Transcript Unit 4 Day 9 – The Biot

Unit 4 Day 9 – The Biot-Savant Law
• Restrictions of Ampere’s Law
• The Biot-Savant Law
• Differences Between Ampere’s Law &
the Biot-Savant Law
• The Magnetic Field Due to a Current in a Straight Wire
• Magnetic Field On-Axis of a Current Loop
• Magnetic Field of a Wire Segment
Restrictions of Ampere’s Law
• Ampere’s Law is restricted to situations where
the symmetry of the given currents allows us to
easily evaluate the integral
 B d l
• Jean Baptiste Biot & Felix Savant overcame this
limitation in 1820 by considering the current
flowing in any path as many infinitesimal current
elements
Biot-Savant Law
• The current flowing in any path, can be considered as
many infinitesimal current elements, each of length dl,
flowing in a magnetic field dB, at any point P
0 I d l  rˆ
dB 
4 r 2
Biot-Savant Law
• The magnitude of dB is:
0 I dl sin 
dB 
4
r2
where θ is the angle between dl
and r
• The total magnetic field at point P is:
0 I d l  rˆ
B  dB 
2  r 2
• This is equivalent to Coulomb’s Law written in differential
form:
1 dq
dE 
40 r 2
Differences between Ampere’s law &
the Biot-Savant Law
• The difference between Ampere’s Law and the
Biot-Savant Law is that in Ampere’s Law
(
 dl magnetic
  0 I encl
), Bthe
field is not necessarily due only to
the current enclosed by the path of integration, as
Ampere suggests
• In the Biot-Savant Law, dB is due entirely to the
current element I·dl. To find the total B, it is
necessary to include all currents
Magnetic Field Due to Current in a
Straight Wire
• To find the magnetic field near
an infinitely long, straight wire,
carrying a current I, the BiotSavant Law gives us:
0 I
B
4
y  
dy sin 
2

r
y  
• The solution of this integral yields:
0 I
B
2R
• This is the same as Ampere’s Law
The Magnetic Field On-Axis of a
Current Loop
• To find the magnetic field On-Axis, of a
Current Loop, applying the Biot-Savant
Law yields:
0 I dl sin 

dB 
,


90
because dl  rˆ
2
4R
 I dl
dB  0 2
4R
0 I
0 I
2R 
B
dl 
2 
2
4R
4R
B
0 I
2R
Magnetic Field of a Loop at a Point
On-Axis
• By is = 0 from the symmetry of
the problem
• Bz is calculated using the BiotSavant Law:
0 2R 2 I
Bz 
4 z 2  R 2 3 2
• For z>>R then the above relationship simplifies to:
0 2R I
Bz 
4 z 3
2
r̂
Magnetic Field of a Wire Segment
• Again:   90 because dl  rˆ
the Biot-Savant Law gives:
 0 I dl
dB 
4R 2
r̂
• Solving the Integral yields:
B
1
8
0 I
R
dB