Top view
 net  iAB sin 
Side view
(28 – 13)
CFnet  0
C
Magnetic torque on a current loop
Consider the rectangular loop in fig.a with sides of lengths a and b which carries
a current i. The loop is placed in a magnetic field so that the normal nˆ to the loop
forms an angle  with B. The magnitude of the magnetic force on sides 1 and 3 is:
F1  F3  iaB sin 90  iaB. The magnetic force on sides 2 and 4 is:
F2  F4  ibB sin(90   )  ibB cos  . These forces cancel in pairs and thus Fnet  0
The torque about the loop center C of F2 and F4 is zero because both forces pass
through point C. The moment arm for F1 and F3 is equal to (b / 2) sin  . The two
torques tend to rotare the loop in the same (clockwise) direction and thus add up.
The net torque    1 + 3 =(iabB / 2) sin   (iabB / 2) sin   iabB sin   iAB sin 
Magnetic dipole moment :
  B
The torque of a coil that has N loops exerted
by a uniform magnetic field B and carrries a
current i is given by the equation:   NiAB
U    B
We define a new vector  associated with the coil
which is known as the magnetic dipole moment of
U  B
U   B
the coil.
The magnitude of the magnetic dipole moment   NiA
Its direction is perpendicular to the plane of the coil
The sense of  is defined by the right hand rule. We curl the fingers of the right hand
in the direction of the current. The thumb gives us the sense. The torque can
expressed in the form:    B sin  where  is the angle between  and B.
In vector form:     B
The potential energy of the coil is: U    B cos      B
U has a minimum value of   B for   0 (position of stable equilibrium)
U has a maximum value of  B for   180 (position of unstable equilibrium)
Note : For both positions the net torque   0
(28 – 14)
The Hall effect
In 1879 Edwin Hall carried out an experiment in which
R
he was able to determine that conduction in metals is due
to the motion of negative charges (electrons). He was also
able to determine the concentration n of the electrons.
He used a strip of copper of width d and thickness . He passed
a current i along the length of the strip and applied a magnetic
field B perpendicular to the strip as shown in the figure. In the
R L
L
presence of B the electrons experience a magnetic force FB that
L
R
pushes them to the right (labeled "R") side of the strip. This
accumulates negative charge on the R-side and leaves the left
side (labeled "L") of the strip positively charged. As a result
of the accumulated charge, an electric field E is generated as
shown in the figure so that the electric force balances the magnetic
force on the moving charges. FE  FB  eE  evd B 
E  vd B (eqs.1). From chapter 26 we have: J  nevd 
vd 
J
i
i


ne Ane
dne
(eqs.2)
(28 – 15)
R L
L
R
E  vd B (eqs.1).
vd  i / dne
(eqs.2)
Hall measured the potential difference V between the left and
the right side of the metal strip. V  Ed (eqs.3)
We substitute E from eqs.3 and vd from eqs.(2) into eqs.1 and get:
V
i
Bi
B

n
(eqs.4)
d
dne
V e
Fig. a and b were drawn assuming that the carriers are electrons.
In this case if we define V  VL  VR we get a positive value.
L
R
(28 – 16)
If we assume that the current is due to the motion
of positive charges (see fig.c) then positive charges accumulate on
the R-side and negative charges on the L-side and thus V  VL  VR
is now a negative number.
By determining the polarity of the voltage develops between the
left and right hand side of the strip Hall was able to prove that
current was composed of moving electrons. From the value of V
using equation 4 he was able to determine the concentration of
the negative charge carriers.
(28 – 17)
The cyclotron particle accelerator
The cyclotron accelerator consissts of two hollow
conductors in the shape of the letter dee (these are
known as the "dees" of the cyclotron. Between the
two dees an oscillator of frequency f osc creates an
oscillating electric field E that exists only in the gap
between the two dees. At the same time a constant
mv
eB
magnetic field B is applied perpendicular to the
r
f 
eB
2 m
plane of the dees.
In the figure we show a cyclotron accelerator for protons. The protons follow circular
mv
eB
orbits of radius r 
and rotate with the same frequency f 
. If the cyclotron
eB
2 m
frequency matches the oscillator frequency thne the protons during their trip through
the gap between the dees are accelerated by the electric field that exists in the gap.
The faster protons travel on increasingly larger radius orbits. Thus the electric
field changes the speed of the protons while the magnetic field changes only
the direction of their velocity and forces them to move on circular (cyclotron) orbits.
hitt
A proton undergoes a circular motion at a
speed of 4 × 106 m/s in a uniform magnetic
field of 4 T. What is the period of the
motion?
A. 16 ns B. 32 ns C. 8.9 ps D. 18 ps
E. 27 ps.
Phy 2049: Magnetism etc.
•
•
•
•
Magnetic fields and magnetic forces
Torque on a current loop
Hall effect
Magnetic field produced by currents
*
*
– Biot-Savart Law
– Ampere law
• Force between two current carrying wires.
Chapter 29
Magnetic Fields Due to Currents
In this chapter we will explore the relationship between an electric
current and the magnetic field it generates in the space around it.
We will follow a two-prong approach, depending on the symmetry
of the problem.
For problems with low symmetry we will use the law of Biot-Savart
in combination with the principle of superposition.
For problems with high symmetry we will introduce Ampere’s law.
Both approaches will be used to explore the magnetic field
generated by currents in a variety of geometries (straight wire, wire
loop, solenoid coil, toroid coil)
We will also determine the force between two parallel current
carrying conductors. We will then use this force to define the SI
unit for electric current (the Ampere)
(29 – 1)
oi ds  r
dB 
4 r 3
A
The law of Biot - Savart
This law gives the magnetic field dB generated by a wire
segment of length ds that carries a current i. Consider
the geometry shown in the figure. Associated with the
element ds we define an associated vector ds that has
magnitude whch is equal to the length ds. The direction
of ds is the same as that of the current that flows through
segment ds.
The magnetic field dB generated at point P by the element dS located at point A
 i ds  r
is given by the equation: dB  o
. Here r is the vector that connects
3
4 r
point A (location of element ds ) with point P at which we want to determine dB.
The constant o  4 10 7 T  m/A  1.26 10 6 T  m/A and is known as
"permeability constant". The magnitude of dB is: dB 
Here  is the angle between ds and r .
oi ds sin 
4
r2
(29 – 2)
Magnetic field generated by a long
straight wire
The magnitude of the magnetic field
generated by the wire at point P
located at a distance R from the wire
is given by the equation:
i
B o
2 R
B
o i
2 R
The magnetic field lines form circles that have their centers
at the wire. The magnetic field vector B is tangent to the
magnetic field lines. The sense for B is given by the right
hand rule. We point the thumb of the right hand in the
direction of the current. The direction along which the
fingers of the right hand curl around the wire gives the
direction of B.
(29 – 3)
B
oi
2 R
Consider the wire element of length ds shown in the figure.
The element generates at point P a magnetic field of
 i ds sin 
magnitude dB  o
Vector dB is pointing
2
4
r
into the page. The magnetic field generated by the whole
wire is found by integration.


oi  ds sin 
B   dB  2 dB 
2

2

r

0
0


r  s2  R2
sin   sin   R / r  R / s 2  R 2


o i
o i 
o i
Rds
s
B

 2
 
3/ 2

2
2
2
2 0  s  R 
2 R  s  R  0 2 R


dx
x  a
2

2 3/ 2

x
a2 x2  a2
(29 – 4)
B
.
Magnetic field generated by
dB
.
a circular wire arc of radius
R at its center C
o i
B
4 R
A wire section of length ds generates at the center C a magnetic field dB
oi ds sin 90 oi ds
The magnitude dB 

The length ds  Rd
2
2
4
R
4 R
i
 dB  o d
Vector dB points out of the page
4 R

o i
 i
d = o
4 R
4 R
0
The net magnettic field B   dB  
Note : The angle  must be expressed in radians
For a circular wire   2 . In this case we get: Bcirc 
o i
2R
(29 – 5)
Magnetic field generated by a circular loop
along the loop axis
The wire element ds generates a magnetic field dB
 i ds sin 90 oi ds
whose magnitude dB  o

2
4
r
4 r 2
We decompose dB into two components:
One (dB ) along the z-axis.The second component (dB )
is in a direction perpendiculat to the z-axis. The sum of
all the dB is equal to zero. Thus we sum only the dB terms.
oi ds cos 
R
r R z
dB  dB cos  
cos  
2
4
r
r
oiR ds oiR
ds
 dB 

B   dB
3/ 2
3
2
2
4 r
4  R  z 
2
B
2
oiR
4  R  z
2

2 3/ 2
 ds  4
oiR
R
2
z

2 3/ 2
 2 R  
oiR 2
2R  z
2

2 3/ 2
(29 – 6)
(29 – 7)
Ampere's Law.
The law of Biot-Savart combined with the
principle of superposition can be used to
 B  ds   i
o enc
determine B if we know the distribution
of currents. In situations that have high
symmetry we can use instead Ampere's
law, because it is simpler to apply.
Ampere's law can be derived from the law of Biot-Savart with which it is
mathematically equivalent. Ampere's law is more suitable for advances
formulations of electronmagnetism. It can be expressed as follows:
The line integral
 B  ds of the magnetic field B along any closed path
is equal to the total current enclosed inside the path multiplied by o
The closed path used is known an an "Amperian loop". In its present form
Ampere's law is not complete. A missing term was added by Clark Maxwell.
The complete form of Ampere's law will be discussed in chapter 32.
 B  ds   i
o enc
Implementation of Ampere's law :
 B  ds .
1. Determination of
The closed path is
divided into n elements s1 , s2 , ..., sn
We then from the sum:
n
n
 B  s   B s cos 
i 1
i
i
i
i 1
i
i
Here Bi is the
magnetic field in the i-th element.
n
 B  ds  lim  B  s
i 1
i
i
as n  
2. Calculation of ienc We curl the fingers
of the right hand in the direction in which
the Amperian loop was traversed. We note
the direction of the thumb.
All currents inside the loop parallel to the thumb are counted as positive.
All currents inside the loop antiparallel to the thumb are counted as negative.
All currents outside the loop are not counted.
(29 – 8)
In this example : ienc  i1  i2
Magnetic field outside a long straight wire
We already have seen that the magnetic field lines
of the magnetic field generated by a long straight
wire that carries a current i have the form of
circles which are cocentric with the wire.
We choose an Amperian loop that reflects the
cylindrical symmetry of the problem. The
loop is also a circle of radius r that has its center
on the wire. The magnetic field is tangent to the
loop and has a constant magnitude B.
 B  ds   Bds cos 0  B  ds  2 rB   i
o enc
B
 oi
oi
2 r
Note : Ampere's law holds true for any closed path.
We choose to use the path that makes the calculation
of B as easy as possible.
(29 – 9)
Magnetic field inside a long straight wire
We assume that the distribution of the current
within the cross-section of the wire is uniform.
The wire carries a current i and has radius R.
We choose an Amperian loop is a circle of
radius r (r  R) that has its center
on the wire. The magnetic field is tangent to the
loop and has a constant magnitude B.
 B  ds   Bds cos 0  B  ds  2 rB   i
o enc
ienc
B
r2
 i 
2 rB  oi 2  B   o 2  r
R
 2 R 
oi
2 R
O
 r2
r2
i
i 2
 R2
R
R
r
(29 – 10)
(29 – 11)
The solenoid
The solenoid is a long tightly wound helical wire
coil in which the coil length is much larger than the
coil diameter. Viewing the solenoid as a collection
of single circular loops one can see that the magnetic
field inside is approximately uniform.
The magnetic field inside the solenoid is parallel to the solenoid axis. The sense of B
can be determined using the right hand rule. We curl the fingers of the right hand
along the direction of the current in the coil windings. The thumb of the right hand
points along B. The magnetic field outside the solenoid is much weaker and can be
taken to be approximately zero.
We will use Ampere's law to determine
the magnetic field inside a solenoid. We
assume that the magnetic field is uniform
inside the solenoid and zero outside. We
assume that the solenolis has n turns per
unit length
B  o ni
We will use the Amperian loop abcd. It is a rectangle with its long side parallel
to the solenoid axis. One long side (ab) is inside the solenoid, while the other (cd)
is outside.
b
c
d
a
a
b
c
d
 B  ds   B  ds   B  ds   B  ds   B  ds
b
b
b
a
a
a
 B  ds   Bds cos 0  B  ds  Bh

 B  ds  Bh
 B  ds   i
o enc
c
d
a
b
c
d
 B  ds   B  ds  B  ds  0
The enclosed current ienc  nhi
Bh  o nhi  B  o ni
(29 – 12)
B
o Ni
2 r
Magnetic field of a toroid :
A toroid has the shape of a doughnut (see figure)
We assume that the toroid carries a current i and that
it has N windings. The magnetic field lines inside the toroid
form circles that are cocentric with the toroid center.
The magnetic field vector is tangent to these lines.
The sense of B can be found using the right hand rule.
We curl the fingers of the right hand
along the direction of the current in the coil windings.
The thumb of the right hand points along B. The magnetic
field outside the solenoid is approximately zero.
We use an Amperian loop that is a circle of radius r (orange circle in the figure)
 B  ds   Bds cos 0  B  ds  2 rB.
Thus: 2 rB  o Ni  B 
The enclosed current ienc  Ni
o Ni
2 r
Note : The magnetic field inside a toroid is not uniform.
(29 – 13)
The magnetic field of a magnetic dipole.
Consider the magnetic field generated by a wire coil of
radius R which carries a current i. The magnetic field
at a point P on the z-axis is given by:
B
 
B( z )  o 3
2 z
oiR 2
2R  z
2

2 3/ 2
Here z is the distance between
P and the coil center. For points far from the loop
(z
B
R) we can use the approximation: B 
oi R 2
3

oiA o 

3
2 z
2 z 3
oiR 2
2z3
Here  is the magnetic
2 pz
dipole moment of the loop. In vector form:
B( z ) 
o 
2 z 3
The loop generates a magnetic field that has the same
form as the field generated by a bar magnet.
(29 – 14)
hitt
A proton undergoes a circular motion at a
speed of 4 × 106 m/s in a uniform magnetic
field of 4 T. What is the radius of the circular
motion?
A. 0.066 m B. 0.01 m C. 5.7 μm D. 12 μm
E. 0.13 m.