EMF 2005
Handout 8: Conductors and Magnetic Fields
1
CONDUCTORS AND MAGNETIC FIELDS
This handout covers:
•
•
•
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The Biot-Savart Law for d B due to a current element
Force on a current-carrying wire in a magnetic field
Torque on a current loop and the magnetic dipole
Ampere's Law
Magnetic field due to a current-carrying
wire: the Biot–Savart Law
Recall: B at a point P due to a point charge, Q, moving with velocity v, is
given by
Q
∧
µoQ 
B =
v × r 
4πr 2 

r
∧
v
P
B (outwards)
r
Let a thin wire carry a current I.
Let a short length d L contain charge dQ moving
with velocity v
I
r
r
The contribution of d L to the magnetic field at P is
dB =
But
∧
µ o dQ 

v
r
×

2 
4πr 

dQ
dL
I =
and v =
dt
dt
P
∧
dL
⇒
dB =
µ oI
4πr 2
∧


d
L
r
×




The Biot-Savart Law
To find the total field at P, we must integrate this expression along the
whole length of the wire.
EMF 2005
Handout 8: Conductors and Magnetic Fields
2
Special case: Magnetic field due
to a straight conductor
θ2
What is B at a point P which is at a
perpendicular distance R from a
straight wire?
Consider a small element d L
I
θ = angle between the wire and the line
joining d L and P
R
P
r = distance from d L to P
Its contribution to B at P is
µ oI
4πr 2
dB =
θ
∧


L
r
d
×




dL
θ1
dL
∧
∧
θ
r
d L × r = dL sin θ
From the right hand rule, the direction of d B is into the page.
The magnitude of dB is
But
R = rsinθ
and
dLsinθ = rdθ
R
So r =
sin θ
Therefore
⇒
dB =
dB =
µ oI
4πr 2
[dL sin θ]
dθ
R
r
and
θ
dL =
Rdθ
dL
dLsinθ = r dθ
θ
sin 2 θ
µ oI sin 2 θ Rdθ
dB =
sin θ
4π R 2 sin 2 θ
µ oI sin θ
4πR
Contribution of element dL to the field at P
EMF 2005
Handout 8: Conductors and Magnetic Fields
3
To find the total field, we INTEGRATE from one end of the wire to the other
– i.e., from θ1 to θ2.
Example: An infinitely long wire:
dB =
µ oI
4πR
π
∫
sin θdθ =
θ1 = 0 and θ2 = 180o.
µ oI
[− cos θ]0π =
4πR
µ oI
[− 1 − 1]
4πR
0
⇒
B =
µ oI
.
2πR
Magnetic field pattern of a straight current-carrying wire:
As we have seen before, the field lines are circular loops.
Are the lines of B clockwise or anticlockwise?
To decide, we can use another form of the
right hand rule:
B
• Let the thumb point in the direction of the current
• The fingers will then curl in the direction of B .
I outwards
EMF 2005
Handout 8: Conductors and Magnetic Fields
4
The force on a current-carrying
wire in a magnetic field
In a current carrying wire, charges are moving. Therefore, if the wire is in a
magnetic field, a force will be exerted on the moving charges and hence on
the wire. This is the principle of the electric motor.
Consider a section of straight
wire carrying current I in a
magnetic field B . Let d L contain
charge dQ moving with velocity v.
I
dL
The force on d L is
θ
d F = dQ(v × B )
But I =
dQ
dt
Therefore
and v =
dL
dt
d F = I(d L × B )
The magnitude of d F is
B
THE MOTOR EQUATION
dF = BIdLsinθ
Force between two parallel wires
Consider two parallel wires, separation R, carrying currents I1 and I2.
Let B1 = field at wire 2 due to wire 1
B2 = field at wire 1 due to wire 2
Consider an element d L in wire 2
By the right hand rule:
dF2
B1 (in) dL2
B1 is into the paper at wire 2
I1
Wire 1
R
I2
Wire 2
EMF 2005
Handout 8: Conductors and Magnetic Fields
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d F2 = I(d L2 × B1 )
Force on dL2, d F2 is
Direction: Towards wire 1
Magnitude:dF2 = B1I2dL2
Recall, at perpendicular distance R from a straight wire, B =
Therefore,
⇒
dF2 =
µ oI1I2 dL 2
2πR
Force per unit length is
If I1 = I2 = I,
then
µ oI
2πR
.
µ II
dF2
= o12
dL 2
2πR
dF2
I2
= (2 x 10 -7 )
dL 2
R
N m-1
SI definition of the Ampere:
1 Ampere is the current which, when flowing in two long straight parallel
wires separated by 1 m, produces a force per unit length between them
of 2 x 10-7 N m-1.
EMF 2005
Handout 8: Conductors and Magnetic Fields
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Torque on a current loop:
The magnetic dipole
Z
Consider a rectangular loop carrying
current I in a uniform magnetic field B .
B
(i) Plane of the loop perpendicular to B:
F2
X
F1
TOP VIEW
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I
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b
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a
b
Z
B (out of page)
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I
X
Note that the current is always
perpendicular to B .
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Y
I
Let the loop be in the X-Y plane with
B in the Z direction.
SIDE VIEW
B
a
Y
Y
F2
F1
b
Note how all the forces cancel
out in this situation – they just
tend to stretch the loop.
X
(ii) Plane of the loop at an angle to B:
F1
F2
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bcosθ
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F1
X
Z
B (out of page)
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F2
a
Y
F2
bsinθ
B
θ
Y
θ
bcosθ
F2
EMF 2005
Handout 8: Conductors and Magnetic Fields
F1 = BI(bcosθ)
7
F2 = BIa (sides a are still perpendicular to B )
Forces F1 cancel out. But forces F2 don't cancel. Although they are still
equal and opposite, they no longer act along the same line, so they form a
TORQUE.
Torque = (Force)(Perpendicular distance)
τ = (BIa)(bsinθ) = µBsinθ where µ = Iab
This looks like the magnitude of a vector cross product, which is exactly
what it is:
τ = µ×B
µ is the MAGNETIC DIPOLE MOMENT VECTOR
Magnitude of µ :
µ = Iab = (Current)(Area)
or
µ = NIab = (Number of turns)(Current)(Area)
if the loop has N turns of wire.
Direction of µ : This is given by the right hand rule:
Let the fingers curl in the direction of current flow.
The thumb then points along µ .
EMF 2005
Handout 8: Conductors and Magnetic Fields
8
Ampere's Law
Recall:
Gauss's Law relates E to the charge distribution that produces it.
Likewise, Ampere's Law relates B to the current distribution that produces
it.
Recall: The magnetic field at perpendicular distance R from a long
straight
wire is
µ oI
B =
2πR
View along the axis (with the
current coming out of the page).
Consider an arbitrary closed path
around the wire:
What is the line integral
around this closed path?
dL
I (out)
2
∫ B ⋅ dL
B
R
For a typical small element d L ,
B ⋅ d L = BdL' = BRdθ =
µ oI
[Rdθ]
2πR
dL
dL'
B
So
⇒
∫
µ I
µI
B ⋅ d L = o dθ = o
2π
2π
∫
∫ B ⋅ dL = µ I
o
2π
∫
0
dθ =
µ oI
2π
2π
This is AMPERE's LAW.
dθ
EMF 2005
Note:
Handout 8: Conductors and Magnetic Fields
9
1.
We carry out the integral along a some path around the
conductor, NOT along the conductor and not necessarily
along the field line.
2.
Ampere's law holds for ANY closed path and ANY dis
tribution
of current.
3.
The currrent, I, is the total current enclosed by the path, i.e.,
flowing through the area bounded by the path
or
I is the current flowing through any "bag-like" surface whose
edge is defined by the path.
Ienclosed
B
B
dL
dL
dL
dL
B
dL
Usually, we will write
4.
∫ B ⋅ dL = µ I
o enclosed
Ampere's law ⇒ B is NOT conservative.
∫ E ⋅ dL = 0
5.
B
but
∫ B ⋅ dL ≠ 0
Ampere's Law in words: The lineintegral of the magnetic
field around a closed path is equal to µo multiplied by the
current enclosed by the path.
EMF 2005
Handout 8: Conductors and Magnetic Fields
10
Procedure for using Ampere's Law
When to use it: When you want to find the magnetic field produced by
some given distribution of current.
1.
Draw a diagram showing the current distribution and the magnetic field
that it produces.
View along the axis of the current so that current is coming out of the
page. This means that B will be in the plane of the page.
2.
Decide on the best CLOSED path for the line integral. Choose one
that makes the integral easy
⇒ make B either parallel or perpendicular to dL
In problems that we shall do, the path will always be either circular or
rectangular.
∫ B ⋅ dL
3.
Work out the line integral
4.
Decide how much current flows THROUGH the area defined by the
path.
5.
Put
∫ B ⋅ dL = µ I
o enclosed
and rearrange the equation to find B as a
function of position.
Examples:
1.
2.
3.
See lecture notes.
B due to a long straight wire.
B due to a conducting cylinder
B due to a long solenoid