1. Lecture 10
Biot-Savart’s Law

2. Lecture Objectives In this lecture you will learn the following:
Study Biot-Savart's law
Calculate magnetic field of induction due to some simple current configurations
Find force between two current carrying conductors.

3. !
that gives the magnetic field at some point in
space due to a current
Biot and Savart conducted experiments on
They arrived at a mathematical expression
nearby magnet
the force exerted by an electric current on a
Introduction
Biot and Savart conducted experiments on the force exerted by an electric current on a nearby magnet!
They arrived at a mathematical expression that gives the magnetic field at some point in space due to a current

4. Statement of Biot-Savart’s Law
Consider a small element dl of a conductor through which current I flows. Let P be any point at distance r from the centre of the element.
Let the line joining the centre of the elemental section to point P make an angle  with dl. From experiments, Biot and Savart found that the intensity dB of the magnetic field at P due to current I in the elemental section is proportional to I, dl, sin : dl
Idl dB P 
r = r r I 
(into page)
where
μo = 410-7 T.m/A
This equation is the called the Biot-Savart’s Law.
B-SL is an inverse square law !
Notice that if dl and ř are parallel, the contribution is zero!

5. Magnetic Field Due To A Straight Conductor of Finite Length
Consider the following setup where a current I is flowing through a straight wire aligned along the x axis. x I y r a P 1 O
dxi
dl = (dx) i L z 2

6. Assume constant current I is constant and flowing in the positive-x direction. Then
Using basic trigonometry: ;
Therefore:

7. By trigonometry: ;
Therefore:
And so:
Therefore, total magnetic field density at point P,

8. Therefore, total magnetic field density at point P,
So, magnetic field at point P has magnitude
and has direction pointing into the page (-ve z direction).

9. Worked Example 1
A current 10 ampere flows through the wire having configuration as shown in the figure. Determine magnetic field at P.

10. Solution
We shall determine magnetic field due to different straight segments of wires. Let us consider the out of plane orientation as positive. Now, for wire segment AC, the point P is at the end of straight wire of length 4 m and is at a perpendicular linear distance of 4 m. The magnetic field at P due to segment AC is perpendicular and out of the plane of drawing. The magnetic field due to segment AC is pointing out of the plane at point P and is given by A C P
4 m
32 m
10 A
where
, giving
, giving
a = 4 m

11. Therefore, P D C
10 A
For the wire segment CD, the point P lies on the extended line passing through the wire. The magnetic field due to this segment, therefore, is zero.
For the wire segment DE, the angles between the line segment and line joining the point P with end points are known by geometry of the figure. The magnetic field due to segment DE is pointing into the plane at point P and is given by

12. where E P
a = m
10 A D
giving

13. Therefore,
For the wire segment EF, the point P lies on the extended line passing through the wire. The magnetic field due to this segment, therefore, is zero. F E P
10 A

14. For the wire segment FA, the point P is at the end of straight wire of length 4 m and is at a perpendicular linear distance of 4 m. The magnetic field at P due to segment FA is perpendicular and out of the plane of drawing. The magnetic field is: F A P
4 m
32 m
10 A
where
, giving
, giving
a = 4 m

15. Therefore,
The net magnetic field at P is :
The net magnetic field is into the plane of drawing.

16. Worked Example
A regular hexagon whose centre is a distance a = 1 cm from the nearest side has current I = 4.00 A flowing around it. The current flows N = 500 times around. What is the total magnetic field at the centre? I a
60

17. Solution
Draw in the two directions from the centre to the corners of one segment
Top angle is one-sixth of a circle, or 60 degrees
Total angles in circle is 180, so other two angles are 60 each
Use formula to get magnetic field – right hand rule says up.
Multiply by all six side, and then by 500 cycles

18. Magnetic Field Due To A Straight Conductor of Infinite Lengthy
For an infinitely long conductor,
dl = (dx) i y P
θ1 = 0° and θ2 = 180° x
giving r z a
cosθ1 = 1 and cosθ2 = -1 2 1 r
dxi x
Therefore, O x I L

19. Worked Example 2 Solution
Calculate the magnetic flux density at a distance of 5 cm from a long straight circular conductor carrying a current of 250 A and placed in air. Draw a curve showing the variation of B from the conductor surface outwards if its diameter is 2 mm.
5 cm B
250 A
Solution
Now, at the conductor surface, r = 1 mm = 10-3 m. Therefore,

20. The variation of B outside the conductor is shown in the figure below.

21. Worked Example 3 Solution
A long, straight wires carries a current of 5.00 A. At one instant, a proton, 4 mm from the wire travels at 1500 m/s parallel to the wire and in the same direction as the current. Find the magnitude and direction of the magnetic force acting on the proton due to the field caused by the current carrying wire. 5A
4mm + v
X X X
B = +z
v = +y
F = -x
Solution

22. Worked Example 4
Calculate magnetic field at point P due to current 10 A flowing through a long wire bent at right angle as shown in the figure. The point P lies at a linear distance 1 m from the corner.
1 m P
10 A x y

23. Solution
The point P lies on the extension of wire segment in x-direction. Here angle between current element and displacement vectors is zero i.e. θ = 0 and sin θ = 0. As such this segment does not produce any magnetic field at point P. On the other hand, the point P lies near one of the end of the segment of wire in y-direction. The wire being long, the magnetic field due to wire segment in y-direction is:
Applying Right hand thumb rule, the magnetic field at P is perpendicular to xy plane and into the plane of drawing (i.e. negative z-direction).

24. Worked Example 5
Find resultant magnetic field at point O, produced by I1, I2 and I3.
I1 = 6 A
I3 = 8 A
I2 = 4 A
10 cm O

25. Solution B1 B2 B3 O

26. Solution
where B1 B2 B3 O
Bresultant
116.56°
Therefore,

27. The Magnetic Force Between Parallel Conductors
Parallel wires carrying currents will exert forces on each other.  Each wire produces a magnetic field, which influences the other wire.  When the currents in both wires flow in the same direction, then the force is attractive.  When the currents flow in opposite directions, then the force is repulsive.
In the diagram shown below, both wires carry a current in the same direction.  Field due to the first wire at the position of the second wire is given by
where is the unit vector out of the page.

28. The force experienced by the second wire in this field is
where
Therefore,

29. Thus the force between the wires carrying current in the same direction is attractive and is
per unit length. For a wire of length l,
newtons
Notes
Parallel conductors carrying currents in the same direction
attract one another.
Parallel conductors carrying currents in the opposite
direction repel one another

30. Worked Example 6 Solution
Calculate the force of repulsion between two busbars 8 feet long under short-circuit conditions when the current in each is 3600 amperes dc; the separation distance is 8 inches.
Solution
In air, we can take µr equal to unity. This gives, under the conditions specified,
N/m

32. Worked Example 7
Making some simplifying assumptions, estimate the magnitude of the magnetic forces between two thin wires carrying a current of 200 A each and separated by 20 mm.
Solution
Each of the two parallel current-carrying wires produces a magnetic field according to
where R is the radial distance between wires.

33. Therefore, the magnetic force experienced by the second wire will be
or the force per unit length will be
Therefore, in our case, the force per unit length will be

34. Worked Example 8
Four long straight parallel wires pass through the x-y plane at a distance of 4.0 cm from the origin. Figure below shows the location of each wire, and its current. Find the net magnetic field at the origin because of these wires. x y
I3 = 10 A
I2 = 20 A
I4 = 20 A
I1 = 30 A

35. Solution
Let’s apply the principle of superposition to find the net magnetic field at the origin. The four fields, one from each wire, are shown in the figure below. The two fields that are along the y-axis, from the two wires on the x-axis, cancel one another. The two fields along the x-axis, from the two wires on the y-axis, add together.
Thus, the net magnetic field at the origin is directed along the positive x-axis, and has a magnitude of: x y
I3 = 10 A
I2 = 20 A
I4 = 20 A
I1 = 30 A B4 B2 B1 B3

36. Field Due To A Circular Coil On Its Axis
Consider the current loop in the figure shown to be in the x-y plane, which is taken perpendicular to the plane of the paper in which the axis to the loop (z-axis) lies. Since all length elements on the circumference of the ring are perpendicular to r, the magnitude of the field at a point P is given by
The direction of the field due to every element is in the plane of the paper and perpendicular to r, as shown.

37. Corresponding to every element dl on the circumference of the circle, there is a diametrically opposite element which gives a magnetic field dB in a direction such that the component of dB perpendicular to the axis cancel out in pairs. The resultant field is parallel to the axis, its direction being along the positive z-axis for the current direction shown in the figure.
The net field is

38. In terms of the distance z of the point P and the radius a, we have
The direction of the magnetic field is determined by the following Right Hand Rule.

39. Worked Example 9
A circular loop of radius 5.0 cm has 12 turns and lies in the x-y plane. It carries a current of 4 A in the direction such that the magnetic moment of the loop is along the z axis. Find the magnetic field on the z axis at (a) z = 0, (b) z = 15 cm, and (c) z = 3 m.
r = 5.0 cm z x y
N = 12 turns
15 cm
3 m

40. Solution
(a) At z = 0,
(b) At z = 15 cm,

41. (c) At z = 3 m,

42. END