Physics 4B Lecture Notes
Chapter 30 - Magnetic Fields Due to Currents
Problem Set #9 - due:
Ch 30 - 3, 12, 17, 19, 26, 37, 38, 47, 51, 54, 57, 64, 72
Lecture Outline
1. The Biot-Savart Rule
2. Ampere's Law
Now that we know the effects of magnetic fields on currents we can discuss the causes of magnetic fields.
If currents feel the magnetic force, Newton's Third Law requires they must exert it as well. The purpose
of this chapter is to learn to calculate magnetic fields caused by currents. The Biot-Savart Rule is a recipe
that will always work. Ampere’s Law, while more fundamental, can be used to find fields only in cases
where the current distribution is highly symmetric.
1. The Biot-Savart Rule
r
dq
dE 2 = k 2 2 ˆr
r
r
r
dF12 = q 1dE 2
+ q1
r
r
+ dq
2
r
s
r
r
r
r I1
r
dF12 = I1 s × dB2
r
ds
I2
The electric force on a charge q1 is found from the field due to all the other charges. We begin by finding
r
dq
the field due to the point charge dq2 , which is dE 2 = k 2 2 rˆ . Next, the force on q1 due to dq2 is found
r
r
r
using definition of electric field, dF12 = q 1dE 2 . Then integrate all the contributions from the rest of the
charge distribtion. The same proceedure would work for the magnetic force on I1 due to a distribution of
r
current I2 . The magnetic force on the current I1 in the magnetic field dB2 caused by a small current
r
r
r
element in I2 is, dF12 ≡ I1 s × dB2 which means that the magnetic equivalent of the electric charge element
r
r
r
dq is, q1 → I1 s× ⇒ dq → Id s× ⇒ dq 2 → I 2ds ×. Using an analogy to the field due to the point charge
r
r
r
r
r
dq2
I 2 ds ×
I2 d s × rˆ
ˆ
ˆ
dq2 , dE 2 = k 2 r → d B2 = k m
r ⇒ B2 = k m ∫
. This is called the Biot-Savart Rule.
r
r2
r2
r µo Idrs × ˆr
B=
4π ∫ r 2
The magnetic constant is usually replaced with km =
The Biot-Savart Rule
µo
T⋅m
≡ 10 −7
.
4π
A
30-1
Physics 4B Lecture Notes
Example 1: Find the magnetic field due to a longr straight wire carrying a current I.
r µ Ids × ˆr
y
Starting with the Biot-Savart Rule, B = o ∫
.
4π
r2
r
r
Note d s × ˆr = sinθdx and the field points out of the page.
r
µo sinθ
a
The magnitude of the field is,B =
I 2 dx.
∫
4π
r
θ
x
r
I is constant, r = acsc θ and x = −acot θ ⇒ dx = acsc 2 θdθ .
ds
π
µ oI π
sin θ
µ
I
µ
I
µ
I
B=
acsc 2 θdθ = o ∫ sinθdθ = o ⋅ 2 = o
4π ∫θ= 0 a 2 csc 2 θ
4πa θ=0
4πa
2πa
I
x
B
At this point the field is out of the page. In general, the field
I
lines form rings around the wire.
It is convenient to know the magnetic field due to a long straight wire.
B=
µo I
The Magnetic Field Due to a Long Straight Wire
2πr
Beyond the Mechanical Universe (vol. 35 Ch 8,9,12)
Example 2: Use the result of example 1 to find the
two parallel currents.
r force
r between
r
Starting with the definition of magnetic field, F ≡ I 2 l × B1 .
Since the direction is right, just find the magnitude,
F = I 2 lB1 . Using the result of example 1,
µ I
µ II
F = I 2 l o 1 ⇒ F = o ⋅ 1 2 ⋅ l just like before.
2πr
2π r
r
I1
F
B1
I2
Example 3: Find the magnetic field on the axis of a circular loop of radius a and magnetic dipole
r
moment µ .
x
r
ds
y
θ
r
r
r
r
ds
a
a
dB
r θ dθ
d
s
z
θ
z
x
y
r µo Idrs × ˆr µ o I d rs × rr
Use the Biot-Savart Rule B =
=
.
4π ∫ r 2
4π ∫ r 3
r
r
Notice that r = −acos θˆi − asin θˆj + zkˆ and d s = adθ −sin θˆi + cosθˆj .
(
)
30-2
Physics 4B Lecture Notes
ˆi
ˆj
kˆ
r r
d s × r = −asin θdθ acos θdθ 0
−acos θ
−asin θ z
r r
d s × r = azcos θdθ ˆi + azsin θdθˆj + a 2 sin2 θ + cos2 θ dθkˆ
[
) ]
(
(
)
3
Since r 3 = a 2 + z 2 2 and is constant,
r µ I
1
2 ˆ 2π 
 ˆ 2π
ˆ 2π
B= o ⋅
3  az i ∫0cosθdθ + az j ∫0sin θdθ + a k ∫0dθ 
4π a 2 + z 2 2
(
)
Doing the integrals,
r µ oI
r µo I
1
a2
2ˆ
B=
⋅
0 + 0 + a k 2π ⇒ B =
⋅
4π a 2 + z 2 3 2
2
a2 + z2
r
r µo
µ
In terms of the dipole moment, B =
⋅
.
2π a 2 + z 2 3 2
(
[
)
]
(
(
)
3
kˆ
2
)
2. Ampere's Law
Beyond the Mechanical Universe (vol. 35 Ch 19,20)
Ampere's Law states that the sum of the magnetic field along any closed path is proportional to the current
that passes through. Mathematically,
r r
B
∫ • ds = µ oienclosed Ampere's Law
y
In order to understand this idea, consider a circular path around a wire that passes
r
ds
through the center and is perpendicular to the plane of the circular path. We know
µ I
that the magnetic field due to the wire is, B = o and it is parallel to ds = rdθ.
2πr
r r
µo I
µ oI
µ oI
So, ∫ B • ds = ∫
⋅ rdθ =
dθ =
⋅ 2π = µ o I consistent with Ampere's
2πr
2π ∫
2π
r
I
z
Law.
If we change the path so that it doesn't contain the current as shown
onrthe right. rThe path rneeds to be
up
four parts,
r broken
r into
r
r
r
r
r
∫ B • ds = ∫ B • ds + ∫ B • ds + ∫ B • ds + ∫ B • d s .
y
r
ds
r
ds
r
ds
a
x
B
y
a
r r
r r
r r
Along the x and y axes B⊥d s ⇒ ∫ B • ds = ∫ B • ds = 0 .
b
r
ds
I
b
x
x
y
µ I
Along b, B = o and it is parallel to ds = bdθ so,
2πb
r r
µo I
µ I
µ I π µ I
∫ B • ds = ∫ 2πb ⋅ bdθ = 2πo ∫ dθ = 2πo ⋅ 2 = 4o .
b
b
b
30-3
Physics 4B Lecture Notes
r r
µ I
µ I
µ I π
µ I
Similarly along a, ∫ B • ds = − ∫ o ⋅ adθ = − o ∫ dθ = − o ⋅ = − o . Putting it all together,
2πa
2π a
2π 2
4
a
a
r r
µ I
µ I
∫ B • ds = 0 + 4o + 0 − 4o = 0 also consistent with Ampere's Law.
Like Gauss's Law, Ampere's Law can be used to find the magnetic field of a sufficiently symmetric
current distribution.
Example 4: Find the magnetic field inside and outside of a wire of radius a carrying a uniform
current I. Sketch the field as a function of r the distance from the center.
r
Outside the wire choose the circular path shown. Along this path the
ds
symmetry
r r requires B must be constant and it can only point along ds so,
a
∫ B • ds = ∫ Bds = B∫ ds = B⋅ 2πr .
The enclosed current is the entire current I. Applying Ampere's Law,
r r
µ I
∫ B • ds = µ oienclosed ⇒ B ⋅2πr = µoI ⇒ B = 2πro . This is the same
result as for an infinitely thin wire.
Inside the wire choose the circular path shown. Along this path the
symmetry requires B must be constant and it can only point along ds so
again,
r r
B
∫ • ds = ∫ Bds = B∫ ds = B⋅ 2πr .
r
r
ds
r
a
πr 2
The enclosed current is i enclosed = 2 ⋅ I. Applying Ampere's Law,
πa
r r
πr 2
µo I  r 
B
•
d
s
=
µ
i
⇒
B
⋅2πr
=
µ
⋅
. The field grows linearly until the
o enclosed
o
∫
2 ⋅I ⇒ B =
πa
2πa  a 
surface of the wire is reached. The graph of B vs. r is shown below.
B
µ oI
2πa
αr
α 1r
a
r
30-4
Physics 4B Lecture Notes
Example 5: Wire 1.00mm in diameter is wound around a 5.00cm diameter 1.00m long tube to
create a solenoid. The wire carries a current of 100mA. Estimate the magnetic field in the center.
Apply Ampere's Law for the path
indicated. The current enclosed is
i enclosed = Ni where N is the number
B
2
1
3
of turns inside the path and i is the
B≈0
4
current in the wire. The path integral
l
of the field can be broken into four
parts,
forr each siderof the path.
r one
r r
r r
r
r
r
B
•
d
s
=
B
•
d
s
+
B
•
d
s
+
B
•
d
s
+
B
∫
∫
∫
∫
∫ • d s . Along side 4 the field is close to zero. Along
1
2
3
4
sides 1 and 3 the field is perpendicular to the path so,
r
r
r
r
∫ B • ds = 0 + ∫ B • d s + 0 + 0 = Bl .
2
Assuming that the solenoid is so long that symmetry requires the magnetic field to be constant
along the path. Using Ampere's Law,
r r
N
−7  1000
−4
B
∫ • ds = µ oienclosed ⇒ Bl = µ oNi ⇒ B = µ o l i = 4πx10  1.00  (0.100) = 1.26x10 T
(
B = µo
)
N
i The Magnetic Field Inside a Long Solenoid
l
Example 6: The tube is bent into a circle to form a toroid. Find the magnetic field at the center.
The current enclosed by the indicated path is i enclosed = Ni where N is
r
d
s
the total number of turns and i is the current in the wire. The magnetic
D
field can only be
a
function
of
r
and
so
it
must
be
constant
along
the
path
r
r r
ri
shown. So, ∫ B • ds = B ⋅2πr . Using Ampere's Law,
r r
N
B
∫ • ds = µ oienclosed ⇒ B2πr = µ oNi ⇒ B = µ o 2πr i .
Assume the inner circumference is equal to the length of the tube,
r = r i + D2 ⇒ 2πr = 2πr i + π D = l + π D .

N
1000

The field is, B = µ o
i = 4πx10−7 
(0.100) = 1.09x10 −4 T


l + πD
1.00 + π (0.0500)
(
)
30-5
Physics 4B Lecture Notes
Chapter 30 - Summary
r µo Idrs × ˆr
Biot - Savart Rule B =
4π ∫ r 2
The Magnetic Field Due to a Long Straight Wire B =
µo I
2πr
r r
Ampere's Law ∫ B • ds = µ o ienclosed
The Magnetic Field Inside a Long Solenoid B = µ o
N
i
l
30-6