Today’s Concept:
What Causes Magnetic Fields
Physics 2112
Unit 14
2
0ˆ
4 r
rsdIBd
=
Unit 14, Slide 1
Unit 14, Slide 2
2
0ˆ
4 r
rsdIBd
=
2
12
120ˆ
4 r
rvqB
=
Biot-Savart Law
B field from one
moving charge
avgqnAvI =But remember from
previous slides
B field from tiny
of current
carrying wire.
sd
Compare to Electric Fields
Unit 14, Slide 3
2
12
120ˆ
4 r
rvqB
=
2
12
12ˆ
4
1
r
rqE
o=
In the same direction as r12 Perpendicular to r12
v out of the
screen
ATm /104 7
0
−=
x
z
BF
qv
Remembering Directions: The Right Hand Rule
BvqF
=
Unit 12, Slide 4
This method
always
works!
A
BAC
=
A B
C
AB
C
Another way of looking at it…..
Unit 14, Slide 5
Works when something is
going in a circle
What is the magnetic field a distance yo away from a infinitely long wire of current I?
Example 14.1 (Infinite wire of current)
2
0ˆ
4 r
rsdIBd
=
Unit 14, Slide 6
➢ Conceptual Plan
➢ Strategic AnalysisDone in prelecture in detail
Integrate
(Similar to E field for infinite line of charge)
Use Biot-Savart Law
Unit 14, Slide 7
Main Idea
2
0ˆ
4 r
rsdIB
=
)(
)sin(*
4 22
0
oyx
dxI
+=
f
Q
r̂
sd
f
sd
rx
y
Current I OUTr
Magnitude:B
Example 14.1 (answer)
r
IB
2
0=•
Unit 14, Slide 8
ATm /104 7
0
−=
Remember:
rE
o
2=
xy
z
BF
qv
Remembering Directions: The Right Hand Rule
Unit 12, Slide 9
vB
F
2
12
120ˆ
4 r
rvqB
=
BvqF
=
v
B
F
A long straight wire is carrying current from left to right. Two identical charges are moving with equal speed. Compare the magnitude of the force on charge a moving directly to the right, to the magnitude of the force on charge b moving up and to the right at the instant shown (i.e. same distance from the wire).
A) |Fa| > |Fb|
B) |Fa| = |Fb|
C) |Fa| < |Fb|
•B
v
I
v
(a)
r r
(b)
Forces are in different directions (careful with all the angles!)
•F
F
Same q, |v|, B and q (=90)
Question
Unit 14, Slide 10
qsinqvBF =
BvqF
=
Two long wires carry opposite current
What is the direction of the magnetic field above, and midway between the two wires carrying current – at the point marked “X”?
A) Left B) Right C) Up D) Down E) Zero
B
x
Question
Unit 14, Slide 11
x
Example 14.3 (From Loop)
Unit 14, Slide 12
yo
A current, I, flows clockwise
through a circular loop of wire.
The loop has a radius a.
What is the magnetic field at a
point P a distance yo above
the plane of the loop in the
center?a
P
x
x
Q
BcosQ
Q
Question
Unit 14, Slide 13
yo
When doing this integral, how
much of ds is perpendicular to r?
A) all of it
B) ds sinQ
C) ds cosQ
D) ds/sinQ
a
P
x
x
Q
BcosQ
Q
r
ds
Q
Force Between Current-Carrying Wires
BLIF
= 212
12122
Id
LIF o
=
Unit 14, Slide 14
I1 r
IB
2
10=X X X X X X X X X X X X X X X X X X X
X X X X X X X X X X X X X
X X X X X X X X X X X X
X X X X X X X X X X
. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . .
I2
F
I towards us • •
Another I towards usF
Conclusion: Currents in same direction attract!
•
I towards us
Another I away from us
F
Conclusion: Currents in opposite direction repel!
dB
Bd
Force Between Current-Carrying Wires
BLIF
= 212 12122
Id
LIF o
=
Unit 14, Slide 15
Example 14.4 (Two Current Carrying Wires)
Two current carrying wires a 10cm apart for a length for 50cm. Wire 1 carries 5A and Wire 2 carries 10A with both current to the left.
What is the magnitude and direction of the force on wire 2due to wire 1?
Unit 14, Slide 16
I1 = 5A
I2 = 10A
10cm
50cm
CheckPoint 1A
What is the direction of the force on wire 2 due to wire 1?
A) Up B) Down C) Into Screen D) Out of screen E) Zero
XB
F
Unit 14, Slide 17
What we thought…..
What is the direction of the force on wire 2 due to wire 1?
A) Up B) Down C) Into Screen D) Out of screen E) Zero
XB
F
Unit 14, Slide 18
A) The current through wire 2 creates a magnetic field pointing out of the screen at wire 1. Using the RHR, the force on wire 1 must be up.
B) One right-hand rule shows that the B field is into the page, so another right-hand rule shows that the force points down.
CheckPoint 1B
What is the direction of the torque on wire 2 due to wire 1?
Uniform force at every segment of wire
No torque about any axis
Unit 14, Slide 19
A) Up B) Down C) Into Screen D) Out of screen E) Zero
CheckPoint 3A
What is the direction of the force on wire 2 due to wire 1?
A) Up B) Down C) Into Screen
D) Out of screen E) Zero
Unit 14, Slide 20
CheckPoint 3B
What is the direction of the torque on wire 2 due to wire 1?
A) Up B) Down C) Into Screen
D) Out of screen E) Zero
Unit 14, Slide 21
LET’S DRAW A PICTURE!
B
B
IF out of screen
Net Force is Zero!
I F into screen
rr
Consider Force on Symmetric Segments
Electricity & Magnetism Lecture 14, Slide 22
But the torque is not!!!
CheckPoint 3B
What is the direction of the torque on wire 2 due to wire 1?
A) Up B) Down C) Into Screen
D) Out of screen E) Zero
Unit 14, Slide 23
Checkpoint 2: Force on a loop
A. the forces are in opposite directions
B. the net forces are the same.
C. the net force on the loop is greater than the net force on the wire segment
D. the net force on the loop is smaller than the net force on the wire segment
E. there is no net force on the loop
A current carrying loop of width a and
length b is placed near a current carrying
wire.
How does the net force on the loop
compare to the net force on a single wire
segment of length a carrying the same
amount of current placed at the same
distance from the wire?
Checkpoint question
Current flows in a loop as shown in
the diagram at the right. The direction
is such that someone standing at
point a and looking toward point b
would see the current flow clockwise.
What is the orientation of the
magnetic field produced by the loop at
points a and b on the axis?
(A)
(B)
(C)
(D)
B on axis from Current Loop
I
Resulting B Field
Current in Wire
Electricity & Magnetism Lecture 14, Slide 26
What about Off-Axis ?
Biot-Savart Works, but need to do numerically
See Simulation!
Unit 14, Slide 27
Two identical loops are hung next to each other. Current flows in the same direction in both.
The loops will:
A) Attract each other B) Repel each other C) There is no force between them
Two Current Loops
Unit 14, Slide 28
Two ways to see this:
1) Like currents attract
N S N S
2) Look like bar magnets
1. ANY CROSS PRODUCT
2. Direction of Magnetic Moment
Thumb: Magnetic Moment
Fingers: Current in Loop
3. Direction of Magnetic Field from WireThumb: Current
Fingers: Magnetic Field
Right Hand Rule Review
BvqF
= BLIF
=
Fr
= B
=
2
0ˆ
4 r
rsdIBd
=
Unit 14, Slide 29
Remembering difference between two formulas
Unit 14, Slide 30
Which one of these situations will
have bigger B field at point P?
I
P
a
aP
I
A) Situation 1 B) Situation 2
C) They will be equal
Remembering difference between two formulas
Unit 14, Slide 31
Which one of these equation will
have bigger B field at a distance a ?
A) equation 1 B) equation 2
C) They will be equal
sin 𝛼 ± sin 𝛽 = 2 sin1
2𝛼 ± 𝛽 cos
1
2𝛼 ∓ 𝛽
𝝁𝒐𝑰
𝟐𝝅𝒂
𝝁𝒐𝑰
𝟐𝒂
Remembering difference between two formulas
Unit 14, Slide 32
𝝁𝒐𝑰
𝟐𝝅𝒂
𝝁𝒐𝑰
𝟐𝒂
I
P
a
aP
I
➢ Conceptual AnalysisEach wire creates a magnetic field at P
B from infinite wire: B = 0I / 2r
Total magnetic field at P obtained from superposition
y
x
.
z
I1 = 1A
Front view Side view
➢ Strategic AnalysisCalculate B at P from each wire separately
Total B = vector sum of individual B fields
I2 = 1A
4cm
4cm
y
P
3cm
Example 14.2
Two parallel horizontal wires are located in the vertical (x,y) plane as shown. Each wire carries a current of I = 1A flowing in the directions shown.
What is the B field at point P?
Unit 14, Slide 33
What is the direction of Bat P?
.
z
y
P
..
z
y
P
.
z
y
P
.90o
Question
Electricity & Magnetism Lecture 14, Slide 34
y
x
.
z
I1 = 1A
Front view Side view
I2 = 1A
4cm
4cm
y
P
3cm
A B C D
z
P
y
Question
Electricity & Magnetism Lecture 14, Slide 35
What is the magnitude of B at P produced by the top current I1? (0 = 4 x 10−7 T−m/A)
A) 4.0 x 10−6 T B) 5.0 x 10−6 T C) 6.7 x 10−6 T
What is r?r = distance from wire axis to P
3cmz
y
.
4cm r
Two parallel horizontal wires are located in the vertical (x,y) plane as shown. Each wire carries a current of I = 1A flowing in the directions shown.
y
x
.
z
I1 = 1A
Front view Side view
I2 = 1A
4cm
4cm
y
P
3cm
r
IB
2
0=
y
x
.
z
I1 = 1A
Front view Side view
I2 = 1A
4cm
4cm
y
P
3cm
Example 14.2
Two parallel horizontal wires are located in the vertical (x,y)
plane as shown. Each wire carries a current of I = 1A
flowing in the directions shown.
What is the B field at point P?
Unit 14, Slide 36
What is the direction of the magnetic field at point P,
which is at the center of a semicircular loop of wire
carrying a current I as shown?
A. Goes in
B. Goes out
C. Goes left
D. Goes right
E. There is no magnetic field at point P.
P
Question
If I = 6A, what is the
magnitude of the
magnetic field at point
P?
Example 14.5 (Curved Loop of Wire)
P
20cm
12cmConceptual Plan
Strategic Analysis
Integrate both loops
Note straight sections cancel out.
2
0ˆ
4 r
rsdIBd
=
Use Biot-Savart Law
All of the current loops below carry the same
current I. Rate them according to the magnetic
field at the red dot, from largest to smallest.
A. A>B>C
B. A>C>B
C. B=C>A
D. C=B=A
E. C>B>A
Question
Good News!!!!!
Unit 14, Slide 41
Remember how we used Gauss’
Law to avoid doing integral in E
field?
We got similar law for B fields!