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)
A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e) which is also traveling perpendicular to the same field. The ratioof their speeds, vproton/valpha, is:
A. 0.5B. 1C. 2D. 4E. 8
A proton (charge e), traveling perpendicular to a magnetic field in a circular orbit, experiences the same force as an alpha particle (charge 2e, mass 4p) which is also traveling perpendicular to the same field. The ratio of their orbital radii rproton/ralpha, is:
A. 0.5 B. 1 C. 2 D. 4 E. 8
A proton (charge e), traveling perpendicular to a magnetic field in a circular orbit, experiences the same force as an alpha particle (charge 2e, mass 4p) which is also traveling perpendicular to the same field. The ratio of their orbital frequency fproton /f alpha is:
This law gives the magnetic field generated by a wire
segment of length that carries a current . Consider
the geometry shown in the figure. Associated with the
element
dB
ds i
ds
The law of Biot - Savart
we define an associated vector that has
magnitude whch is equal to the length . The direction
of is the same as that of the current that flows through
segment .
ds
ds
ds
ds
A
3
The magnetic field generated at point P by the element located at point A
is given by the equation: . Here is the vector that connects 4
point A (location of element ) w
o
dB dS
i ds rdB r
r
ds
7 6
2
ith point P at which we want to determine .
The constant 4 10 T m/A 1.26 10 T m/A and is known as
sin" ". The magnitude of is:
4Here is the angle
o
o
dB
i dsdB dB
r
permeability constant
between and . ds r
34oi ds r
dBr
(29 – 2)
The magnitude of the magnetic field
generated by the wire at point P
located at a distance from the wire
is given by the equation:
2
oiBR
R
Magnetic field generated by a long
straight wire
2
oiBR
The magnetic field lines form circles that have their centers
at the wire. The magnetic field vector is tangent to the
magnetic field lines. The sense for is given by the
. We po
B
B right
hand rule
int 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)
2
Consider the wire element of length shown in the figure.
The element generates at point P a magnetic field of
sinmagnitude Vector is pointing
4 the page. The magnetic field
o
ds
i dsdB dB
r
into
20 0
2 2 2 2
generated by the whole
wire is found by integration.
sin2
2
sin sin / /
oi dsB dB dB
r
r s R R r R s R
3/ 2
3/ 2 2 2
2 2 22 2
2 20 0
2 2 2o o oi i
dx x
a x a
iRd
x
s sB
R Rs Rs R
a
2oiBR
(29 – 4)
. .dB
B
R
Magnetic field generated by
a circular wire arc of radius
at its center C
2 2
A wire section of length generates at the center C a magnetic field
sin 90The magnitude The length
4 4
Vector points of the page4
The
o o
o
ds dB
i ids dsdB ds Rd
R Ri
dB d dBR
out
0
net magnettic field =4
The angle must be expressed in radians
For a circular wire 2 . In this case we get:
4
2
o
ocirc
o i
R
iB
iB dB
R
R
d
Note :
(29 – 5)
4oiBR
2 2
The wire element generates a magnetic field
sin 90whose magnitude
4 4
We decompose into two components:
One (
o o
ds dB
i ids dsdB
r r
dB
Magnetic field generated by a circular loop
along the loop axis
d ) along the z-axis.The second component (d )
is in a direction perpendiculat to the z-axis. The sum of
all the d is equal to zero. Thus we sum only the d terms.
B B
B B
2
3/ 22 2
2 22
3/ 23 2 2
3/ 2 3/ 22 2 2 2
cos cos cos
4
4 4
24 4 2
o
o o
o oo
i ds Rr R z dB dB
r riR iRds ds
dB B dBr R z
iR iRB ds R
R z R z
iR
R z
(29 – 6)
The law of Biot-Savart combined with the
principle of superposition can be used to
determine if we know the distribution
of currents. In situations that have high
symmetry we can use inst
B
Ampere's Law.
ead 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:
o
The line integral of the magnetic field along any closed path
is equal to the total current enclosed inside the path multiplied by
B ds B
The closed path used is known an an " ". 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
Amperian loop
chapter 32.
o encB ds i
(29 – 7)
o encB ds i
1 2
1 1
Determination of . The closed path is
divided into n elements , , ...,
We then from the sum:
cos Here is the
mag
n
n n
i i i i i ii i
B ds
s s s
B s B s B
Implementation of Ampere's law :
1.
1
netic field in the i-th element.
lim as
Calculation of We curl the fingers
of the right hand in the direction in which
the Amperian loop was traversed. We note
the
n
i ii
enc
B ds B s n
i
2.
direction of the thumb.
All currents inside the loop to the thumb are counted as .
All currents inside the loop to the thumb are counted as .
All currents outside the loop are not count
pa
antiparal
rallel positive
lel negative
1 2
ed.
In this exampl e : enci i i (29 – 8)
We already have seen that the magnetic field lines
of the magnetic field generated by a long straight
wire that carries a current have the form of
circles
i
Magnetic field outside a long straight wire
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 that has its center
on the wire. The magnetic fi
r
eld is tangent to the
loop and has a constant magnitude .
cos0 2
Ampere's law holds true for any closed path.
We choose to use the path that makes the
2
o enc
o
o
B
B ds Bds B ds rB i
iB
r
i
Note :
calculation
of as easy as possible.
B
(29 – 9)
We assume that the distribution of the current
within the cross-section of the wire is uniform.
The wire carries a current and has radius .
We choose an Amper
i R
Magnetic field inside a long straight wire
2 2
2 2
ian loop is a circle of
radius ( ) that has its center
on the wire. The magnetic field is tangent to the
loop and has a constant magnitude .
cos0 2 o enc
enc
r r R
B
B ds Bds B ds rB i
r ri i i
R R
2
2 222 o
o
iB r
R
rrB i
R
R
2oi
R
r
B
O (29 – 10)
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 ca
The solenoid
n see that the magnetic
field inside is approximately uniform.
The magnetic field inside the solenoid is parallel to the solenoid axis. The sense of
can be determined using the right hand rule. We curl the fingers of the right hand
along the direction of the c
B
urrent in the coil windings. The thumb of the right hand
points along . The magnetic field outside the solenoid is much weaker and can be
taken to be approximately zero.
B
(29 – 11)
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 turns per
u
n
nit length
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
a
B ds B ds
cos0 0
The enclosed current
c d a
b c d
b b b c d a
a a a b c d
enc
o en o oc
B ds B ds B ds
B ds Bds B ds Bh B ds B ds B ds
B ds Bh i nhi
B ds i Bh nh B nii
oB ni
(29 – 12)
A toroid has the shape of a doughnut (see figure)
We assume that the toroid carries a current and that
it has windings. The magnetic field lines inside the toroid
form cir
i
N
Magnetic field of a toroid :
cles that are cocentric with the toroid center.
The magnetic field vector is tangent to these lines.
The sense of can be found using the right hand rule.
We curl the fingers of the right hand
along the
B
direction of the current in the coil windings.
The thumb of the right hand points along . The magnetic
field outside the solenoid is approximately zero.
B
We use an Amperian loop that is a circle of radius (orange circle in the figure)
cos0 2 . The enclosed current
Thus: 2
The magnetic field i 2o
enc
o
r
B ds Bds B ds
Ni
rB i N
rB B
i
Nr
i
Note :
nside a toroid is not uniform.
2oNiBr
(29 – 13)
2
2
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:
2
oiRBR
The magnetic field of a magnetic dipole.
3/ 22
2
3
2
3 3 3
Here is the distance between
P and the coil center. For points far from the loop
( ) we can use the approximation: 2
Here is the magnetic2 2 2
dipole m
o
o o o
zz
iRz R B
z
i R iAB
pz z z
3
oment of the loop. In vector form:
( )2
The loop generates a magnetic field that has the same
form as the field generated by a bar magnet.
oB zz
3( )
2oB zz
(29 – 14)
An electron and a proton each travel with equal speeds around circular orbits in the same uniform magnetic field. The field is into the page. Because the electron is less massive than the proton and because the electron is negatively charged and the proton is positively charged:
A. the electron travels clockwise around the smaller circle and the proton travels counter- clockwise around the larger circleB. the electron travels counterclockwise around the smaller circle and the proton travels clockwise around the larger circleC. the electron travels clockwise around the larger circle and the proton travels counterclockwise around the smaller circleD. the electron travels counterclockwise around the larger circle and the proton travels clock- wise around the smaller circleE. the electron travels counterclockwise around the smaller circle and the proton travels counterclockwise around the larger circle
The diagram shows a straight wire carrying current i in a uniform magnetic field. The magnetic force on the wire is indicated by an arrow but the magnetic field is not shown. Of the following possibilities, the direction of the magnetic field is:
A. opposite the direction of the currentB. opposite the direction of ~FC. in the direction of ~FD. into the pageE. out of the page
i
F