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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.
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 ratio
of their speeds, vproton/valpha, is:
A. 0.5
B. 1
C. 2
D. 4
E. 8
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 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
r µ i dsr × rr
dB = o
4π r 3
The law of Biot - Savart
r
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
r
element ds we define an associated vector ds that has
magnitude whch is equal to the length ds. The direction
r
of ds is the same as that of the current that flows through
segment ds.
A
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:
r
r
The magnetic field dB generated at point P by the element dS located at point A
r
r
r µ i ds × r
r
is given by the equation: dB = o
. Here r is the vector that connects
4π r 3
r
point A (location of element ds ) with point P at which we want to determine dB.
The constant µo = 4π ×10 −7 T ⋅ m/A = 1.26 ×10 −6 T ⋅ m/A and is known as
r
µ i ds sin θ
"permeability constant". The magnitude of dB is: dB = o
4π r 2
r
r
Here θ is the angle between ds and r .
(29 – 2)
B=
Magnetic field generated by a long
straight wire
µo i
2π R
Consider the wire element of length ds shown in the figure.
The element generates at point P a magnetic field of
r
µ i ds sin θ
magnitude dB = o
Vector dB is pointing
4π
r2
into the page. The magnetic field generated by the whole
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:
µi
B= o
2π R
B=
wire is found by integration.
φ
µo i
2π R
B=
×
∞
∞
−∞
0
r = s +R
2
The magnetic field lines form circles that have their centers
r
at the wire. The magnetic field vector B is tangent to the
r
magnetic field lines. The sense for B is given by the right
µ oi ∞ ds sin θ
∫ dB = 2∫ dB = 2π ∫
2
r2
0
sin θ = sin φ = R / r = R / s 2 + R 2
∞
B=
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
r
direction of B.
(29 – 3)
(x

µo i ∞
Rds
µi 
s
µ oi
= o 
 =
2π ∫0 ( s 2 + R 2 )3/ 2 2π R  s 2 + R 2  0 2π R
dx
2
+ a2 )
3/ 2
=
x
a 2 x2 + a 2
(29 – 4)
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r
B
r
dB
.
Magnetic field generated by a circular loop
Magnetic field generated by
a circular wire arc of radius
R at its center C
.
B=
along the loop axis
r
The wire element ds generates a magnetic field dB
µ i ds sin 90° µoi ds
whose magnitude dB = o
=
4π
r2
4π r 2
r
We decompose dB into two components:
µoiφ
4π R
One (dB ) along the z-axis.The second component (dB⊥ )
r
A wire section of length ds generates at the center C a magnetic field dB
The magnitude dB =
→ dB =
µ oi
dφ
4π R
µoi ds sin 90° µoi ds
=
4π
R2
4π R 2
is in a direction perpendiculat to the z-axis. The sum of
The length ds = Rdφ
all the dB⊥ is equal to zero. Thus we sum only the dB terms.
r
Vector dB points out of the page
µo i
µ iφ
dφ = o
4π R
4π R
0
→ dB =
Note : The angle φ must be expressed in radians
For a circular wire φ = 2π . In this case we get: Bcirc =
dB = dB cos α =
r = R2 + z 2
φ
The net magnettic field B = ∫ dB = ∫
µo i
2R
B=
(29 – 5)
(29 – 7)
Ampere's Law.
The law of Biot-Savart combined with the
ds
µ oiR ds µoiR
=
4π r 3
4π ( R 2 + z 2 )3/ 2
µoiR
4π ( R 2 + z 2 )
r
3/2
∫ ds = 4π
o enc
i =1
i =1
r
Here Bi is the
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.
The complete form of Ampere's law will be discussed in chapter 32.
In this example :
ienc = i1 − i2
(29 – 8)
Magnetic field outside a long straight wire
We already have seen that the magnetic field lines
Magnetic field inside a long straight wire
We assume that the distribution of the current
of the magnetic field generated by a long straight
within the cross-section of the wire is uniform.
The wire carries a current i and has radius R.
wire that carries a current i have the form of
circles which are cocentric with the wire.
We choose an Amperian loop is a circle of
We choose an Amperian loop that reflects the
cylindrical symmetry of the problem. The
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.
r r
∫ B ⋅ ds = ∫ Bds cos 0 = B ∫ ds = 2π rB = µoienc
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.
r r
∫ B ⋅ ds = ∫ Bds cos 0 = B ∫ ds = 2π rB = µoienc = µoi
We choose to use the path that makes the calculation
r
of B as easy as possible.
(29 – 9)
(29 – 6)
i =1
Ampere's law is not complete. A missing term was added by Clark Maxwell.
µoi
2π r
Note : Ampere's law holds true for any closed path.
3/ 2
2. Calculation of ienc We curl the fingers
r r
r
∫ B ⋅ ds of the magnetic field B along any closed path
→B=
µoiR 2
2 ( R2 + z2 )
magnetic field in the i-th element.
n r
r r
r
∫ B ⋅ ds = lim ∑ Bi ⋅ ∆si as n → ∞
law, because it is simpler to apply.
The closed path used is known an an "Amperian loop". In its present form
( 2π R ) =
We then from the sum:
n r
n
r
∑ Bi ⋅ ∆si = ∑ Bi ∆si cos θi
of currents. In situations that have high
symmetry we can use instead Ampere's
is equal to the total current enclosed inside the path multiplied by µ o
3/ 2
1. Determination of
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
R
r
B = ∫ dB
µoiR
( R2 + z 2 )
cos α =
Implementation of Ampere's law :
r r
∫ B ⋅ ds . The closed path is
r
r
r
divided into n elements ∆s1 , ∆s2 , ..., ∆sn
r
∫ B ⋅ ds = µ i
principle of superposition can be used to
r
determine B if we know the distribution
r r
∫ B ⋅ ds = µoienc
µ oi ds cos α
r2
4π
ienc = i
B
µo i
2π R
O
π r2
r2
=i 2
π R2
R
2π rB = µoi
R
r
r2
 µi 
→ B =  o 2 r
R2
 2π R 
(29 – 10)
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We will use Ampere's law to determine
(29 – 11)
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 = µo ni
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.
r
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
r
points along B. The magnetic field outside the solenoid is much weaker and can be
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)
r r b r r c r r d r r a r r
is outside. ∫ B ⋅ ds = ∫ B ⋅ ds + ∫ B ⋅ ds + ∫ B ⋅ ds + ∫ B ⋅ ds
a
B=
µo Ni
2π r
c
d
b
r r b
∫ B ⋅ ds = ∫ Bds cos 0 = B ∫ ds = Bh
r r d r r a r r
∫ B ⋅ ds = ∫ B ⋅ ds = ∫ B ⋅ ds = 0
a
b
→
r
a
a
r
∫ B ⋅ ds = Bh
r
r
∫ B ⋅ ds = µ i
o enc
taken to be approximately zero.
b
b
c
c
d
The enclosed current ienc = nhi
→Bh = µo nhi → B = µo ni
(29 – 12)
Magnetic field of a toroid :
The magnetic field of a magnetic dipole.
A toroid has the shape of a doughnut (see figure)
We assume that the toroid carries a current i and that
Consider the magnetic field generated by a wire coil of
it has N windings. The magnetic field lines inside the toroid
at a point P on the z-axis is given by:
form circles that are cocentric with the toroid center.
The magnetic field vector is tangent to these lines.
r
The sense of B can be found using the right hand rule.
B=
We curl the fingers of the right hand
along the direction of the current in the coil windings.
r
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)
r r
∫ B ⋅ ds = ∫ Bds cos 0 = B ∫ ds = 2π rB. The enclosed current ienc = Ni
Thus: 2π rB = µo Ni → B =
µo Ni
2π r
Note : The magnetic field inside a toroid is not uniform.
(29 – 13)
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 circle
B. the electron travels counterclockwise around the smaller circle and
the proton travels clockwise around the larger circle
C. the electron travels clockwise around the larger circle and the
proton travels counterclockwise around the smaller circle
D. the electron travels counterclockwise around the larger circle and
the proton travels clock- wise around the smaller circle
E. the electron travels counterclockwise around the smaller circle and
the proton travels counterclockwise around the larger circle
radius R which carries a current i. The magnetic field
r
r
µ µ
B( z ) = o 3
2π z
µ oiR 2
2 ( R2 + z 2 )
3/ 2
Here z is the distance between
P and the coil center. For points far from the loop
(z
R ) we can use the approximation: B ≈
µo iπ R 2
µo iR 2
2 z3
µ iA µ µ
= o 3= o 3
2π z
2π z
Here µ is the magnetic
B=
2 pz 3
dipole moment of the loop. In vector form:
r
r
µ µ
B( z ) = o 3
2π z
The loop generates a magnetic field that has the same
form as the field generated by a bar magnet.
(29 – 14)
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:
i
F
A. opposite the direction of the current
B. opposite the direction of ~F
C. in the direction of ~F
D. into the page
E. out of the page
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The magnitude of the magnetic
field at point P, at the center of the
semicircle shown, is given
by:
i
A. 2µ0i/R
B. µ0i/R
C. µ0i/4πR
D. µ0i/2R
E. µ0i/4R
P
4