1. ELEC 3105 Basic EM and Power Engineering
Biot-Savard Law
Applications of Biot-Savard Law

2. Biot-Savard Law
Usually we deal with current localized to wires rather than spread over space.
When the current is contained by thin wires, “we can use the magnetic vector potential to derive an integral from which the magnetic field can be found directly from the wire location and currents”. I P
Origin

3. Biot-Savard Law KISS Rule
We will find the magnetic field at some point P located at vector from the origin due to a short segment of wire located at from the origin, carrying current I, where the current is in the direction of the vector , as shown below.
find here P I
Origin
small segment of wire at I

4. Biot-Savard Law This expression is The Biot-Savard Law
Consider a small segment of wire of overall length P
Same result as postulate 2 for the magnetic field I

5. Biot-Savard Law This expression is The Biot-Savard Law
Consider a small segment of wire of overall length
The magnetic field produced by all the short current elements along the line is the Biot-Savard law. P
An integral is another way of saying “principle of superposition” applies to magnetic fields. I

6. Biot-Savard Law & Current Density J (A/m2)
find here P I
Origin 𝑅 I
The current in the differential volume element dxdydz is:

7. Biot-Savard Law & Surface Current Density K (A/m)
find here 𝐾 P I
Origin 𝑅 dA I
The current in the differential surface element dA is:

8. ELEC 3105 Basic EM and Power Engineering
START Applications of Biot-Savard Law
BIOT from RAMA

9. Magnetic field produced by a circular current ring

10. Magnetic field produced by a circular current ring
We can use the Biot-Savard Law
We will first consider the magnetic field present at the center of a current carrying ring. y x

11. Magnetic field produced by a circular current ring
All contributions from each line segment point in the same direction: in the direction of the axis of the ring. y x
I out of page
Side view
Top view
I into page

12. Magnetic field produced by a circular current ring
Side view
Top view
Starting from the Biot-Savard Law for a single segment. x
We get:

13. Magnetic field produced by a circular current ring
Side view
Top view x
Direction:
As shown by arrows in side view

14. Magnetic field produced by a circular current ring
Summation over all segments around the ring y
Top view x
Magnetic field at the center of the circular current ring

15. Magnetic field produced by a circular current ringx y
3-D view z
We will now consider the magnetic field present on the central axis of a current carrying ring.
We can use the Biot-Savard Law

16. Magnetic field produced by a circular current ring
All contributions from each line segment point in different directions: We must consider components of z
3-D view z y x
I out of page
I into page

17. Magnetic field produced by a circular current ringz
Horizontal components will cancel in pairs
around the segments of the current ring
I out of page
I into page
Z components will add around the
segments of the current ring

18. Magnetic field produced by a circular current ring
Horizontal components will cancel in pairs
around the segments of the current ring
Z components will add around the
segments of the current ring

19. Magnetic field produced by a circular current ring
3-D view
I out of page
I into page z
Can use Biot-Savard Law
We need only concern ourselves with the magnitude of since we know the final direction of is along the z direction.

20. Magnetic field produced by a circular current ring
3-D view
I out of page
I into page z
Expression for magnetic field produced by one current segment of the ring.

21. Magnetic field produced by a circular current ring
I out of page
I into page z
3-D view z y x
Summation around ring:
Magnetic field on the central axis of the circular current ring

22. Magnetic field produced by a circular current ring We only considered the central axis field The full field is obtained when we will discuss the magnetic dipole
magnetotherapy

23. Magnetic field of a long FINITE solenoid

24. Magnetic field of a FINITE solenoid
Current out of page
Axis of solenoid P L
Current into page
finite coil of wire carrying a current I
Evaluate B field here
Radius of solenoid is a.
Cross-section cut through solenoid axis

25. Magnetic field of a FINITE solenoid
Segment of the solenoid coil
arc length 1 2 3 4 5
Current out of page
Axis of solenoid

26. Magnetic field of a FINITE solenoid
The solenoid consists on N turns of wire ”rings of current” each carrying a current I.
The number of turns per unit length of solenoid can be expressed as:

27. Magnetic field of a FINITE solenoid
The current in a segment (dI) can be expressed as the current in one ring (I) multiplied by the number of turns per unit length (N/L) and multiplied by the segment length .

28. Return to this slide for a momentz
3-D view y x
Each ring of the solenoid contributes a magnetic field. Can use this expression for our finite solenoid. Sum over the rings.
Magnetic field on the central axis of the circular current ring

29. Magnetic field of a FINITE solenoid
Segment of the solenoid coil
arc length 1 2 3 4 5
Current out of page
Axis of solenoid

30. Magnetic field of a FINITE solenoid
sub in 1 2 3 4 5

31. Magnetic field of a FINITE solenoidz L
We can now sum (integrate) the expression for over the angular extent of the coil. I.e. sum over all the rings of the finite length solenoid.

32. Magnetic field of a INFINITE solenoidz L
INFINITE SOLENOID RESULT
infinite solenoid result

33. Magnetic field at one end of a FINITE solenoidz L
Magnetic field is ½ that of center

35. Magnetic flux Φ 𝑚 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐵 ∙ 𝑑𝐴
Φ 𝑚 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐵 ∙ 𝑑𝐴
Goes by the name of magnetic flux density vector 𝐵

36. Magnetic flux: CLOSED SURFACE
Φ 𝑚 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐵 ∙ 𝑑𝐴 =0
Magnetic field lines loop back on themselves
No magnetic source
No magnetic sink
No magnetic monopole
No magnetic charge

37. Divergence of a Magnetic Field
For a closed surface
Φ 𝑚 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐵 ∙ 𝑑𝐴 =0
No magnetic monopole
No magnetic charge
Divergence theorem
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐵 ∙ 𝑑𝐴 = 𝑉𝑜𝑙𝑢𝑚𝑒 𝛻 ∙ 𝐵 𝑑𝑣𝑜𝑙
𝛻 ∙ 𝐵 =0
𝛻 ∙ 𝐷 = 𝜌 𝑣
𝛻 ∙ 𝐻 =0
Electric charge exist

38. Magnetic flux
Given that the radial magnetic field is 𝐵 = 𝜇 𝑜 2.39× 𝑐𝑜𝑠 𝜙 𝑟 𝑟 , calculate the magnetic flux Φ 𝑚 through the surface defined by − 𝜋 4 ≤𝜙≤+ 𝜋 4 and 0≤𝑧≤1.
We have the expression for the magnetic field AND the limits for the integral Both specified in cylindrical coordinates.
Solution
Φ 𝑚 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐵 ∙ 𝑑𝐴
Remainder of solution presented in class

39. Magnetic flux
In cylindrical coordinates 𝐵 = 2.0 𝑟 𝜙 , calculate the magnetic flux Φ 𝑚 through the surface defined by 0.5≤𝑟≤2.5 and 0≤𝑧≤2.
We have the expression for the magnetic field AND the limits for the integral Both specified in cylindrical coordinates.
Solution
Φ 𝑚 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐵 ∙ 𝑑𝐴
Remainder of solution presented in class