Torque on a Current Loop, 2


There is a force on sides 2
& 4 since they are
perpendicular to the field
The magnitude of the
magnetic force on these
sides will be:



F2 = F4 = I a B
The direction of F2 is out of
the page
The direction of F4 is into
the page
Torque on a Current Loop, 3


The forces are equal
and in opposite
directions, but not
along the same line of
action
The forces produce a
torque around point O
Torque on a Current Loop,
Equation


The maximum torque is found by:
b
b
b
b
τ max  F2  F4  (I aB )  (I aB )
2
2
2
2
 I abB
The area enclosed by the loop is ab, so τmax =
IAB

This maximum value occurs only when the field is
parallel to the plane of the loop
Torque on a Current Loop,
General



Assume the magnetic
field makes an angle of
q < 90o with a line
perpendicular to the
plane of the loop
The net torque about
point O will be τ = IAB
sin q
Use the active figure to
vary the initial settings
and observe the
resulting motion
PLAY
ACTIVE FIGURE
Torque on a Current Loop,
Summary



The torque has a maximum value when the
field is perpendicular to the normal to the
plane of the loop
The torque is zero when the field is parallel to
the normal to the plane of the loop
  IA  B where A is perpendicular to the
plane of the loop and has a magnitude equal
to the area of the loop
Direction



The right-hand rule can
be used to determine
the direction of A
Curl your fingers in the
direction of the current
in the loop
Your thumb points in
the direction of A
Magnetic Dipole Moment

The product I A is defined as the magnetic
dipole moment,  , of the loop



Often called the magnetic moment
SI units: A · m2
Torque in terms of magnetic moment:
   B

Analogous to   p  E for electric dipole
Chapter 30
Sources of the Magnetic Field
Biot-Savart Law – Introduction


Biot and Savart conducted experiments on
the force exerted by an electric current on a
nearby magnet
They arrived at a mathematical expression
that gives the magnetic field at some point in
space due to a current
Biot-Savart Law – Set-Up


The magnetic field is dB
at some point P
The length element is
ds

The wire is carrying a
steady current of I
Please
replace with
fig. 30.1
Biot-Savart Law –
Observations


The vectordB is perpendicular to both ds and
to the unit vector rˆ directed from ds toward P
The magnitude of dB is inversely proportional
to r2, where r is the distance from ds to P
What does this tell you about
the magnetic field, dB ?
1.
2.
3.
33%
It goes like the
scalar dot product
of ds and rˆ
It goes like ds X rˆ
dB is usually zero
33%
33%
0 of 30
1
1
2
3
4
5
6
7
8
9
10
21
22
23
24
25
26
27
28
29
30
11
12
13
14
15
2
16
17
18
3
19
20
Biot-Savart Law –
Observations, cont


The magnitude of dB is proportional to the
current and to the magnitude ds of the length
element ds
The magnitude of dB is proportional to sin q,
where q is the angle between the vectors ds
and rˆ
Biot-Savart Law – Equation

The observations are summarized in the
mathematical equation called the Biot-Savart
law:
μo I ds  ˆr
dB 
4π r 2

The magnetic field described by the law is the
field due to the current-carrying conductor

Don’t confuse this field with a field external to the
conductor
Permeability of Free Space


The constant o is called the permeability of
free space
o = 4p x 10-7 T. m / A
Total Magnetic Field


dB is the field created by the current in the
length segment ds
To find the total field, sum up the
contributions from all the current elements I ds
μo I ds  ˆr
B
4π  r 2

The integral is over the entire current distribution
Biot-Savart Law – Final Notes


The law is also valid for a current consisting
of charges flowing through space
ds represents the length of a small segment
of space in which the charges flow

For example, this could apply to the electron
beam in a TV set
B Compared to E

Distance


The magnitude of the magnetic field varies as the
inverse square of the distance from the source
The electric field due to a point charge also varies
as the inverse square of the distance from the
charge
B Compared to E , 2

Direction


The electric field created by a point charge is
radial in direction
The magnetic field created by a current element is
perpendicular to both the length element ds and
the unit vector rˆ
B Compared to E, 3

Source


An electric field is established by an isolated
electric charge
The current element that produces a magnetic
field must be part of an extended current
distribution

Therefore you must integrate over the entire current
distribution
Which variable can be pulled
out of the integral?
1.
2.
3.
4.
25%
ds
sinθ
r2
None of them
25%
25%
25%
0 of 30
1
2
3
4
5
6
7
8
9
10
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24
25
26
27
28
29
30
11 1
12
13
214
15
16 3 17
18
4
19
20
How are θ and Φ related?
1.
2.
3.
4.
25%
Φ = θ – π/2
Φ=θ
Φ = π/2 – θ
Φ = θ + π/2
25%
25%
2
14
3
25%
0 of 30
1
1
2
3
4
5
6
7
8
9
10
21
22
23
24
25
26
27
28
29
30
11
12
13
15
16
17
18
4
19
20
B for a Long, Straight
Conductor, Special Case

The field becomes
μo I
B
2πa
B for a Long, Straight
Conductor, Direction




The magnetic field lines are
circles concentric with the
wire
The field lines lie in planes
perpendicular to to wire
The magnitude of the field
is constant on any circle of
radius a
The right-hand rule for
determining the direction of
the field is shown
B for a Curved Wire Segment


Find the field at point O
due to the wire
segment
I and R are constants
μo I
B
θ
4πR

q will be in radians
What about the contribution from the
wires coming in and going out?
33%
33%
33%
1. They are distant
enough to neglect
their contribution
2. ds X rˆ = 0
3. The two currents
cancel each other
0 of 30
1
1
2
3
4
5
6
7
8
9
10
21
22
23
24
25
26
27
28
29
30
11
12
13
14
15
2
16
17
18
3
19
20
B for a Curved Wire Segment


Find the field at point O
due to the wire
segment
I and R are constants
μo I
B
θ
4πR

q will be in radians
B for a Circular Loop of Wire

Consider the previous result, with a full circle

q = 2p
μo I
μo I
μo I
B
θ
2π 
4πa
4πa
2a

This is the field at the center of the loop
B for a Circular Current Loop


The loop has a
radius of R
and carries a
steady current
of I
Find the field
at point P
What can we pull out of the
integral this time?
1.
2.
3.
4.
25%
r2
Sin θ
ds
nothing
25%
25%
25%
0 of 30
1
2
3
4
5
6
7
8
9
10
21
22
23
24
25
26
27
28
29
30
11
1
12
13
2
14
15
16
3
17
18
4
19
20
B for a Circular Current Loop


The loop has a radius
of R and carries a
steady current of I
Find the field at point P
Bx 
μo I a2
2 a  x
2
2

3
2
Comparison of Loops



Consider the field at the center of the current
loop
At this special point, x = 0
Then,
Bx 

μo I a2

2 a x
2
2

3
2
μo I

2a
This is exactly the same result as from the curved
wire