The Biot Savart Law.ppt

B.Sc. Sem-II

1. ShriGovindrao Munghate Arts and Science
College , Kurkheda
Department of Physics
Topic –The Biot-SavartLaw
Prof. B.V. Tupte
Head, Department of Physics
Shri Govindrao Munghate Arts and Science
College Kurkheda.

2. •Biot and Savart recognized that a conductor
carrying a steady current produces a force on a
magnet.
•Biot and Savart produced an equation that gives
the magnetic field at some point in space in
terms of the current that produces the field.
•Biot-Savart law says that if a wire carries a steady
current I, the magnetic field dB at some point P
associated with an element of conductor length
ds has the following properties:
–The vector dB is perpendicular to both ds (the
direction of the current I) and to the unit
vector r
hatdirected from the element ds to the
point P.

4. –The magnitude of dB is inversely
proportional to r
2
, where r is the
distance from the element ds to the
point P.
–The magnitude of dB is proportional
to the current I and to the length ds
of the element.
–The magnitude of dB is proportional
to sin q, where qis the angle
between the vectors ds and r
hat.
•Biot-Savart law:  
2
o
rπ4
rˆxdsIμ
dB




5. •m
ois a constant called the permeability of
free space; m
o=4·x 10
-7
Wb/A·m (T·m/A)
•Biot-Savart law gives the magnetic field at a
point for only a small element of the
conductor ds.
•To determine the total magnetic field B at
some point due to a conductor of specified
size, we must add up every contribution
from all elements ds that make up the
conductor (integrate)! 





2
o
r
rˆxds
π4
Iμ
dB

6. •The direction of the magnetic field
due to a current carrying element
is perpendicular to both the
current element ds and the radius
vector r
hat.
•The right hand rule can be used to
determine the direction of the
magnetic field around the current
carrying conductor:
–Thumb of the right hand in the
direction of the current.
–Fingers of the right hand curl around
the wire in the direction of the
magnetic field at that point.

7. Magnetic Field of a Thin Straight
Conductor
•Consider a thin, straight wire
carrying a constant current I
along the x axis. To determine
the total magnetic field B at the
point P at a distance a from the
wire:

8. •Use the right hand rule to determine that
the direction of the magnetic field
produced by the conductor at point P is
directed out of the page.
•This is also verified using the vector cross
product (ds x r
hat): fingers of right hand in
direction of ds; point palm in direction of
r
hat(curl fingers from ds to r
hat); thumb
points in direction of magnetic field B.
•The cross product (ds x r
hat) = ds·r
hat·sin q;
r
hatis a unit vector and the magnitude of a
unit vector = 1.
•(ds x r
hat) = ds·r
hat·sin q  ds·sin q

9. •Each length of the conductor ds is also a small
length along the x axis, dx.
•Each element of length ds is a distance r from P
and a distance x from the midpoint of the
conductor O. The angle qwill also change as r and
x change.
•The values for r, x, and qwill change for each
different element of length ds.
•Let ds = dx, then ds·sin q becomes dx·sin q.
•The contribution to the total magnetic field at
point P from each element of the conductor ds is:2
o
r
θsindx
π4
Iμ
dB






10. •The total magnetic field B at point P can be
determined by integrating from one end of
the conductor to the other end of the
conductor.
•The distance a from the midpoint of the
conductor O to the point P remains
constant.
•Express r in terms of a and x.
•Express sin qin terms of a and r. 
2
1
22
xa
a
r
a
θsin

  
2
1
22222
xarxar 

11. •For an infinitely long wire:
•From the table of integrals:  
   











































2
3
22
o
2
3
22
o
2
1
22
22
o
2
o
2
o
xa
dx
π4
aIμ
xa
dxa
π4
Iμ
B
xa
a
xa
dx
π4
Iμ
B
r
θsindx
π4
Iμ
r
θsindx
π4
Iμ
dB    




2
1
2222
3
22
xaa
x
xa
dx

12.    
   


aπ2
Iμ
B
aπ4
Iμ2
11
aπ4
Iμ
aπ4
Iμ
B
aπ4
Iμ
B
aa
aπ4
Iμ
B
xa
x
aπ4
aIμ
xaa
x
π4
aIμ
B
o
ooo
2
1
22
1
2
o
2
1
222
1
22
o
2
1
22
2
o
2
1
222
o























































































13. •For a conductor with a finite length:
•From the table of integrals:  
   


































x
x
x
x
x
x
x
x
x
x
x
x
2
3
22
o
2
3
22
o
2
1
22
22
o
2
o
2
o
xa
dx
π4
aIμ
xa
dxa
π4
Iμ
B
xa
a
xa
dx
π4
Iμ
B
r
θsindx
π4
Iμ
r
θsindx
π4
Iμ
dB    




2
1
2222
3
22
xaa
x
xa
dx

14.    
   
  
 
 
2
1
22
o
2
1
22
o
2
1
222
1
22
o
2
1
222
1
22
o
x
2
1
22
2
o
x
2
1
222
o
xaaπ2
xIμ
B
xa
x2
aπ4
Iμ
B
xa
x
xa
x
aπ4
Iμ
B
xa
x
xa
x
aπ4
Iμ
B
xa
x
aπ4
aIμ
xaa
x
π4
aIμ
B
xx












































































15. •When the angles are provided:
–express r in terms of a and the angle q:
–Because angles are involved, we need to
change dx to dq:
-Take the derivative of x:θcsca
θsin
a
r
r
a
θsin  θcota
θtan
a
x
x
a
θtan 



 dθθcscadx
dθθcscadx
2
2



16. •To determine the magnitude of the magnetic field
B, integrate: 
   
12
oθ
θ
o
θ
θ
o
θ
θ 22
2
o
θ
θ 2
2
o
2
θ
θ
o
θ
θ
θcosθcos-
aπ4
Iμ
θcos-
aπ4
Iμ
B
dθθsin
aπ4
Iμ
B
θcsca
dθθsinθcsca
π4
Iμ
B
θcsca
θsindθθcsca
π4
Iμ
B
r
θsindx
π4
Iμ
dB
2
1
2
1
2
1
2
1
2
1
2
1

































17. •The magnetic field lines are
concentric circles that
surround the wire in a plane
perpendicular to the wire.
•The magnitude of B is constant
on any circle of radius a.
•The magnitude of the
magnetic field B is
proportional to the current
and decreases as the distance
from the wire increases.

18. Magnetic Field of a Current Loop
•To determine the magnetic field B at the
point O for the current loop shown:

19. •The magnetic field at point O due to the
straight segments AA'and CC'is zero
because ds is parallel to r
hatalong path AA'
and ds is antiparallel to r
hatalong path CC'.
•For the curved portion of the conductor
from A to C, divide this into small elements
of length ds.
•Each element of length ds is the same
distance R away from point O.0180sin0;0sin
θsinrˆdsrˆxds



20. •Each element of length ds contributes
equally to the total magnetic field B at
point O.
•The direction of the magnetic field B at
point O is down into the page.
•At every point from A to C, ds is
perpendicular to r
hat, therefore:
•Integrate from A to C:ds90sin1dsθsinrˆdsrˆxds   








C
A
2
o
2
C
A
o
C
A R
ds
π4
Iμ
r
rˆxds
π4
Iμ
dB

21. •Pull the constant R out in front of the
integral and integrate from A to C:
•The distance s is the arc length from A to C;
arc length s = R·q. Revising the equation:2
o
C
A
2
o
Rπ4
sIμ
ds
Rπ4
Iμ
B





  Rπ4
θIμ
Rπ4
θRIμ
Rπ4
sIμ
B
o
2
o
2
o










22. Magnetic Field on the Axis of a Circular
Current Loop
•Consider a circular loop of wire of radius R in the
yz plane and carrying a steady current I:

23. •To determine the magnetic field B at a
point P on the axis a distance x from the
center of the loop:
–Divide the current loop into small elements
of length ds.
–Each element of length ds is the same
distance r to point P on the x axis.
–Each element of length ds contributes
equally to the total magnetic field B at point
P. 
2
o
r
rˆxds
π4
Iμ
dB 




24. •Express r in terms of R and x:
•Each element of length ds is perpendicular
to the unit vector r
hatfrom ds to point P.
•Substituting into the integral equation:222
xRr  ds90sin1dsθsinrˆdsrˆxds  





22
o
xR
ds
π4
Iμ
dB

25. •Notice that the direction of the magnetic
field contribution dB from element of length
ds is at an angle qwith the x axis.

26. •At point P, the magnetic field contribution
from each element of length ds can be
resolved into an x component (dB
x) and a y
component (dB
y).
•The dB
ycomponent for the magnetic field
from an element of length ds on one side
of the ring is equal in magnitude but
opposite in direction to the dB
y
component for the magnetic field
produced by the element of length ds on
the opposite side of the ring (180º away).
These dB
ycomponents cancel each other.

27. •The net magnetic field B at point P is the
sum of the dB
xcomponents for the
elements of length ds.
•The direction of the net magnetic field is
along the x axis and directed away from
the circular loop.θcos
xR
ds
π4
Iμ
dB
θcos
xR
ds
π4
Iμ
dB
θcosdBdB
22
o
22
o
x

















28. •Express R
2
+ x
2
in terms of an angle q:
•Substituting into the integral equation: 
2
1
22
xR
R
r
R
θcos

  
2
1
22
22
o
22
o
xR
R
xR
ds
π4
Iμ
dB
θcos
xR
ds
π4
Iμ
dB
















29. •Pull the constants out in front of the
integral:
•The sum of the elements of length ds
around the closed current loop is the
circumference of the current loop; s = 2··R 
 












ds
xRπ4
RIμ
B
xR
Rds
π4
Iμ
dB
2
3
22
o
2
3
22
o

30. •The net magnetic field B at point P is given
by: 
 
 
2
3
22
2
o
2
3
22
2
o
2
3
22
o
xR2
RIμ
B
xRπ4
RIμπ2
B
Rπ2
xRπ4
RIμ
B











31. •To determine the magnetic field strength B
at the center of the current loop, set x = 0:  

R2
Iμ
B
R2
RIμ
R2
RIμ
R2
RIμ
B
R2
RIμ
xR2
RIμ
B
o
3
2
o
6
2
o
3
2
2
o
2
3
2
2
o
2
3
22
2
o



















32. •For large distances along the x axis from
the current loop, where x is very large in
comparison to R:  

3
2
o
6
2
o
3
2
2
o
2
3
2
2
o
2
3
22
2
o
x2
RIμ
B
x2
RIμ
x2
RIμ
B
x2
RIμ
xR2
RIμ
B
















33. •The magnetic dipole mof the loop is the
product of the current I and the area A of
the loop: m= I··R
2 
 
2
3
22
o
2
3
22
2
o
2
xRπ2
μμ
B
xR2
RIμ
B
π
μ
RI






