EEE 411- LECTURE 8

MAGNETOSTATICS

L.Sitoloma (M.Eng, B.Eng, MEIZ, REng)

CONTENT
Forces due to magnetic fields & Ampere’s force law
Force on a moving point of charge
Force on a Differential Current Element
Force btn Differential Current Elements
Magnetic Dipole Moments
Magnetic Boundary Conditions
Tangential Components
Normal Components
8.0 Force on a Moving Point Charge

According to the discussion in the previous chapter, a static electric field E exerts a
force on a static or moving charge Q. Thus according to Coulomb’s law, the force Fe
exerted on an electric charge can be obtained. The force is related to the electric
field intensity E as
 Magnetic Force
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 Law Lorentz Force Equation
F  ma Equating F  q  E  u  B  q u  B 

q 1 m
a uB  2a x  3a y  3a z 2
m 2 sec
 Magnetic Force

A 10. nC charge with velocity 100. m/sec in the 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.

F  q  E  u  B  q u  B 
 V m Wb 
F  q  E  u  B   10 x10 C  800 a x  100 a z 12 2
9
  4 Na x
 m s m ay
 
8.1 Force on a Differential Current Element
8.2 Force btn Differential Current Elements
• When discussing the electrostatic field, we learnt
that a point charge exerts force on another point
charge, separated by distance R.
• If these charges are of the same type (i.e both
positive or negative), then they repel each other.
But when two charges are of different type, then
they attract each other.
• Now consider that two current carrying conductors
are placed parallel to each other. Each of this
conductor produces its own flux around it.
• So when such conductors are placed close to each • If the directions of both the currents are same,
other, there exists a force due to the in direction of then the conductors experience a force of
two fluxes. attraction as shown in Fig. 8.2 (a)
• The force between such parallel current carrying • and if the directions of the two currents are
conductors depends on the directions of the two opposite to each other, then the conductors
currents. experiences a force of repulsion as shown in the
Fig.8.2 (b).
Fig 8.3 Force btn Different Current Elements
 Example: Magnetic Force
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?

Magnetic force between two current elements  o I1 I 2 L

when current flow is in the same direction F12   ay
2 y

Magnetic force per unit length F12  o I1 I 2

 ay
L 2 y
 Magnetic Force

Given: I=2A, distance between lines (a) 10 cm , (b) 100 cm

F12  o I1 I 2
Magnetic force per unit length
 ay
L 2 y
Case (a) y = 10 cm

7
F12 (4  10 )(2)(2)
 2
a y  8  N/m
L 2 (10  10 )
 Magnetic Force

Given: I=2A, distance between lines (a) 10 cm , (b) 100 cm

F12  o I1 I 2
Magnetic force per unit length
 ay
L 2 y
Case (a) y = 10 cm

7
F12 (4  10 )(2)(2)
 2
a y  0.8  N/m
L 2 (100  10 )
8.3 Magnetic Dipole Moments
8.4.MAGNETIC BOUNDARY CONDITIONS
8.4.1 Boundary Conditions for Normal Component
8.4.2 Boundary Conditions for Tangential Component

𝐛 𝟐 𝐜 𝐝 𝟏 𝐚
𝐇. 𝐝𝐋 = 𝐚
𝐇. 𝐝𝐋 + 𝐛
𝐇. 𝐝𝐋 + 𝟐
𝐇. 𝐝𝐋 + 𝐜
𝐇. 𝐝𝐋 + 𝐝
𝐇. 𝐝𝐋 + 𝟏
𝐇. 𝐝𝐋 =𝐈 ...(10)
Figure 8.13
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