1. Chapter 3 Magnetostatic
ECT1026 Field Theory
Chapter 3 Magnetostatic
Lecture 3-3
Application of Biot-Savart Law
By Dr Mardeni Roslee

2. Application of Biot-Savart Law
ECT1026 Field Theory
Lecture 3-3
Application of Biot-Savart Law
Magnetic Field of a
Pie-Shaped Loop
Circular Loop
Linear Conductor

3. B ? Example 3.2-1: Magnetic Field of a Linear Conductor
ECT1026 Field Theory
Example 3.2-1: Magnetic Field of a Linear Conductor
Using Biot-Savart law to find the magnetic flux density B at a point in the plane that bisects a long straight wire (or point P) of length 2L carrying a direct current I. P L
B ?

4. ECT1026 Field Theory
How to solve ?
Step-1: Cylindrical coordinate system - Imagine the wire lies along the z-axis
Step-2: Let the bisecting plane be the xy-plane z ˆ
(r,,z) I ˆ x y

5. ECT1026 Field Theory
How to solve ?
Step-1: Cylindrical coordinate system - Imagine the wire lies along the z-axis
Step-2: Let the bisecting plane be the xy-plane z ˆ z ˆ P r ^ L
- L
r = r r
Simplified I r ˆ x y r ^  P
(r,,z)

6. ECT1026 Field Theory
How to solve ?
Step-3: Start by considering the B generated by a current element dl= z dz at a distance R from the current element ^ z ˆ P r ^ L
- L
r = r r I R = - z r ˆ
Therefore =
z dz
(r r – z z ) ˆ
dl  R  ^ dz z R
r  z dz
r z ^
=  r dz
dl  R =  r dz ^  
dl  R = ^
 r dz R
R = R R ^

7. I dB = mo dl × R R 4 4p ˆ dB = mo I 4 R rdz dB = rdz mo I 
ECT1026 Field Theory
How to solve ? R 2 mo
dB = 4p
dl × R I ^
Step-4: Biot-Savart Law z ˆ P r ^ L
- L
r = r r I
dl  R = ^
 r dz R
Since dz z R r z R
dB =
mo I 4 R 3
rdz ^
dB =
(z2 + r2)
3/2
rdz
mo I 4  ^
R2 = z2 + r2

8. ∫ ∫ B = dB ˆ moIr  B = 4 mo Ir  B = 4 mo IL  B = How to solve ?
ECT1026 Field Theory
How to solve ? +L
B = ∫ dB
Step-5: Integration -L z ˆ P r ^ L
- L
r = r r I  ^
B =
(z2 + r2)
3/2 dz ∫ -L +L
moIr 4 dz z R  ^
B =
mo Ir 4
r2 (z2 + r2)
1/2 z -L +L  ^
B =
r (L2 + r2)
1/2 1
mo IL 2

9. For an infinitely long wire such that L >> r ˆ ^
B =
r (L2 + r2)
1/2 1
mo IL 2
For an infinitely long wire such that L >> r z ˆ P r ^ L
- L
r = r r I  ^
B =
mo I
2r  ^
B =
(L2 + r2)
1/2 L
mo I
2r
L >> r,  ^
B =
(L2 + 02)
1/2 L
mo I
2r

10. Divide the circular loop into two segments:
ECT1026 Field Theory
Example 3.2-2: Magnetic Field on a Pie-Shaped Loop
A circular loop of radius a carries a steady current I.
Determine the magnetic field B at a the origin O of the loop.
Divide the circular loop into two segments:
(ii) AC
(i) OA & OC
B ?

11. dl is parallel or anti-parallel to R hence dl  R = 0 dl
ECT1026 Field Theory
Step 1: Segment OA and OC
For the straight segments OA and OC,
the magnetic field at point O is zero.
As all points along these segments,
dl is parallel or
anti-parallel to R
hence dl  R = 0 O C A ^ dl ^ R ^
Parallel R ^
Anti-parallel
AB = AB sin 
sin 0 = sin 180 = 0

12. dl =a d A C dl =  a d r = - R a or r = - R dl  r = z a d ^ Thus,
ECT1026 Field Theory
r  z
0 dl 0 ^
= z dl
Step 2 - Segment AC:
Along segment AC, dl is perpendicular to R
 dl  R = - dl  r = z dl = z a d. ^ ^ ^ ^ ^
dl =a d A C ^ d
dl =  a d ^
(opposite direction) ^
r = - R a or r = - R
dl  r = z a d ^ ^
Thus,

13. Circle:  = 2  2r  =    r = (/2)×2r dl  (d/2)×2r = r d
circumference
Circle:  = 2  r
(Full circle)
 =    r = (/2)×2r
(Half circle) r d
dl  (d/2)×2r = r d r
(radius=r)
dl =a d (radius=a)
Thus,

14. ∫ ò ò moI B = 4 C A ^ B Step 2 - Segment AC:
ECT1026 Field Theory
Step 2 - Segment AC:
Along segment AC, dl is perpendicular to R , and
dl  R = z dl = z a d. ^
B =
moI 4 l R2
dl  R ∫ ^ A C ˆ m I ò ad f z
(R=a)
radius = 2 4 p a ^ m I ò ˆ = z df z 4p a B

15. Example 3.2-3: Magnetic Field of a Circular Loop
ECT1026 Field Theory
Example 3.2-3: Magnetic Field of a Circular Loop
A circular loop of radius a carries a steady current I. Determine the magnetic field B at a point on the axis of the loop.
Answer:

16. z Any element dl on the circular loop is
ECT1026 Field Theory
Our task is to obtain an expression for B at point P(0,0,z)
Any element dl on the circular loop is
perpendicular to the distance vector R
At the same distance R from point P, with R = √ a2 + z2 z
The magnitude of dB due to element dl is given by:

17.  dB is in the r-z plane, therefore it has components fo dBr and dBz
ECT1026 Field Theory
Direction of dB is in the plane containing R and dl
dB is in the r-z plane, therefore it has
components fo dBr and dBz
the z-components of B due to dl and dl’ add
because they are in the same direction, but
Their r-components cancel because their are in opposite directions
Consider element dl’ located diametrically
opposite to dl 
The net magnetic field is along z-axis only

18. Hence, the net magnetic field is along z-axis only
ECT1026 Field Theory
Hence, the net magnetic field is along z-axis only
For a fixed point P(0,0,z) on the axis of the loop,
all quantities are constant, except for dl
(Length of circular loop)
circumference

19. Since,  At the center of the loop (z = 0) At point very far away from
ECT1026 Field Theory
Since, 
At the center of the loop (z = 0)
At point very far away from
the loop such that z2 >> a2