Chapter 29 Magnetic Fields
• F is on a moving charged particle.
• Each magnet has two poles,
north pole and south pole,
regardless the size and shape
of the magnet.
• Like poles repel each other,
unlike poles attract each other.
• Compass: north pole points to
the north of the Earth.
• Magnetic poles are always
found in pair. (“Monople” ?)
• A piece of iron can be magnetized in a strong magnetic field.
• F = 0, if B || v. (sin = 0)
• Direction of B is the direction in which the north pole of a
compass needle points at that location. Magnetic field lines are
from N pole to S pole outside a magnet, and continue inside the
magnet.
• F is to both B and v (cross
product)
• | F | = qvBsin
T=
Wb
N
N
=
2 =
m
C m / s A m
A common non-SI unit: gauss (G),
1 T = 104 G
• The direction of the force can be
determined by the “Right-Hand Rule”.
29.1 The magnetic field
• Magnetic field B is a vector, (like E ).
• The magnitude of B is defined in
terms of the magnetic force:
F = qv B
Unit: tesla, it is also called weber per square
meter (Wb/m2)
F
B
q
v
• Magnetic forces on a positive charge
and a negative charge are in opposite
direction.
Difference between E and B :
(1) The electric force is always in the
direction of the E ; The magnetic force
is to B .
(2) The electric force acts on a charged
particle regardless its velocity; The magnetic force only acts
on a moving charged particle.
(3) For a magnetic force, F • ds = (F • v)dt = 0 . Therefore, a
steady B can only change the direction of the velocity vector,
but it cannot change the speed or kinetic energy of the
charged particle.
z
F
Example: A proton moves with a speed
of 8 106 m/s along the x axis. The
magnetic field is 2.5 T as shown.
(a) Calculate initial magnetic force.
+e
v
60°
F = qvBsin
x
= (1.6 10 19 C)(8 10 6 m / s)(2.5T)(sin 60°)
= 2.8 10 12 N
F is in the +z direction.
(a) Calculate the acceleration of the proton.
F
2.8 10 12 N
a=
=
= 1.7 1015 m / s 2
27
m p 1.67 10 kg
a is in the +z direction.
y
B
29.2 Motion of a Charged Particle In a Magnetic Field
• Magnetic force is to the velocity of the particle.
• The work done on the particle by the field is 0.
• A charged particle moves in a circle whose plane is to B.
• F acts like a central force, which changes the direction of v, but
not the magnitude.
q>0
From the Newton’s 2nd Law:
mv 2
v
v
F
F = qvB =
r
F
mv
r=
qB
Angular frequence:
F
v qB
= =
v
r m
The period of its motion:
2r 2 2m
T=
=
=
v
qB
Example: A proton is moving in a circular orbit of radius 14 cm.
B = 0.35 T, to the velocity of proton. Find the orbital speed of
the proton.
qBr (1.6 10 19 C)(0.35T)(14 10 2 m)
v=
=
= 4.7 10 6 m / s
27
1.67 10 kg
m
Velocity Selector:
* A charged particle having a velocity vector that has a component
parallel to a uniform magnetic field moves in a helical path.
• v|| (velocity component parallel to B) remains constant.
• v(velocity component perpendicular to B) produce a circular
mv
motion with a radius r = .
qB
v= E/B
Only charged particles with the
desired velocity (v = E/B) can pass
through a velocity selector.
Mass Sectrometer:
m/q = rB0/v
29.3 Applications Involving Charged Particles Moving in a
Magnetic Field
In the presence of both electric field E and magnetic filed B,
the total electro-magnetic force (called the Lorentz force) acting on
a charge is
F = qE + qv B
m/q = rB0B/E
Mass/charge ratio can be determined.
29.4 Magnetic Force on a Current-Carrying Conductor
Case Study 1:
A curved wire carrying a current I.
B
The total force on the wire,
b
F = I ds B
a F = IL' B
r
L’ is the length perpendicular to B field.
Force on a charge moving at speed vd,
Fi = (q i v d B)
dF = I ds B
b
Total force on a wire F = I ds B
a
where ds is in the direction of I.
ds
I
a
B
B
An closed loop carrying a current I.
q
dF = (qv d B) nAds = (nqv d A) ds B
L'
Case Study 2:
A
Force on a segment of wire,
carrying current I,
b
vd
ds
F = I( ds) B 0
ds
“The total magnetic force on any
closed current loop in a uniform
magnetic field is zero.”
I
a
Example:
(a) Find the net force on the loop
This is the maximum torque:
max = IabB
B
b
B
must = 0
(b) Force on the straight portion.
Fs = IBL,
The direction is out of paper.
L
I
(c) Force on the curved portion.
I
In general,
a
a
= F1 sin + F2 sin 2
2
= IabBsin F1
a/2
= IABsin = IABsin F2
Fc + Fs = 0
Fc = IBL, the direction is into the paper.
B
Where, A = ab is the area of the
loop.
F1
29.5 Torque on a Current Loop in a Uniform Magnetic Field
a/2
Force on the sides of length b:
F1 = F2 = IbB
Force on the sides of length a:
F= 0
Torque:
a
a
= F1 + F2
2
2
a
a
= (IbB) + (IbB) = IabB
2
2
has the maximum (IAB) when = 90°.
=0 when = 0°.
a
sin 2
B
F2
Define:
(1) A , a vector to the plane of the loop, | A | = are of the loop.
The direction of A is determined by the right-hand rule.
(2) Mangetic moment of the loop: μ = I A (unit: A•m2)
Then, the torque: = μ B
Example: A rectangular coil, 5.4 8.5 cm,
has 25 turns of wire. Current is 15 mA.
(a) Find the magnitude of it magnetic
moment.
μ = nIA
= (25)(15 10 3 A)(5.4 8.5 10 4 m2 )
= 1.72 10 3 Am2
(b) Suppose B = 0.35 T, and it is || to the
plate of the loop. What is the torque?
= μB
= (1.72 10 3 Am 2 )(0.35T )
= 6.02 10 4 Am2T = 6.02 10 4 N m
(c) What is the torque when = 60°?
= μBsin = (6.02 10 4 N m)(sin 60°)
= 5.21 10 4 N m