1. Magnetic Field and Magnetic Forces
Chapter 17
Magnetic Field
and
Magnetic Forces

2. Opposite poles : attract each other Like poles: repel each other
Magnetism
(Rotation axis) S N
South
North
South magnetic pole
South geographic pole
Earth’s magnetic field
North magnetic pole
North geographic pole
North
north pole
south pole
(a vector field)
The needle of a compass aligns with the magnetic field
Earth is a magnetic. The axis of earth’s magnetic is not parallel to its geographic axis 
Magnetic declination
Opposite poles : attract each other
Like poles: repel each other

3. Attract each other Attract each other Repel each otherN S F
Attract each other S N F
Attract each other S N F
Repel each other N S F
Repel each other

4. Magnetic Field
In addition to the electric field, a moving charge or a current in space can create a magnetic field.
An electric force (F = Q0E) will exert on other charge (Q) present in the electric field (E). Similarly, the magnetic field also exerts a magnetic force on other moving charge or current present in the magnetic field.
Oersted’s Experiment N S
I = 0 (no current) E W
I  0 I N S E W
I  0 I N S E W

5. Magnitude of magnetic force is proportional to:
magnitude of the charge
magnitude or strength of the magnetic field
velocity of the moving particle (for electric force, it is the same no matter the charge is moving or not) or the component of velocity perpendicular to the field.
A charged particle at rest will have no magnetic force.
The direction of magnetic force (F) is not the same as the direction of magnetic field (B). Instead, the magnetic force is always perpendicular to both direction of magnetic field (B) and the velocity (v).
Direction of B: the north pole direction of a compass needle.
For a magnet, the direction of B is pointing out of its north pole and into its south pole. S N

6. Magnetic force on a moving charged particle:
where:
F : magnetic force [N]
Q : magnitude of charge [C]
v : velocity of the charge [m/s]
B : magnetic field [T or Ns/Cm or N/Am (A: ampere)]
1 N/Am = 1 tesla = 1 T [Nikola Tesla (1857 – 1943)]
Right-hand rule
Magnetic field (B) + Q
Force (F)
Velocity (v)  v
Positive charge
Right hand rule
cross product

7. For a negative charge, the force is opposite to the case of the positive charge.
Magnetic field (B) _ -Q
Force (F)
Velocity (v)  v
negative charge

8. Total magnetic flux through a surface A:
Sum of magnetic flux thru areas of all elements
Where:
A: magnetic flux (a scalar) [weber (Wb)]
B : magnetic field [T]
A : surface area [m2] dA 
1 Wb = 1 T m2 = 1 Nm / A
Gauss’s Law for Magnetism
Total magnetic flux through a closed surface = 0
In = Out
Magnetic field B is also called magnetic flux density

9. Motion of Charged Particles in a Magnetic Field
A charge particle under the action of a magnetic field only moves with a constant speed. The motion is determined by Newton’s laws of motion.
Circular motion of a positive charge in a uniform magnetic field (B):
m : mass of the particle
v : constant velocity
R : radius of the circular orbit v1 v2 P1 P2  R s x F v + R
“x” denotes that the magnetic field is pointing into the plane

10. Magnetic Force on a Conductor
Magnetic force on a straight wire: x F v + L Q A
Conductor with current I I L  F B
Where:
F : magnetic force
I : total current
L : length of the wire segment
Magnetic force on an infinitesimal wire (not straight wire):
Divide the wire into infinitesimal straight line

11. SOURCES OF MAGNETIC FIELD
“Source point” is referred to the location of a charge (Q) moving with a constant velocity (v) in a magnetic field.
“Field point” is referred to the location or point where the field is to be determined, e.g. location of point “k”.
Magnetic field of a point charge moving with a constant velocity: + B v k Q 
B = 0
where:
B : magnetic field
Q : point charge
v : velocity of the charge
r : distance from the charge to the field point x I B
+ Q
charge is moving into the plane
0 = 4  10-7 Ns2/C2
1 Ns2/C2 = 1 Wb/Am = 1 Tm/A = 1 n/A2

12. Law of Biot and Savart for Magnetic Field of a Current Element (B):
Applying the principle of superposition, the magnetic fields of a number of moving charges can be calculated.
Total magnetic field due to a number of moving charges = vector sum of the electric fields due to the individual charges
Law of Biot and Savart for Magnetic Field of a Current Element (B): B k  dB
dB = 0 dL
where
dL : represents a short segment of a current-carrying conductor
I : current in the segment
nQ : total charges
vd : drifting velocity
A : cross-section area of segment

13. Magnetic field of a straight current-carrying conductor:x y -L +L dL dB I 
 - 

14. Use principle of superposition of magnetic fields: Btotal = B1 + B2
Example 17.1:
The figure shows an end view of two parallel wires carrying the same current I in opposite directions. Determine the magnitude and direction of magnetic flux B at point A. x 2L
Wire 1
Wire 2 I B1 B2 A y
Solution:
Use principle of superposition of magnetic fields:
Btotal = B1 + B2
Point A is closer to wire 1 than to wire 2, the field magnitude B1 > B2
x : into the plane
 : out of the plane x I B
Use right hand rule, B1 is in the – y-direction and B2 is in the + y-direction. As B1 > B2, Btotal is in the – y-direction and the magnitude is:

15. Ampere’ Law
The line integral of magnetic field intensity around a single closed path is equal to the algebraic sum of currents enclosed. r I B dL
Integration path enclosing the conductor r1 r2 1 2 3 4 B
Integration path not enclosing the conductor

16. Example 17.2
A long, straight wire with radius of a, and the wire carries a current I0, which is distributed uniformly over its cross section. Find the magnetic field both inside and outside the wire.

17. By applying Ampere’s Law,
Solution:
By applying Ampere’s Law,
∮B·dL = ∮BdL = B∮dL = B(2πr)
For r ≤ a, (inside the wire)
πr2
πa2 r2 a2
I = I0
=> I = I0
where I0 is the total current over the cross-section of the wire. r2 a2
Ampere’s Law
∴ B(2πr) = μ0( ) I0
So, we have
B =
μ0 I 0 2π r a2
( r ≤ a )
For r > a, (outside the wire)
μ0 I 0
2πr
B =
( r > a )

18. In-Class Exercise 17.1: (Magnetic field of a circular current loop)
Determine the magnitude and direction of the magnetic field at point P due to the current in the semicircular section of wire shown in figure I R P
Solution:
There is no magnetic field at the center of the loop from the straight sections. The magnetic field from the semicircle is just half that of a complete loop:
Into the page.

19. Magnetic Field of a Circular Current Loop

20. The Solenoid
A long wire wound in the form of a helical coil is known as a solenoid.

21. 1 y2 (y2+R2) 1/2 (y2+R2) 3/2 R2 (y2+R2) 3/2
From the figure, the current for a length increment dy is
Also, the magnetic field dB due to the current dI in dy can be found as, 1
(y2+R2) 1/2 y2
(y2+R2) 3/2 R2
(y2+R2) 3/2

22. 2 L 2 L 2 L 2 L μ0 I N μ0 I N So, dB = ( ) cos θ dθ μ0 I Nθ2 θ1
=> B = ( ) ∫ cos θ dθ
μ0 I N
2 L
= ( ) (sin θ2 – sin θ1 )
B = (sin θ2 – sin θ1 ) j
μ0 n I 2 N L
where n = , number of turns per unit length
This formula represents the magnetic field along the centroid axis of a finite solenoid.
For infinite long solenoid, it is assumed that θ1 = -π/2 and θ2 = π/2.
B = μ0 n I j

23. Example :
A solenoid has 300 turns wound around a cylinder of diameter 1.20 cm and length 14.0 cm. If the current through the coils is A, what is the magnitude of the magnetic field inside and near the middle of the solenoid.
Solution
Since the length of the solenoid is quite large in comparative with its diameter, the magnetic field near its middle is approximately uniform. It is therefore reasonable to consider it as a case of infinite solenoid
B = μ0 n I j.
The number of turns per unit length (n) is
n = N/L = (300 turns) / (0.14 m)
= 2.14 × 103 turns / m
Therefore, the magnetic field inside and near the middle of the solenoid is,
B = μ0 n I j
= (4π × 10-7 Tm/A) (2.14 × 103 turns / m) (0.410 A)
= 1.10 × 10-3 T

24. Magnetic fields of a finite solenoid