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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-pronged 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)
B
.
Magnetic Field Generated by
dB
.
a Circular Wire Arc of Radius
R at Its Center C
 0i
B
4 R
A wire section of length ds generates at the center C a magnetic field dB.
0i ds sin 90 0i ds
The magnitude dB 

. The length ds  Rd
2
2
4
R
4 R
i
 dB  0 d .
Vector dB points out of the page.
4 R

i
 i
The net magnetic field B   dB   0 d = 0 .
4 R
4 R
0
Note : The angle  must be expressed in radians.
For a circular wire,   2 . In this case we get: Bcirc 
0i
2R
.
(29-5)
Ampere's Law
The law of Biot-Savart combined with the
principle of superposition can be used to
 B  ds   i
0 enc
determine B if we know the distribution
of currents. In situations that have high
symmetry we can use Ampere's
law instead, 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 advanced
formulations of electromagnetism. 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 0 .
The closed path used is known as 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.
(29-7)
 B  ds   i
0 enc
Implementation of Ampere's Law :
 B  ds .
1. Determination of
The closed path is
divided into n elements s1 , s2 , ..., sn .
We then form 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 ith 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 concentric 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
0 enc
B
 0i
0i
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 that is a circle of
radius r (r  R) with 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
0 enc
ienc
B
r2
 i 
2 rB  0i 2  B   0 2  r
R
 2 R 
0i
2 R
O
 r2
r2
i
i 2
2
R
R
R
r
(29-10)
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
(29-11)
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 solenoid has n turns per
unit length.
B  0 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
0 enc
c
d
a
b
c
d
 B  ds   B  ds  B  ds  0
The enclosed current ienc  nhi.
Bh  0 nhi  B  0 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 concentric 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  0 Ni  B 
The enclosed current ienc  Ni.
0 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 that carries a current i. The magnetic field
at a point P on the z -axis is given by
B
 
B( z )  0 3
2 z
0iR 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 
0i R 2

0iR 2
2z3
0iA 0 

. Here  is the magnetic
3
3
2 z
2 z
2 pz 3
dipole moment of the loop. In vector form:
B( z ) 
0 
.
3
2 z
The loop generates a magnetic field that has the same
form as the field generated by a bar magnet.
(29-14)
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