torque on a current loop, 2 there is a force on sides 2 & 4 since they are perpendicular to the...
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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
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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
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Torque on a Current Loop, Equation
The maximum torque is found by:
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
2 42 2 2 2max (I ) (I )
I
b b b bτ F F aB aB
abB
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Torque on a Current Loop, General
Assume the magnetic field makes an angle of
< 90o with a line perpendicular to the plane of the loop
The net torque about point O will be τ = IAB sin
Use the active figure to vary the initial settings and observe the resulting motion
PLAYACTIVE FIGURE
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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
where is perpendicular to the plane of the loop and has a magnitude equal to the area of the loop
I A B
A
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Direction
The right-hand rule can be used to determine the direction of
Curl your fingers in the direction of the current in the loop
Your thumb points in the direction of
A
A
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Magnetic Dipole Moment
The product I 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:
Analogous to for electric dipole
A
B
p E
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Chapter 30
Sources of the Magnetic Field
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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
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Biot-Savart Law – Set-Up
The magnetic field is at some point P
The length element is
The wire is carrying a
steady current of I
Please replace with fig. 30.1
dB
ds
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Biot-Savart Law – Observations
The vector is perpendicular to both and to the unit vector directed from toward P
The magnitude of is inversely proportional to r2, where r is the distance from to P
dBr̂
dB
ds
ds
ds
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What does this tell you about the magnetic field, ?
1 2 3
33% 33%33%dB
ds
r̂ds
r̂
dB
1. It goes like the scalar dot product of and
2. It goes like X
3. is usually zero
0 of 30
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
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Biot-Savart Law – Observations, cont
The magnitude of is proportional to the current and to the magnitude ds of the length element
The magnitude of is proportional to sin where is the angle between the vectors and
ds
r̂ds
dB
dB
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The observations are summarized in the mathematical equation called the Biot-Savart law:
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
Biot-Savart Law – Equation
24oμ d
dπ r
s rB
ˆI
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Permeability of Free Space
The constant o is called the permeability of free space
o = 4 x 10-7 T. m / A
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Total Magnetic Field
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
The integral is over the entire current distribution
dB
24oμ d
π r
s rB
ˆI
ds
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Biot-Savart Law – Final Notes
The law is also valid for a current consisting of charges flowing through space
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
ds
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Compared to
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
E
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Compared to , 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 and the unit vector r̂
ds
B
E
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Compared to , 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
B
E
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Which variable can be pulled out of the integral?
1 2 3 4
25% 25%25%25%1. ds
2. sinθ
3. r2
4. None of them
0 of 30
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21 22 23 24 25 26 27 28 29 30
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How are θ and Φ related?
1 2 3 4
25% 25%25%25%
1. Φ = θ – π/2
2. Φ = θ
3. Φ = π/2 – θ
4. Φ = θ + π/2
0 of 30
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
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for a Long, Straight Conductor, Special Case
The field becomes
2
IoμB
πa
B
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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
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for a Curved Wire Segment
Find the field at point O due to the wire segment
I and R are constants
will be in radians4
IoμB θ
πR
B
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What about the contribution from the wires coming in and going out?
1 2 3
33% 33%33%
ds
r̂
1. They are distant enough to neglect their contribution
2. X = 0
3. The two currents cancel each other
0 of 30
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
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for a Curved Wire Segment
Find the field at point O due to the wire segment
I and R are constants
will be in radians4
IoμB θ
πR
B
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for a Circular Loop of Wire
Consider the previous result, with a full circle = 2
This is the field at the center of the loop
24 4 2
o o oμ μ μB θ π
πa πa a
I I I
B
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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
B
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What can we pull out of the integral this time?
1 2 3 4
25% 25%25%25%1. r2
2. Sin θ
3. ds
4. nothing
0 of 30
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
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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
2
32 2 22
ox
μ aB
a x
I
B
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Comparison of Loops
Consider the field at the center of the current loop
At this special point, x = 0 Then,
This is exactly the same result as from the curved wire
2
32 2 2 22
o ox
μ a μB
aa x
I I