Department of  

ECE

Electromagnetic Fields
Subject code:PCC-EE-216-G
Semester: VI
CONTENTS
• Magnetostatics
• Biot Savarts Law
• Amperes Circuit law
• Application of Ampere Circuit Law
• Magnetic Flux density
• Maxwell’s equation- Ampere Circuit law
• Maxwell equation for static feild
• Magnetic Scalar & Vector Potential
• Magnetic Force
• Magnetic Dipole
• Magnetic Boundary Condition
• Magnetic Energy
 UNIT III
Static Magnetic Fields
Analogy between Electric
and magnetic field
 Biot Savarts Law
• Biot-Savart's law states that the magnetic field
intensity dH produced at a point P, by the differential
current clement Idl is proportional to the product Idl
and the sine of the angle a between the element and
the line joining P to the element and is inversely
proportional to the square of the distance K between P
and the element.
 Ampere Circuit
Law
• Ampere's circuit law states that the line integral of the tangential
component of H around a closed path is the same as the net current
Ienclosed by the path.
Application of Ampere circuit
law
Application of Ampere circuit law
Application of Ampere circuit law
Application of Ampere circuit law
MAGNETIC FLUX DENSITY—
MAXWELL'S
EQUATION
Maxwell's Equations for static
EM fields
Magnetic scalar and vector
potentials
 Magnetic Force
The second term in the Lorentz Force Equation represents magnetic force Fm(N) on a
moving charge q(C) is given by

Fm  q u  B

where the velocity of the charge is u (m/sec) within a field of magnetic flux density B
(Wb/m2). The units are confirmed by using the equivalences Wb=(V)(sec) and J=(N)
(m)=(C)(V).

The magnetic force is at right angles to the magnetic field.

The magnetic force requires that the charged particle be in motion.

It should be noted that since the magnetic force acts in a direction normal to the
particle velocity, the acceleration is normal to the velocity and the magnitude of the
velocity vector is unaffected.

Since the magnetic force is at right angles to the magnetic field, the work done by
the
magnetic field is given by

W  F dL  FdL cos 90 0
 Magnetic Force
D3.10: At a particular instant in time, in a region of space where E = 0 and B = 3ay
Wb/m2, a 2 kg particle of charge 1 C moves with velocity 2ax m/sec. What is the
particle’s acceleration due to the magnetic field?
Given: q= 1 nC, m = 2 kg, u = 2 ax (m/sec), E = 0, B = 3 ay Wb/m2.
Newtons’ Second Lorentz Force Equation
Law
F  q  E  u  B  q u 
F  ma Equating
q 1 m
a uB 2ax 3aBy  z 2

m 2 3a sec

To calculate the units: C m Wb  kg m   N m  J   V sec   m 2


kg sec m 2
 N sec   J   C V  
2

Wb
P3.33: A 10. nC charge with velocity 100. m/sec in thesec z direction enters a region
where the electric field intensity is 800. V/m ax and the magnetic flux density 12.0
Wb/m2 ay. Determine the force vector acting on the charge.
Given: q= 10 nC, u = 100 az (m/sec), E = 800 ax V/m, B = 12.0 ay Wb/m2.

 Vx m Wb 
 800
F  q E  u  B  10x109 C m a 100
s
z
m2 a   4
a 12 x
 y 
Na
 Magnetic Force on a current
Element
Consider a line conducting current in the presence of a magnetic field. We wish to
find the resulting force on the line. We can look at a small, differential segment dQ of
charge moving with velocity u, and can calculate the differential force on this charge
from
dF  dQ u 
B u velocity
The velocity can also be written
dQ segment
dt
dL
u
Therefore
dQ
dF  dL  B
dt
Now, since dQ/dt (in C/sec) corresponds to the current I in the line, we have

dF  IdL  (often referred to as the motor equation)

B
We can use to find the force from a collection of current elements, using the integral
F12   I dL
2 2  B1 .
Magnetic Force – An infinite current
Element
Let’s consider a line of current I in the +a direction on the z-axis. For current element
z
IdLa, we have

IdLa  Idza a z .

dF12  I2 dL2  B1.

The magnetic flux density B1 for an infinite length line of

B1   oH 1
current is I I
H1 21a  B 1  o21a 

 
We know this element produces magnetic field, but the field cannot exert magnetic
force on the element producing it. As an analogy, consider that the electric field of a
point charge can exert no electric force on itself.

What about the field from a second current element IdLb on this line?
From Biot-Savart’s Law, we see that the cross product in this particular case will be
zero, since IdL and aR will be in the same direction. So, we can say that a straight line of
current exerts no magnetic force on itself.
 Magnetic Force – Two current
Elements
dF12asecond
Now let us consider  of1 current parallel to the first.
I 2 dLline
2
The force dF12 from the magnetic field of line 1 acting on a differential section of line 2
B
is
The magnetic flux density B1 for an infinite length line of
a = -a

current is recalled from equation to be x

ρ=y
o I 1 
B 1  2 a
 we see that ρ = y and a = -ax. Inserting this in
By inspection of the figure
the above equation and considering that dL2 = dzaz, we have

F12  I2 dL 2  B1  I 2dza z  o I1 a 
   I dza  2I - a
2  2 z o 1 x

 
II
F12  o 1 2 a y  dz
2
y 
 Magnetic Force on a current Element
To find the total force on a length L of line 2 from the field of line 1, we must
integrate dF12 from +L to 0. We are integrating in this direction to account for the
direction of the current.
0
II
F12  o 1 2 a y  dz
2 y

L a = -ax
o I 1 I 2 L
 ay ρ=y
2
y
This gives us a repulsive force.

Had we instead been seeking F21, the magnetic force acting on line 1 from the field of
line 2, we would have found F21 = -F12.

Conclusion:
1)Two parallel lines with current in opposite directions experience a force of
repulsion.
2) For a pair of parallel lines with current in the same direction, a force of
attraction
would result.
 Magnetic Force on a current Element
In the more general case where the two lines are not parallel, or not straight, we
could use the Law of Biot-Savart to find B1 and arrive at

o
F12  I2I1   dL 2  dL1  a12 
4 2
R12

This equation is known as Ampere’s Law of Force between a pair of current carrying
circuits and is analogous to Coulomb’s law of force between a pair of charges.
 Magnetic Force
D3.11: A pair of parallel infinite length lines each carry current I = 2A in the same
direction. Determine the magnitude of the force per unit length between the two
lines if their separation distance is (a) 10 cm, (b)100 cm. Is the force repulsive or
attractive? (Ans: (a) 8 mN/m, (b) 0.8 mN/m, attractive)

F12   o I 1 I 2 L a
Magnetic force between two current elements
y
when current flow is in the same direction 2
y
Magnetic force per unit length F12 o I1 I 2

2 a
y
L
y
Case (a) y = 10 cm
F12   (4 107 )(2)(2)
ay  8
L 2 (10 N/m
102 )
Case (a) y = 10 cm
F12   (4 107 )(2)(2)
a y  0.8
L 2 (100 N/m
102 )
 Electromagnetic Force
The electromagnetic force is given by Lorentz Force Equation (After Dutch physicist
Hendrik Antoon Lorentz (1853 – 1928))

F  q  E  u  B

The Lorentz force equation is quite useful in determining the paths charged particles
will take as they move through electric and magnetic fields. If we also know the
particle mass, m, the force is related to acceleration by the equation
F  ma.

The first term in the Lorentz Force Equation represents the electric force Fe acting on a
charge q within an electric field is given by.

Fe  qE

The electric force is in the direction of the electric field.

 MAGNETIC TORQUE AND MOMENT
The torque T (or mechanical moment of force) on the loop is the Vector
product of the force F and the moment arm r.
A MAGNETIC DIPOLE
 Magnetostatic Boundary Conditions
Will use Ampere’s circuital law and Gauss’s law to derive normal and
tangential boundary conditions for magnetostatics.

Ampere’s circuit law:

 H dL  Ienc Path 1

Path 2
Path 4
The current enclosed by the path is Path 3

I enc 
 KdW  Kw.
We can break up the circulation of H into four integrals:
b c d a

 H dL         H dL Kw.
a b w b c d

Path 1:  H dL   H T1 a T dLaT  HT1 w.

a 0

c 0 h / 2

 H dL   H N 1 a N dLaN  dLa N    H N1  H N 2  h
Path 2:
 H N 2a N 2
b h / 2 0
 Magnetostatic Boundary Conditions
d 0

Path 3:  H dL   H T 2 aT dLaT  H T 2 w.

c w
a 0
h / 2
h
Path 4:  H dL   H N 2 a N dLa 
N  H N1a N dLa N   H N1  H N 2 
d h / 2 0
2

Now combining our results (i.e., Path 1 + Path 2 + Path 3 + Path 4), we obtain
ACL:
 H dL  I enc

 H dL  H T1  H T 2  w Equating
I enc 
 KdW  Kw
Tangential BC: K
H T1  H T 2
A more general expression for the first magnetostatic boundary condition
can be written as

a 21  H1  H 2   K

where a21 is a unit vector normal going from media 2 to media 1.

 Magnetostatic Boundary
Conditions
Special Case: If the surface current density K = 0, we get
If K = 0
H K
HT1 T 2
H T1  H T 2
The tangential magnetic field intensity is
continuous across the boundary when the
surface current density is zero.

Important Note:
B
H
We know that B  o  rH (or) o  r
Using the above relation, we obtain

H T1  H T 2 BT1

BT2
o 1 o  2
Therefore, we can say that
BT1  BT2
The tangential component of the magnetic flux density B is not continuous across
the boundary.
 Magnetostatic Boundary Conditions
Gauss’s Law for Magnetostatic
fields:

 B dS = 0
To find the second boundary condition, we center a Gaussian pillbox
across the interface as shown in Figure.

We can shrink h such that the flux out of the side of the pillbox is
negligible. Then we have

 B dS   B N1
aN dSaN   BN 2 a N dS(a N )

  BN 1  BN 2  S  0.

Normal BC: .
B N1  BN 2
 Magnetostatic Boundary
Conditions
Normal BC: BN1  BN 2
Thus, we see that the normal component of the
magnetic flux density must be continuous across the
boundary.
Important Note:

We know that B  o  r H
Using the above relation, we obtain

BN1  BN 2 o 1H N1  o  2 H N2

Therefore, we can say that H N1  H N2

The normal component of the magnetic field intensity is not continuous across the
boundary (but the magnetic flux density is continuous).
 Magnetostatic Boundary
Conditions
Example 3.11: The magnetic field intensity is given as H1 = 6ax + 2ay + 3az (A/m) in a
medium with r1 = 6000 that exists for z < 0. We want to find H2 in a medium with
r2 = 3000 for z >0.

Step (a) and (b): The first step is to break H1 into its normal component (a) and its
tangential component (b).
Step (c): With no current at the interface, the tangential component is the same
on
both sides of the boundary.
Step (d): Next, we find BN1 by multiplying HN1 by the permeability in medium 1.
Step (e): This normal component B is the same on both sides of the boundary.
Step (f): Then we can find HN2 by dividing BN2 by the permeability of medium 2.
MAGNETIC ENERGY