Download Chapter 21: Electric Charge and Electric Field

Chapter 26: Magnetism: Force and Field
Magnetism
 Magnets
Magnetism
 Magnetic
forces
Magnetism
 Magnetic
field of Earth
Magnetism
 Magnetic
monopoles?
Perhaps there exist magnetic charges, just like electric charges.
Such an entity would be called a magnetic monopole (having + or
magnetic charge).
How can you isolate this magnetic charge?
Try cutting a bar magnet in half:
S
N
S
N
S
N
Even an individual
electron has a
magnetic “dipole”!
• Many searches for magnetic monopoles—the existence of which
would explain (within framework of QM) the quantization of
electric charge (argument of Dirac)
• No monopoles have ever been found:
Magnetism
 Source
of magnetic field
What is the source of magnetic fields, if not magnetic charge?
Answer: electric charge in motion!
e.g., current in wire surrounding cylinder (solenoid) produces
very similar field to that of bar magnet.
Therefore, understanding source of field generated by bar magnet
lies in understanding currents at atomic level within bulk
matter.
Orbits of electrons about nuclei
Intrinsic “spin” of
electrons (more
important effect)
Magnetism
 Magnetic
force: Observations
Magnetism
 Magnetic
force (Lorentz force)
Magnetism
 Magnetic
force (cont’d)
Components of the magnetic force
Magnetism
 Magnetic
force (cont’d)
Magnetic force
B
x x x x x x
B

x x x x x x
v
x x x x x x
q
F
v

 q
F
B
v
q
F=0
Magnetism
 Magnetic
force (cont’d)
Units of magnetic field
Magnetism
 Magnetic
force vs. electric force
Magnetic Field Lines and Flux
 Magnetic
field lines
Magnetic Field Lines and Flux
 Magnetic
field lines
S
N
Magnetic Field Lines and Flux
 Magnetic
field lines (cont’d)
Magnetic Field Lines and Flux
 Magnetic
field lines (cont’d)
Electric Field Lines
of an Electric Dipole
Magnetic Field Lines of a bar
magnet
S
N
Magnetic Field Lines and Flux
 Magnetic
field lines (cont’d)
Magnetic Field Lines and Flux
 Magnetic
field lines (cont’d)
Magnetic Field Lines and Flux
 Magnetic
flux
  
d B  B dA  B cos dA  B  dA  B  nˆdA
 
magnetic flux through a surface
 B   B dA   B cos dA   B  dA
  B  BA

  B  B  nˆ dA
 
 B  dA
 B cosdA

  B   B  nˆdA
B
Area A
  
B // A, A  Anˆ
n̂

B
n̂
B
Magnetic Field Lines and Flux
 Magnetic
Units:
flux (cont’d)
1 weber  1 Wb  1 Tm 2  1 (N/A)m  1 Nm/A
 Gauss’s
law for magnetism
No magnetic monopole has been observed!
A=C/s, T=N/[C(m/s)]
-> Tm2=Nm/[C/s]=Nm/A
 
 B  dA  0 (magnetic flux throu gh any closed surface)
Motion of Charged Particles in a Magnetic Field
 Case
1: Velocity perpendicular to magnetic field
υ perpendicular to B
The particle moves at constant speed υ in a circle in the plane
perpendicular to B.
F/m = a provides the acceleration to the center, so
FL
q υB

a
m
m
hence
v
R
F
x
B
q B

a
m
R
2
and
m
R
qB
Motion of Charged Particles in a Magnetic Field
 Case
1: Velocity perpendicular to magnetic field
Motion of Charged Particles in a Magnetic Field
 Case
1: Velocity perpendicular to magnetic field
Velocity selector
Motion of Charged Particles in a Magnetic Field
 Case
1: Velocity perpendicular to magnetic field
Mass spectrometer
Motion of Charged Particles in a Magnetic Field
 Case
1: Velocity perpendicular to magnetic field
Mass spectrometer
Motion of Charged Particles in a Magnetic Field
 Case
1: Velocity perpendicular to magnetic field
Mass spectrometer
Motion of Charged Particles in a Magnetic Field
 Case
1: Velocity perpendicular to magnetic field
Mass spectrometer
Motion of Charged Particles in a Magnetic Field
 Case
2: General case (v at any angle to B)




Begin by separating the two components of v into v// and v with respect to B.


 
 
The Lorentz force : F  qv  B  q (v//  v )  B


Now the cross product of any two parallel vectors is zero, so

 
The Lorentz force : F  q v  B




Note that F is perpendicu lar to B and to v// , so v// stays constant.


This results in a circular motion wit h the radius R  mv /( qB),

while there is a constant v// (a helical motion).
Motion of Charged Particles in a Magnetic Field
 Case
2: General case (cont’d)
Since the magnetic field does not exert
force on a charge that travels in its direction,
the component of velocity in the magnetic
field direction does not change.
Magnetic Force on a Current-Carrying Conductor
 Magnetic
force on a current (straight wire)
Magnetic Force on a Current-Carrying Conductor
 Magnetic
force on a current (straight wire) (cont’d)
Magnetic Force on a Current-Carrying Conductor
 Magnetic
force on a current (curved wire)
Magnetic Force on a Current-Carrying Conductor
 Magnetic
force on a current: Example1
Magnetic Force on a Current-Carrying Conductor
 Magnetic
force on a current: Example1 (cont’d)
Magnetic Force on a Current-Carrying Conductor
 Magnetic
force on a current: Example1 (cont’d)
Magnetic Force on a Current-Carrying Conductor
 Magnetic
force on a current: Example1 (cont’d)
Magnetic Force on a Current-Carrying Conductor
 Magnetic
force on a current: Example1 (cont’d)
Magnetic Force on a Current-Carrying Conductor
 Magnetic
force on a current: Example2
Magnetic Force on a Current-Carrying Conductor
 Magnetic
force on a current: Example2 (cont’d)
Magnetic Force on a Current-Carrying Conductor
 Magnetic
force on a current: Example2 (cont’d)
Force and Torque on a Current Loop
 Plane
of loop is parallel to the magnetic field
Force and Torque on a Current Loop
 Plane
of loop : general case
if =90o
Force and Torque on a Current Loop
 Plane
of loop and magnetic moment
Force and Torque on a Current Loop
 Plane
of loop : magnetic moment (cont’d)
The same magnetic dipole moment
formulae work for any shape of
planar loop.
Any such loop can be filled by a
rectangular mesh as in the sketch.
Each sub-loop is made to carry the
current NI. You will now see that all
the interior wires have zero current
and are of no consequence.
Nevertheless, each sub-loop
contributes to μ in proportion to its
area.
Force and Torque on a Current Loop
 Plane
of loop : magnetic moment (cont’d)
   
Analogous to the electric dipole p :   p  E
Force and Torque on a Current Loop
 Potential
energy of a magnetic dipole
Work done by the torque when the magnetic
moment is rotated by df :
dW  df  B sin fdf
In analogy to the case of an electric dipole
in Chapter 22, we define a potential energy:
dU  dW   B sin fdf
f2
f2
f1
f1
W    df   ( B sin f )df  B cos f2  B cos f1
 (U 2  U1 )
 
U (f )   B cos f     B
Potential energy of a magnetic
dipole at angle f to a magnetic
field
Applications
 Galvanometer
We have seen that a magnet can exert a torque on a loop of current
– aligns the loop’s “dipole moment” with the field.
– In this picture the loop (and hence the
needle) wants to rotate clockwise
– The spring produces a torque in the
opposite direction
– The needle will sit at its equilibrium
position
Current increased
 μ = I • Area increases
 Torque from B increases
 Angle of needle increases
Current decreased
 μ decreases
 Torque from B decreases
 Angle of needle decreases
Applications
 Motor
Slightly tip the loop
Restoring force from the magnetic
torque
Oscillations
Now turn the current off, just as the loop’s μ is aligned with B
Loop “coasts” around until its μ is ~antialigned with B
Turn current back on
Magnetic torque gives another kick to the loop
Continuous rotation in steady state
Applications
 Motor
(cont’d)
Even better
Have the current change directions every half rotation
Torque acts the entire time
Two ways to change current in loop:
1. Use a fixed voltage, but change the circuit (e.g., break
connection every half cycle
 DC motors
2. Keep the current fixed, oscillate the source voltage
AC motors
VS I
t
Applications
 Motor
(cont’d)
flip the current direction
Applications
 Hall
charges accumulate
(in this case electrons)
effect
-
-
-
+
+
+
Measuring
Hall voltage
(Hall emf)
In a steady state
qEH =qvdB
Charges move sideways until the Hall field EH grows to balance
the force due to the magnetic field: E H   υd  B
So, EH  d B and VH  EH w  d Bw
As previously, I  nqd A  nqd tw and J  I / A
JB
nq 
EH
n can be measured
Applications
 Electromagnetic
rail gun
Exercises
 Exercise
1
If a proton moves in a circle of radius 21 cm perpendicular to a B
field of 0.4 T, what is the speed of the proton and the frequency of
motion?
qB
f

1
2m
v
x
x
1.6 1019 C  0.4T
f 
2 1.67 1027 kg
r
1.6  0.4
f

108 Hz  6.1106 Hz
x
x
6.28 1.67
2
f  6.1106 Hz
qBr
v
m
1.6 1019 C  0.4T  0.21m
v
1.67 1027 kg
v
1.6  0.4  0.21
108
1.67
m
s
 8.1106
m
s
v  8.1106
m
s
Exercises
 Exercise
2
Example of the force on a fast moving proton due to the earth’s
magnetic field. (Already we know we can neglect gravity, but can
we neglect magnetism?)
Let v = 107 m/s moving North.
What is the direction and magnitude of F?
Take B = 0.5x10-4 T and v B to get maximum effect.
F  qvB  1.6 1019 C 107 ms  0.5 104 T
FM  8 10 17 N
(a very fast-moving proton)
volts
FE  qE  1.6 1019 C 100 meter
FE  1.6 10
17
F
v
N
B
N
vxB is into the
paper (west).
Check with globe
Magnetic Field of a Moving Charge
 Magnetic
r̂  r
r
field produced by a moving charge
0 q υ  rˆ
B
4 r 2
r̂  r
Note the factor of μ0 /4π the constant of proportionality needed just as
1/(4πε0) is needed in electrostatics.
0  4 107 Tm/A  4 107 N/A2
r
Magnetic Field of a Current Element
 Magnetic
field produced by a current element
0 q υ  rˆ
B
4 r 2
ds
r̂  r
For an element ds of a conductor carrying a current I there are n A ds
charges with drift velocity υd (using priciple of superposition).
number of charge q
0 qnA ds
dL υd  rˆ
dB 

2
4
r
ds  rˆ
0 I dL
dB 
4 r 2
r
Magnetic Field of a Current Element
 Biot-Savart
law
0 I dL
ds  rˆ
dB 
4 r 2
Note: ds is dL in the textbook.
ds
Note that ds is in the direction of I, but has a magnitude which is
ds the length of wire considered.
Deduced by Biot and Savart c. 1825 from experiments with coils
Magnetic Field of a Current Element
 Biot-Savart
law (cont’d)
0 I ds ×rˆ
dB 
;
2
4 r
dB P
r

ds
rˆ 
r
r
The magnitude of the field dB is:
I
0 I ds sin 
dB 
4 r 2
Total magnetic field at P is found by summing over all the current
elements ds in the wire.
B   dB
Magnetic Field of a Straight Current
Carrying Conductor
 A straight
wire of length L
• A thin straight wire of length L carries constant current I .
• Calculate the total B field at P.
0 I ds  rˆ
dB 
4 r 2
y
P
r
r̂
 dB
ds  rˆ always points out of the page
It has magnitude ds sin 
R

x
ds
I
x
So the magnitude of dB is given by:
0 I ds sin  0 I dx sin 
dB 

2
4
r
4
r2
Magnetic Field of a Straight Current
Carrying Conductor
 A straight
wire of length L (cont’d)
0 I ds sin  0 I dx sin 
dBz 

2
4
r
4
r2
y
P
r
r̂
 dB
R
 sin 
r
R

x
ds
I
;
r
x2  R2
0 I
R dx
dBz 
4  x 2  R 2 3/ 2
x
L/2
0 IR L / 2
0 IR
0 I 
dx
x
L

Bz 


2
2 3/ 2
2
2
2 1/ 2 | L / 2


L
/
2
4
(x  R )
4 R ( x  R )
4R  L2 / 4  R 2




Magnetic Field of a Straight Current
Carrying Conductor
 A straight
wire of length L (cont’d)
y
P
r
r̂
0 I 
L

Bz 
4R  L2 / 4  R 2
 dB
R





In the limit (L/R) →∞
x
ds
I
x

L

 2
2
L
/
4

R

Magnetic field by a long straight wire

L/R

2

2
( L / R) / 4  1

0 I
Bz 
2 R
Magnetic Field of a Straight Current
Carrying Conductor
 A straight
wire of length L (cont’d)
B
B
I
B
B
Magnetic Field of a Straight Current
Carrying Conductor
 Example:
A long straight wire
Iron filings
Magnetic Field of a Current Element
 Example
Calculate the magnetic field at point O due to
the wire segment shown. The wire carries
uniform current I, and consists of two straight
segments and a circular arc of radius R that
subtends angle .
A´
A
ds
r̂

C
R
O
I
C´ The magnetic field due to segments A´A and
CC´ is zero because ds is parallel to r̂ along
these paths.
Along path AC, ds and r̂ are perpendicular.
0 I ds ×rˆ
dB 
4 r 2
0 I
0 I
B
ds 
2 
4 R
4 R 2
d s  rˆ  ds
0 I ds
dB 
4 R 2
0 I
 R d  4 R
Note: B field at the centre of a
loop, =2
B
0 I
2R
0 I
 d  4 R 
Force Between Parallel Conductors
 Two
parallel wires
At a distance a from the wire with current I1 the
magnetic field due to the wire is given by
0 I1
B1 
2 a
F2  I 2 L  B1
0 I1
0 I1 I 2
F2  I 2 LB1  I 2 L

L
2 a
2 a
Force Between Parallel Conductors
 Two
parallel wires (cont’d)
Parallel conductors carrying current in the same direction attract
each other. Parallel conductors carrying currents in opposite
directions repel each other.
Force Between Parallel Conductors
 Definition
of ampere
0 I1
B1 
2 a
F2
F2  I 2 L  B1
0 I1 I 2

L
2 a
The chosen definition is that for a = L = 1m, The ampere is made to
be such that F2 = 2×10−7 N when I1=I2=1 ampere
This choice does two things (1) it makes the ampere (and also the volt)
have very convenient magnitudes for every day life and (2) it fixes the
size of μ0 = 4π×10−7. Note ε0 = 1/(μ0c2).
All the other units follow
almost automatically.
Magnetic Field of a Circular Current Loop
 Magnetic
field
produced by a loop current

 0 Ids  rˆ
Use B   4 r 2
to find B field at the center of a loop of wire.
I
R
Loop of wire lying in a plane. It has radius R and
total current I flowing in it.
First find

ds  rˆ is a vector coming out of the paper at the

ds
same angle anywhere on the circle. The
angle is constant.
r̂

ds
B
r̂
 0I
2R

ds  rˆ
 0 Ids  0 I
0 I
B   dB 

ds 
2R
2
2 
2

4 R
4 R
4 R
Magnitude of B field at center of
loop. Direction is out of paper.
i
R
k̂
  0I
B
kˆ
2R
Magnetic Field of a Circular Current Loop
 Example 1:
Loop of wire of radius R = 5 cm and current I = 10 A. What is B at the
center? Magnitude and direction
I
B
 0I
2R
N 10 A
B  4 107 2
A 2(.05m)
B  1.2 10 6 10 2 T
B  1.2 104 T  1.2 Gauss
Direction is out
of the page.
Magnetic Field of a Circular Current Loop
 Example
2:
What is the B field at the center of a segment or circular
arc of wire?
0 I
B
ds

2 
ds
4 R
I
r̂
Total length of arc is S.
0
R
P
B
0 I
S where S is the arc length S =R0
2
4 R
0 is in radians (not degrees)
Why is the contribution to the B field at P equal to zero from
the straight section of wire?
Ampere’s Law
 Ampere’s
law : A circular path
• Consider any circular path of radius R
centered on the wire carrying current I.
• Evaluate the scalar product B·ds around this
path.
• Note that B and ds are parallel at all points
along the path.
• Also the magnitude of B is constant on this
path. So the sum of all the B·ds terms around
the circle is
 B  ds  B  ds  B  2 r 
Previously from the Biot-Savart’s law we had
On substitution for B
 B  ds  
0
I
0 I
B
2 r
Ampere’s Law
Ampere’s Law
 Ampere’s
z
^
k
law : A general path
^
x
r

y
^

Let us look at the integral along any shape of closed
path in 3D. The most general ds is
r̂  r d τˆ dz k̂k
ds  dr u
Where unit vectors are used for the radial
^
^
r and the tangential directions  and for z
^
along the wire k. In this system we have
tangential component of ds
0 I r d
B  ds 
2 r
0 I ˆ
B
τ
2 r
0 I

d .
2
For any path which encloses the wire

only.
0 I
B  ds 
2
 d
For any path which does not enclose the wire
 d
 2
 d
0
Ampere’s Law
 Ampere’s
law :
 
 B  ds  0 I
• This law holds for an arbitrary closed path that is threaded by a
steady current.
• I is the total current that passes through a surface bounded by the
closed path.
Ampere’s Law

Electric field vs. magnetic field
• Electric Field
• General: Coulomb’s Law
• High symmetry: Gauss’s Law
• Magnetic Field
• General: Biot-Savart Law
• High Symmetry: Ampère’s
Law
Applications of Ampere’s Law
 Magnetic
field by a long cylindrical conductor
A long straight wire of radius R carries a steady current I that is uniformly
distributed through the cross-section of the wire. Outside R.
 B  ds  B  ds  B  2 r   
I
0 tot
In region where r < R choose a circle of radius r centered on the wire as a path of
integration. Along this path, B is again constant in magnitude and is always
parallel to the path.
• Now Itot ≠ I.
• However, current is uniform over the cross-section
of the wire.
• Fraction of the current I enclosed by the circle of
radius r < R equals the ratio of the area of the circle
of radius r and the cross section of the wire R2.
I tot
2
I
r
2
 j r 2 

r
 2I
2
R
R
0 I tot 0 I r
B

2 r
2 R 2
for r  R
Applications of Ampere’s Law
 Magnetic
field by a long cylindrical conductor
0 I r
B
2 R 2
0 I
B
2 r
for r  R
for r  R
B
R
r
Applications of Ampere’s Law
 Magnetic
field by a circular current
Consider circular current carrying loop.
Calculate B field at point P, a dist x from
the centre of the loop on the axis of the
loop.
ds  rˆ
0 I dL
dB 
4 r 2
Again in this case vector I ds is tangent to
loopand perp to vector r from current element
to point P. dB is in direction shown, perp to
vectors r and I ds. Magnitude dB is:
ds
ds
0 I dL
0 I dL
dB 

2
4 r
4 ( x 2  R 2 )
Ids
Applications of Ampere’s Law
 Magnetic
field by a circular current (cont’d)
ds
ds
Ids
0 I dL
0 I dL
dB 

2
4 r
4 ( x 2  R 2 )
• Integrate around loop, all
components of dB perp to axis (e.g.
dBy). integrate to zero.
• Only dBx , the components parallel
to axis contribute.

dBx  dB sin   dB 


Field due to entire loop
obtained by integrating:
 0 I dL
ds 

2
2 

2
2 
x  R  4 x  R 
R
Bx 
 dBx


2
2 
x R 
R
Applications of Ampere’s Law
 Magnetic
field by a circular current (cont’d)
But I, R and x are constant
Bx 
 dBx


Bx 
ds
0
IR dL
4 ( x 2  R 2 ) 3 2
 IR (2 R)
Bx  0
4 ( x 2  R 2 ) 3 2
 dBx
0
IR

4 ( x 2  R 2 ) 3 2
Ids
ds
 dL
0
R2 I

2 ( x2  R2 ) 3 2
B on the axis of
a current loop
Applications of Ampere’s Law
 Magnetic
Bx 
0
field by a circular current (cont’d)
R2 I
2 ( x2  R2 ) 3 2
x >>R
Limits: x 0
0 2 R 2 I
0 I
Bx 
B
4 x3
2R

0 2 
4 x3
  I  R 2  IA mag moment of loop
Compare case of electric field on axis of electric
dipole far from dipole
Ex 
2p
4 0 x3
vs.
0 2 R 2 I
Bx 
4 x3

0 2 
4 x3
Applications of Ampere’s Law
 Magnetic
field by a solenoid
When the coils of the solenoid are closely spaced, each turn can be regarded as
a circular loop, and the net magnetic field is the vector sum of the magnetic
field for each loop. This produces a magnetic field that is approximately
constant inside the solenoid, and nearly zero outside the solenoid.
I
Applications of Ampere’s Law
 Magnetic
field by a solenoid (cont’d)
The ideal solenoid is approached when the coils are very close together
and the length of the solenoid is much greater than its radius. Then we
can approximate the magnetic field as constant inside and zero outside the
solenoid.
I
Applications of Ampere’s Law
 Magnetic
field by a solenoid (cont’d)
Use Ampère’s Law to find B inside an ideal solenoid.
 B  ds   B  ds   B  ds   B  ds   B  ds
12
23
 B  ds  BL  0  0  0
B  0 I
34
 BL
41
 0 NI
N
 0 I times number of turns per unit length
L
Applications of Ampere’s Law
 Magnetic
field by a toroid
A toroid can be considered as a solenoid “bent” into a circle as shown. We
can apply Ampère’s law along the circular path inside the toroid.
 Bd s   I
0 encl
 B  d s  B  ds  B(2 r )
I encl  NI
N is the number of loops in the toroid, and
I is the current in each loop
0 NI
B
2 r
Exercises
 Problem
1
The wire semicircles shown in Fig. have radii a and b. Calculate the net
magnetic field that the current in the wires produces at point P.
I
Since point P is located at a symmetric position
I
with respect to the two straight sections where the
b
current I moves (anti)parallel to the radial direction.
So there is no contributions from these segments.
a
The contribution from the semicircle of radius a is
P
a half of that from a complete circle of the same radius:
1 0 I
Ba 
2 2a
(out of page)
Similarly the contribution from the semicircle of radius b is:
1 0 I
Bb 
2 2b
(into page)
From principle of superposition, the net magnetic field at point P is:
1   I  1 1   I  a 
B  Ba  Bb   0     0 1   (out of page)
2  2  a b  4a  b 
Exercises
 Problem
2
Long, straight conductors with square cross sections and
each carrying current I are laid side-by-side to form an
infinite current sheet. The conductors lie in the xy-plane,
are parallel to the y-axis and carry current in the +y
direction. There are n conductors per unit length measured
along the x-axis. (a) What are the magnitude and direction
of the magnetic a distance a below the current sheets?
(b) What are the magnitude and direction of the magnetic
field a distance a above the current sheet?
y
x
z
Exercises
 Problem
2 (cont’d)
B
a) Below sheet, all the magnetic
field contributions from different
B
wires add up to produce a magnetic
L
field that points in the positive x-direction. Components in the z-direction
cancel. Using Ampere’s law, where we use the fact that the field is antisymmetric above and below the current sheets, and that the legs of the path
perpendicular provide nothing to the integral. So, at a distance a beneath the
sheet the magnetic field is:
 
 nI
I encl  nLI ,  B  ds  B(2 L)  0 nLI  B  0 in the positive x - direction.
2
b) The field has the same magnitude above the sheet, but points in the negative
x-direction.