Week of Feb 2 – 4
No labs or tests!
Week of Feb 9 – 11
Tutorial and Test 2
Chapters 19, 20, 21
Monday, February 1, 2010
29
Force on a charge moving in a magnetic field
�F
F = qvB sin !
Force is at right angles to both �v and �B
�B
!
q
Magnetic field
�v
Force on a current in a
magnetic field
F = !q v B sin " =
�
�
!q
(v !t) B sin "
!t
I
�v
Δq
L
F = ILB sin !
Monday, February 1, 2010
30
Clicker Question
The drawing shows three possible directions for the velocity v of a
positively charged particle moving through a uniform electric field E
(directed out of the screen) and a uniform magnetic field B. In which
direction, 1, 2, or 3, is the magnitude of the net force (the combined
magnetic and electric forces) the greatest?
a) 1
b) The magnitude of the net force is the same for all three velocities.
c) 2
d) 3
Answer d): The electric and
magnetic forces are in the
same direction.
Monday, February 1, 2010
31
Clicker Question.
Which one or more of the following choices is (are) true regarding how a
magnetic field acts on a charged particle. A magnetic field __________
1. can increase the speed of a moving particle.
2. can cause a moving particle to accelerate.
3. always exerts a force on a moving particle.
4. can exert a force on a stationary particle.
a) 1 and 2
b) 2
c) 3
d) 2 and 4
e) 2 and 3
Answer 2): The force is always at right angles to both v and B, so
does no work, but does cause acceleration (change of direction).
There is no force if the charge is stationary.
Monday, February 1, 2010
32
Prob. 21.78/30: The triangular loop of wire carries a current, I = 4.7 A
and B = 1.8 T. Find the force acting on each side and the net force.
! F1
. F2
L
h
900
θ = 550
AC: 24.2 N
AB: 0 N
BC: 24.2 N
F3 = 0
Monday, February 1, 2010
33
21.41/35: B = 0.05 T, vertical
m = 0.2 kg = mass of bar
m
I
L
v constant
A current I flows through the bar. The bar slides down the rails
without friction at constant speed.
What is I? In what direction does it flow?
• The component of the weight down the plane must be equal to the
component of the magnetic force up the plane...
14.1 A
Monday, February 1, 2010
34
Charges inside the moving rod experience a
force due to the magnetic field...
–––
++
+
I
�v
I
I
I
Conductor
The moving conductor acts as a generator.
The basis of electromagnetic induction (next chapter)
Monday, February 1, 2010
35
Force on a current-carrying conductor
An alternating current in the voice coil causes the cone of the
loudspeaker to move in and out and generate sound – an electrical
signal is converted into a sound wave.
Monday, February 1, 2010
36
The voice coil of a speaker has a diameter d = 0.025 m, contains 55
turns of wire and is placed in a magnetic field of 0.1 T. The current
in the coil is 2 A.
a) Find the magnetic force that acts on the coil and cone.
b) The voice coil and cone have a combined mass of 0.02 kg. Find
their acceleration.
B = 0.1 T
m = 0.02 kg
d = 0.025 m
F = 0.86 N
a = 43 m/s2
Monday, February 1, 2010
37
Magnetohydrodynamics
(MHD)
Ions within the water
allow a current to flow
when a potential
difference is applied.
The moving ions feel a
force due to the
magnetic field and that
force pushes the water
out through the back of
the boat.
The boat feels an equal
and opposite reaction
force to the front.
Propulsion with no moving parts.
Monday, February 1, 2010
38
The Torque on a Current-Carrying Coil
! = angle
between normal
to coil and B
L
From above
w
w
L
Area of coil, A = Lw
The force on each vertical arm of the coil is: F = ILB
Fw
Fw
The torque is: τ =
sin φ +
sin φ = I(Lw)B sin φ = IAB sin φ
2
2
(A = Lw)
If the coil has N turns: ! = NIAB sin "
Monday, February 1, 2010
39
If the coil has N turns: ! = NIAB sin "
The formula is valid for any shape of coil, not just rectangular.
NIA is known as the “magnetic moment” of the coil.
! = 90◦
Monday, February 1, 2010
! = 0o
40
Variation of torque with φ
Torque: ! = NIAB sin "
1
sin(pi*phi/180)
0.8
Torque, τ 0.6
0.4
0.2
00
0
1800
3600
5400
!
-0.2
-0.4
-0.6
-0.8
-1
0
100
200
300
400
500
600
700
Torque changes sign! – rotation reverses, no good for an electric motor!
Monday, February 1, 2010
41
Reverse the current at the right times...
...the torque is always in the same direction and the motor continues to rotate
1
sin(pi*phi/180)
0.8
Torque, τ0.6
0.4
0.2
00
0
1800
3600
5400
!
-0.2
-0.4
-0.6
-0.8
-1
0
Monday, February 1, 2010
100
200
300
400
500
600
700
42
Electric motor
! = 0◦
1
2
1
2
Commutator reverses direction of current when φ = 0o, 180o...
Torque continues in the same direction
Monday, February 1, 2010
43
Prob. 21.47/41: The loop is free to rotate about the z-axis.
Find the torque on the loop and whether the 35º angle will increase or
decrease.
I = 4.4 A
B = 1.8 T
170 N.m
Monday, February 1, 2010
! = 90º – 35º
44
Prob. 21.50/44: A square coil and a rectangular coil are made from
the same length of wire. Each contains a single turn. The long sides
in the rectangle are twice as long as the short sides.
Find the ratio of torques the coils experience in the same magnetic
field.
• Find the relation between the lengths of the sides of the square
and the rectangle, given that the perimeters are of the same
length.
• Work out the areas of the two shapes and relate to the torques.
a = 2b
L
b
L
"square/"rect = 9/8
Monday, February 1, 2010
45