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Lecture #1
Topics to be covered:
(i) Magnetostatic Field Intensity and Field Sources
(ii) Biot-Savart’s Law
Reference: Hayt and Buck, page 210-218
Motivation:
To understand the concept of magnetic field intensity and its method of
calculation using Biot-Savart’s Law
Magnetic (Magnetostatic) Field Intensity
Magneto + static
Magnetic field intensity source
Symbol of . Unit of Ampere/meter (A/m).
Field sources: Steady current.
Type of field sources: (i) Filament current
(ii) Surface current
(iii) Volume current
H
Filament Current
I
dl Filamentary current element: dlI
Surface Current
sJ
dSSurface current element: dSJS
Surface current density : unit (A/m)
(A.m)
(A.m)
Volume Current
JVolume current element:
dvJ
dv
Those elements (filament, surface and volume) will create magnetic field.
But, how ???
(A.m)
volume current density : unit (A/m2)
Recall
Previously; in electric field calculation
Charge density; Electric fieldCoulomb’s Law
vSlQ ρρρ ,,,
Current density; JJI S ,,
Similarly, in magnetic field calculation
Magnetic fieldBiot-Savart’s
Law
Biot-Savart’s Law
The law was co-originated by Felix Savart, a professor at College de France and Jean Baptiste Biot, a French physicist in 1820.
I
R
Observations:
rradiuszI
H
ˆˆ
ˆ
→→→ φ
2
1R
H
IH
∝
∝
Biot-Savart’s Law
Hence, Biot-Savart’s Law can be stated as:
)/(4
ˆ
,2 mA
RadlIH
llength
R∫×
=π
(filament current)
)/(4
ˆ
,2 mA
RdSaJH
Ssurface
RS∫×
=π
)/(4
ˆ
,2 mA
RdVaJH
Vvolume
R∫×
=π
(surface current)
(volume current)
Biot-Savart’s Law
I
B
At any point, let B:
dl
R
Ra
)/(4
ˆ
,2 mA
RadlIH
llength
R∫×
=π
Consider a filamentary current;
Biot-Savart’s Law-Filamentary Current
Consider a finite length of filamentary current:
)',,( zr φ
y
x
z
φ
z’
r
Step 3:
Integrate over filament current length to get total H
Rdl
Step 1:Select current element
(0,0,z)
a
b
Step 2:
Identify dH
Hd
I
Biot-Savart’s Law-Filamentary Current
Step 1: Current element dlI
Step 2: Identify dH
24ˆ
RadlIHd R
π×
=zdzdl ˆ=
( )( ) rrzzzR ˆˆ' +−−=Hence,
( )( )2322 '4
ˆ
zzr
IrdzHd−+
=π
φ
Using:
( ) ( )∫+
=+ 2
12222
322 xcc
x
xc
dx
Biot-Savart’s Law-Filamentary Current
Step 3:
( )( ) ( )( )φ
πˆ
'
'
'
'4 2
1222
122 ⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
−+
−−
−+
−=
zar
za
zbr
zbr
IH
Alternatively:
( ) )/(ˆsinsin4 21 mA
rIH φααπ
+=1α
2αFor infinite length; ( )mAr
IH /ˆ2
φπ
=
r
Example 1
Consider AB in figure below as part of an electric circuit. Find H at the origin due to AB.
x
y
1
1 A
B
6 A
0
Example 2A square conducting loop of side 2a lies in the z=0 plane and carries a current I in the counterclockwise direction. Find a magnetic field intensity at center of the loop.
Example 3
An infinitely long conductor is bent into an L shape as shown in figure below. If a direct current of 5 A flows in the conductor, find the magnetic field intensity at (a) (2,2,0), (b) (0,0,2)
x
y
5 A0
5 A
Example 4The y and z axes, respectively, carry filamentary currents 10 A along and 20 A along . Find H at (-3,4,5).
y z−
Example 5
Find the expression for the H field along the axis of the circular current loop (radius of a), carrying a current I.
Next Lecture
Please have a preliminary observation on the following topics:
(i) Ampere’s Circuital Law
Please refer to Hayt and Buck, page 218-225
