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Chapter 19Magnetic Force
on Chargesand Current-
Carrying Wires
F B v B q qv sin
This force acts in the direction perpendicularto the plane defined by the vectors v and Bas indicated by the right-hand rule!
WARNING: cross-product(NOT simple multiplication!)
DON’T FORGET:Forceshave
directions!
Thumb points in thedirection of the current.
Fingers point in thedirection of the magneticfield.
Palm faces the directionof the magnetic force.
F v Bq sin
BF
vq sin
BF
v q sin
N
C m / s
N
A m
Tesla Weber/m2
1 Tesla is a very strong magnetic field.
So, magnetic fields are often measured in the cgs (i.e., centimeter, gram, second) unit of Gauss.
1 Tesla = 104 Gauss
The Earth’s magnetic field is about 0.5 Gauss near the surface.
A proton moves at right angles to a magnetic field of 0.1 T with a speed of2.0 X 107 m/s. Find the magnitude of the acceleration.
F v B
q sin
( . )( . )( . )(sin )
.
o16 10 2 0 10 01 90
32 10
19 7
13
C m / s T
N
F a
aF
m
m
3 2 1019 10
1314 2.
.N
1.67 10-27 kgm / s
Remember when we talked about the motionof charges in a wire….
E
A
Vd t
The charges move with an average velocity vd.
The magnetic force on a wire with N chargecarriers moving with velocity vd in a uniform magnetic field B should just be the vector sum of the force on each individual charge.
Since the average velocity is the same for all Ncharge carriers, the magnetic force acts in thesame direction (on average) on all the chargecarriers. Therefore...
F v BdNq sin
Where is the angle between the long dimensionof the wire and the magnetic field. The forceacts in the direction perpendicular to both.
F v BdNq sin
N = Number of charge carriers
n = number of charge carries per unit volume
A = cross-sectional area of wire
L = length of wire
N nAL
I n Aqvd
F B I L sin
F v BdnALq sin
But recall our definition of current in a wire...
Substituting, we get the simpler expression:
So, the maximum value of the magnetic force on a current carrying wire occurs when the wire is perpendicular to the magnetic field and has the value...
F Bmax I L
Hey, Mr. Sluggo.Magnetic Force ona wire is namedafter me!
So what happens if we put a loop of wire carrying current I in a magnetic field?
B
I
The currents in the top and bottom of this loopare anti-parallel and parallel to the magnetic field.Therefore, sin = 0. So the magnetic forces onthe top and bottom of the loop are 0!
F B I L sin
B
I
On the left side of the loop, we use the right-hand rule to determine that the force is out (toward us).
On the right side of the loop, we use the right-hand rule to determine that the force is in (away from us).
F B I L sin
In each case, = 90o, sin = 1, so
So, the magnitude of the force on the left sideof the loop is the same as the magnitude onthe right side of the loop.
What, therefore, is going to happen to theloop in this magnetic field?
F B I L
B
I
The looprotates! The magnetic
force producesa TORQUE onthe current loop,causing it torotate. In thiscase, the looprotatescounterclockwiseas viewed fromabove...a/2 a/2
b
left F B B F
HGIKJd Ld bI I
a
2 counter - clockwise
right F B B F
HGIKJd Ld bI I
a
2 counter - clockwise
B
I
The looprotates! The magnetic
force producesa TORQUE onthe current loop,causing it torotate. In thiscase, the looprotatescounterclockwiseas viewed fromabove...a/2 a/2
b
total left right B B I I (Area of loop)ab
B
I
The looprotates! The magnetic
force producesa TORQUE onthe current loop,causing it torotate. In thiscase, the looprotatescounterclockwiseas viewed fromabove...a/2 a/2
b
B I A sin
Where is the angle between the normalto the loop and the magnetic field.
Top View:
BI
normal
x
The normal is the direction perpendicularto the plane of the loop of wire.
If the loop has N turns, then the torquebecomes...
N I A sinB
A small circular coil of 20 turns of wire lies in a uniform magnetic of 0.5 T so that the normal to the plane of the coil makes an angle of 60o with the direction of B. The radius of the coil is 4 cm, and it carries a current of 3 A. What is the magnitude of the torque on the coil?
x x x x x
x x x x x
x x x x x
x x x x x
side viewN = 20 turns B = 0.5 TI = 3 A A = (.04)2
= 60o
60o
top view
( ) . )(5
.
20 ( T)( A m ) sin
Nm
205 3 10 60
013
3 o
N I A sinB
x x x x x
x x x x x
x x x x x
x x x x x
side view
I60o
top view
Ix
If the current flows clockwise around the loop (viewed face on), which way does the loop rotate?
The force on the left side of the loop is toward the top of the top view diagram.
CLOCKWISE!
The force on the right side of the loop is toward the bottom of the top view diagram.
These two forces combine, both creating atorque in the same direction, causing the loopto rotate
x x x x x
x x x x x
x x x x x
x x x x x
side view
I60o
top view
Ix
