The direction is into the page (use the right-hand rule).
9
Circuits (MTE 120) (Spring 2010)
Example #5
at P.
Find the magnetic filed Solution
The wire can be divided into four segments:
B1=0, because the angle θ between ds and r is equal to 0; (sin(0)=0)
B2=0, because the angle θ between ds and r is equal to 180; (sin(0)=0)
^ U 4 ']
4 ']
4 ']
_ U 4 '
4 '
4 '
`ab 1 1 4
'] '
If B is into the page (+)
If B is out of the page (-)
10
Circuits (MTE 120) (Spring 2010)
Force Between Two Parallel Currents
The magnetic filed due to ia:
| |
cd e
2
The expression of the force on length of current-carrying wire tells us that the force on wire b
is:
cd
x f cd P |
|
f cd |P | cd 90 e P
2
eP . Applying the right-hand
x The direction of f eP is in the direction of the cross product f eP points directly toward wire a as shown.
rule, we find that 11
Circuits (MTE 120) (Spring 2010)
Ampere’s Law
If the distribution o f the magnetic field due to current-carry wire has symmetry, we can apply
Ampere’s law to find the magnetic filed.
Ampere’s Law: The line integral of B around any closed loop, called Amperian loop, equals times the
current encircled by that loop:
g [Zhijhk[l
. is to be integrated around an
. The circle on the integral sign means that the dot product imaginary closed loop, called Amperian loop.
Curled-straight right-hand rule for Ampere’s law:
Curl your right hand around the Amperian loop, with fingers pointing in the direction of the
integration. A current through the loop in the direction of your outstretched thump is assigned
a pulse (+) sign and a current in the opposite direction is assigned a minus (-) sign.
For the above equation:
[Zhijhk[l ] Example #6:
Use Ampere’s law to find the magnetic field B at a distance r from an infinite straight wire
carrying a current i.
Solution:
12
Circuits (MTE 120) (Spring 2010)
g . g K0 g B has the same magnitude at every point along the loop then:
g m is equal to the circumference of the Amperian loop:
g 2 2
This is the same formula derived (with considerably more effort) using Biot-Savart law.
Example #7:
Use Ampere’s law to find the magnetic field B inside an infinite-straight wire carrying a current i
(the current is uniformly distributed over the cross section) and at a distance x from the center
of the wire. The radius of the cross section of the wire is a. (x<a)
Solution
g . [Zh
13
Circuits (MTE 120) (Spring 2010)
g . g 2 n [Zh
[Zh ?
Since the current is uniformly distributed, we can use the current density (J) to find the current
encircled by the Amperian loop:
p
[Zh
n
p n 2 n [Zh n
n q 2 From the previous example (example#6) if xr then:
n r 2 n
14
n