Static Magnetic Field(Electromagnetic Fields and Waves)

1. ECE-2205
Electromagnetic Fields & Waves
Md. Ebtidaul Karim

2. Lorentz’s Force Equation
• When a small test charge q is placed in an electric field E, it experiences an electric
force 𝐹𝑒. Here 𝐹𝑒 = 𝑞𝐸
• But magnetic force 𝐹𝑚 follows the below characteristics, Such as:
The magnitude 𝐹𝑚 is proportional to q.
The direction of 𝐹𝑚 at any point is at right angles to the velocity vector of the test
charge as well as to a fixed direction at any point.
The magnitude 𝐹𝑚 is also proportional to the component of the velocity at right
angles to this fixed direction.

3. Lorentz’s Force Equation
• Now, 𝐹𝑚 = 𝑞𝒗 × 𝑩 , here B is the magnetic flux density and v is the velocity
of the charge moving in the field.
• Now, The flux density B is the number of magnetic lines of flux that pass
through a certain point on a surface. The SI unit is T (tesla), which is weber
per square meter (
𝑊𝑏
𝑚2)
• Why we use cross product ??

4. Lorentz’s Force Equation

5. Lorentz’s Force Equation
• So total electromagnetic force on charge q is,
𝐹 = 𝐹𝑒 + 𝐹𝑚
or, 𝐹 = 𝑞𝐸 + 𝑞𝑣 × 𝐵
or, 𝐹 = 𝑞(𝐸 + 𝑣 × 𝐵)
• It is also called fundamental postulate of electromagnetic model.

6. Fundamental Postulates of Magneto statics
• 𝛻. 𝐵 = 0......(1) Why???
• 𝛻 × 𝐵 = 𝜇𝑜𝐽......(2) Here, 𝜇𝑜=4𝜋 × 10−7 is the permeability of free
space.
• Permeability is the measure of the ability of a material to support the
formation of a magnetic field within itself.
• Now we get from (2), 𝛻. 𝐽 = 0

7. Fundamental Postulates of Magneto statics
• Taking the volume integral of (1) and applying divergence theorem we
obtain, 𝑠
𝐵. 𝑑𝑠 = 0 .....(3)
• Comparing (3) with Gauss law we deny the existence of isolated magnetic
charge. We also can say that there are no magnetic flow sources and magnetic
flux lines always close upon themselves.
• It is known as law of conservation of magnetic flux.

8. Fundamental Postulates of Magneto statics
There is no existence of isolated
positive and negative charge in
magnetic field
Fig : Broken Bar Magnet

9. Ampere’s Circuital Law
• Integrating both sides of (2) over an open surface and applying Stokes’s
theorem we obtain, 𝑐
𝐵. 𝑑𝑙 = 𝜇𝑜𝐼.......(4)
• It is a form of Ampere’s Circuital Law.
• It states that circulation of magnetic flux density in free space around any
closed path is equal to 𝜇𝑜 times the total current flowing through the surface
bounded by the path.

10. Ampere’s Circuital Law
• Ampere’s Circuital law is most useful in determining the magnetic field
caused by a current when cylindrical symmetry exists.
• That is, it is relevant when there is a closed path around the current over
which the magnetic field is constant.

11. Postulates
Differential Form Integral Form
𝛻. 𝐵 = 0
𝑆
𝐵. 𝑑𝑠 = 0
𝛻 × 𝐵 = 𝜇0𝐽
𝐶
𝐵. 𝑑𝑙 = 𝜇0𝐼

12. Example 6.1
• An infinitely long, straight conductor with a circular cross section area of
radius b carries a steady current I. Determine the magnetic flux density both
inside and outside the conductor.
 Follow Class Note

13. Vector Magnetic Potential
• 𝐵 = 𝛻 × 𝐴, Here A is the vector magnetic potential.
• Now, 𝛻 × 𝛻 × 𝐴 = 𝛻 𝛻. 𝐴 − 𝛻2
𝐴 [ Bac-cab rule]
• So, 𝛻 𝛻. 𝐴 − 𝛻2𝐴 = 𝜇0𝐽
• As 𝛻. 𝐴 = 0 then 𝛻2𝐴 = −𝜇0𝐽. It is known as poisson’s equation.
• Now we can write, 𝐴 =
𝜇0
4𝜋 𝑣
𝐽
𝑅
𝑑𝑣, It is also known as Coulomb gauge.
• From Stokes’s we can write, 𝜑 = 𝑠
𝛻 × 𝐴 . 𝑑𝑠 = 𝑐
𝐴. 𝑑𝑙

14. Biot-Savart Law
• 𝐴 =
𝜇0
4𝜋 𝑐
𝑑𝑙
𝑅
. Now, 𝐵 = 𝛻 × 𝐴 =
𝜇0
4𝜋 𝑐
𝛻 ×
𝑑𝑙
𝑅
• Now applying vector identity, 𝐵 =
𝜇0
4𝜋 𝑐
[
1
𝑅
𝛻 ×
𝑑𝑙
𝑅
+ (𝛻
1
𝑅
) × 𝑑𝑙]
• Now,
1
𝑅
= −𝑎𝑅
1
𝑅2
• So, 𝐵 =
𝜇0
4𝜋 𝑐
𝑑𝑙×𝑎𝑅
𝑅
, It is known as Biot-Savart Law.
• [Detail derivation in class lecture]

15. Math
• Example 6-4: A direct current I flows in a straight wire of length 2L. Find the magnetic flux
density B at a point located at a distance r from the wire in the bisecting plane: (a) by
determining the vector magnetic potential A first, and (b) by applying Biot-Savart's law.

16. Math
• Example 6-7: Find the magnetic flux density at a distant point of a small
circular loop of radius b that carries current I

17. Magnetic Dipole
• We select the center of the loop to be the origin of spherical coordinates as
shown in the figure.
• Source co-ordinates are primed.

18. Magnetic Dipole

19. Scalar Magnetic Potential

20. Magnetic Field Intensity
• Because the application of an external magnetic field causes both an
alignment of the internal dipole moments and an induced magnetic moment
in a magnetic material. we expect that the resultant magnetic flux density in
the presence of a magnetic material will be different from its value in free
space.

21. Magnetic Material Classification
• Diamagnetism:
 Diamagnetic materials have negative magnetic susceptibility.
Relative permeability slightly less than unity.
When a diamagnetic material is placed in a magnetic field, magnetization vector M
is in the opposite direction of applied field.
Negative susceptibility means that diamagnetic material is trying to expel the applied
field from the material.
When it is placed in a non-uniform magnetic field it experiences a net force towards
smaller field.

22. Magnetic Material Classification
• A substance shows diamagnetism whenever the
constitutes atoms in the material have closed
subshells and shells. Because they do not have any
permanent magnetic moment in the absence of
applied field.
• Covalent crystals and many ionic crystals are
diamagnetic material.
• Magnetic susceptibility of silicon crystal is 𝜒𝑚 =
− 5.2 × 10−6

23. Magnetic Material Classification
• Para magnetism:
• Here each individual atom possesses a
permanent magnetic moment, but due to
thermal agitation there is no average
moment per atom and no magnetization.
• In the presence of an applied field molecular
magnetic moment take various alignment
with the field. Degree of alignment and
hence magnetization increases with the
increase of applied field strength.

24. Magnetic Material Classification
• Magnetization decreases with the increase of temperature, because at higher
temperature molecular collision is more, which destroys the alignments of
molecular magnetic moments.
• Para magnetic materials shows small positive
magnetic susceptibility.
• When it is placed in a non-uniform magnetic field,
the induced magnetization M is along B and there
is net force towards greater field.

25. Magnetic Material Classification
• Ferromagnetism:
• Magnetic susceptibility is typically
positive and very large.
• The relationship between the
magnetization M and applied
magnetic field 𝜇0𝐻 is highly non-
linear.
• At sufficiently high fields, the
magnetization M of the ferromagnet
saturates.

26. Magnetic Material Classification
• Crystal Domain has a magnetic ordering as all the atomic magnetic moments
have been aligned parallel to each other.
• Ferromagnetism occurs below a critical temperature called the Curie
temperature 𝑇𝐶.
• At temperature above 𝑇𝐶, ferromagnetism is lost and the material becomes
paramagnetic.

27. Magnetic Material Classification
• Anti-ferromagnetism:
• They have small but positive susceptibility.
• They can nit possess any magnetization in
the absence of an applied field.
• Anti ferromagnetic materials possess a
magnetic ordering in which the magnetic
moments of the alternating atoms in the
crystal align in opposite direction.

28. Magnetic Material Classification
• As a result there is no net magnetization.
• Anti ferromagnetism occurs below a critical temperature known as Neel
temperature, 𝑇𝑁.
• Above 𝑇𝑁, anti ferromagnetic material becomes paramagnetic.

29. Magnetic Material Classification
• Ferrimagnetism:
• The origin of ferrimagnetism is based on
magnetic ordering.
• Here in the figure all A atom have their spin
aligned in one direction and all B atoms have
their spins aligned in opposite direction.
• As the magnetic moment of A is greater than
that of B atom, there is net magnetization M
in the crystal.

30. Magnetic Material Classification
• Unlike the anti ferromagnetic case, the oppositely directed magnetic
moments have different magnitudes and do not cancel each other.
• The net result is that crystal can possess magnetization even in the absence
of an applied field.
• Example of ferromagnetic material is 𝐹𝑒3𝑂4

31. M Versus H Behavior

32. Magnetic Energy
• In fig 6.22 of book,𝐶1, 𝐶2 are the closed loops with bounding surface 𝑆1 and
𝑆2 respectively.
• In 𝐶1 due to current 𝐼1 a magnetic field 𝐵1 will be created.
• Some of the magnetic flux due to 𝐵1will link with 𝐶2. Here mutual flux
𝜙12 = 𝑠1
𝐵1. 𝑑𝑠2
• Now we can write, 𝜙12 = 𝐿12𝐼1 , where 𝐿12 is the mutual inductance
between loop.

33. Magnetic Energy
• In electromagnetism and electronics, inductance is the property of an
electrical conductor by which a change in electric current through it induces
an electromotive force (voltage) in the conductor.
• It is more accurately called self-inductance.
• The same property causes a current in one conductor to induce an
electromotive force in nearby conductors; this is called mutual inductance.

34. Magnetic Energy
• Consider a single closed loop with a self inductance 𝐿1, in which current is
initially zero. Now if current 𝑖1 increases from zero to 𝐼1 then according to
physics, an EMF will be induced in the loop that will oppose the current
change.
• Let 𝑣1 = 𝐿1
𝑑𝑖1
𝑑𝑡
be the voltage across the inductance.
• The amount of work done to overcome the induced EMF 𝑊1 =
1
2
𝐼1𝜙1

35. Magnetic Energy
• Now we consider two closed loops 𝐶1, 𝐶2 that carries current 𝑖1 and 𝑖2
respectively.
• At first we keep 𝑖2 = 0 and increase 𝑖1 from zero to 𝐼1. Here work required
in 𝐶1 is
1
2
𝐼1𝜙1 and in 𝐶2 is zero.
• Now, if we keep 𝑖1 at 𝐼1 and increase 𝑖2 from zero to 𝐼2, due to mutual
coupling some of the magnetic flux due to 𝑖2 will link with loop 𝐶1, giving
rise to induced EMF. It must be overcome by a voltage 𝑣21 = 𝐿1
𝑑𝑖1
𝑑𝑡
in order
to keep 𝑖1 constant.

36. Magnetic Energy
• Here, 𝑊21 = 𝐿21𝐼1𝐼2
• Again in loop 𝐶2, 𝑊22 =
1
2
𝐼2𝜙2
• So total 𝑊 =
1
2
𝐿1𝐼1
2
+ 𝐿21𝐼1𝐼2 +
1
2
𝐿2𝐼2
2

37. Hall Effect
The production of a potential
difference across an electrical
conductor when a magnetic field
is applied in a direction
perpendicular to that of the flow
of current is known as hall
effect.

38. Hall Effect

39. Hall Effect
• Let magnetic field 𝐵 = 𝑎𝑧𝐵𝑜
• Current moving in y direction is 𝐽 = 𝑎𝑦𝐽𝑜 = 𝑁𝑞𝑢 where N is the number of
charge carriers per unit volume, moving with velocity u and q is the charge
of each carrier.
• If the conductor is n-type semiconductor , the charge carriers are electron
and q is negative.
• Magnetic force tends to move the electrons in the positive x-direction,
creating a transverse electric field.

40. Hall Effect
• In steady state net force on the charge carrier is zero.
• 𝐸ℎ + 𝑢 × 𝐵 = 0 or, 𝐸ℎ = −𝑢 × 𝐵 Here 𝐸ℎ is the hall field.
• Now, 𝐸ℎ = − −𝑎𝑦𝑢0 × 𝑎𝑧𝐵0 = 𝑎𝑥𝑢0𝐵0
• A transverse potential appears across the sides of the material. Here 𝑉ℎ =
0
𝑑
𝐸ℎ𝑑𝑥 = 𝑢0𝐵0𝑑
• Here 𝑉ℎ is the hall voltage.

41. Forces and Torque on Current Carrying
Conductor
• Let us consider an element of conductor 𝑑𝑙 with a cross-sectional area 𝑆. If
there are 𝑁 charge carrier per unit volume moving with velocity 𝑢 in the
direction of 𝑑𝑙 then the magnetic force on the differential element is,
𝑑𝐹𝑚 = −𝑁𝑒𝑆 𝑑𝑙 𝑢 × 𝐵 = −NeS 𝑢 𝑑𝑙 × 𝐵
• Now, 𝑑𝐹𝑚 = 𝐼𝑑𝑙 × 𝐵
• Magnetic force on a closed circuit of contour 𝐶 that carries a current 𝐼 in
magnetic field 𝐵 is then, 𝐹𝑚 = 𝐼 𝑐
𝑑𝑙 × 𝐵

42. Forces and Torque on Current Carrying
Conductor
• We have two circuit carrying currents 𝐼1 and 𝐼2 respectively.
• In the presence of the magnetic field 𝐵21, which was caused by the current
by the current 𝐼2 in 𝐶2, the force 𝐹21 on circuit 𝐶1 can be written as
𝐹21 = 𝐼1 𝐶
𝑑𝑙1 × 𝐵21
• Here 𝐵21 =
𝜇0𝐼2
4𝜋 𝐶1
𝑑𝑙2×𝑎𝑅21
𝑅21
2

43. Forces and Torque on Current Carrying
Conductor
• Putting the value of 𝐵21 in 𝐹21, 𝐹21 =
𝜇0
4𝜋
𝐼1𝐼2 𝐶1 𝐶2
𝑑𝑙1×(𝑑𝑙2×𝑎𝑅21)
𝑅21
2
• It is known as ampere’s law of force which is comparable with Coulomb’s
law of force in chapter 3
• Now the force 𝐹21 on circuit 𝐶2, in the presence of magnetic field set up by
the current 𝐼1 in the circuit 𝐶1 is
𝐹12 =
𝜇0
4𝜋
𝐼1𝐼2
𝐶1 𝐶2
𝑑𝑙2 × (𝑑𝑙1 × 𝑎𝑅12
)
𝑅12
2

44. Forces and Torque on Current Carrying
Conductor
• Since 𝑑𝑙2 × (𝑑𝑙1 × 𝑎𝑅12
) ≠ 𝑑𝑙1 × (𝑑𝑙2 × 𝑎𝑅21
); 𝐹21 ≠ −𝐹12 So it do not
follow Newton’s third law.
• Now using BAC-CAB law,
𝑑𝑙1×(𝑑𝑙2×𝑎𝑅21)
𝑅21
2 =
𝑑𝑙2(𝑑𝑙1.𝑎𝑅21)
𝑅21
2 −
𝑎𝑅21(𝑑𝑙1.𝑑𝑙2)
𝑅21
2
• Double line integral of first term of the right hand side of the above
equation gives:

45. Forces and Torque on Current Carrying
Conductor
• 𝐶1 𝐶2
𝑑𝑙2(𝑑𝑙1.𝑎𝑅21)
𝑅21
2 = 𝐶2
𝑑𝑙2 𝐶1
(𝑑𝑙1.𝑎𝑅21)
𝑅21
2 = 𝐶2
𝑑𝑙2 𝐶1
𝑑𝑙1 (−𝛻1
1
𝑅21
) =
0
• So 𝐹21 =
𝜇0
4𝜋
𝐼1𝐼2 𝐶1 𝐶2
𝑎𝑅21(𝑑𝑙1.𝑑𝑙2)
𝑅21
2
• Now assuming 𝑎𝑅12
= −𝑎𝑅21
we get, 𝐹21 = −𝐹12