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LECTURE 14
- Fundamental Laws for Calculating B-field
- Biot-Savart Law (“brute force”)
- Ampere’s Law (“high symmetry”)
- Example: B-field of an Infinite Straight Wire
- from Biot-Savart Law
- from Ampere’s Law
- Other examples
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Our Study of Magnetism
- Lorentz Force Equation
- Motion in a uniform B-field
- Forces on charges moving
in wires
- Magnetic dipole
- Today: fundamentals of how currents
generate magnetic fields
F = q
E + q
v ×
B
r
c
mv
qB
d
F = Id
l ×
B
μ =
Ai
τ =
μ ×
B
U = −
μ ⋅
B
Calculation of Electric Field
What are the analogous equations for the
Magnetic Field?
"Brute force"
Coulomb’s Law
"High symmetry"
Gauss’ Law
E =
0
q
r
2
r
ε
0
E • d
A = Q
inside
Calculation of Magnetic Field
"Brute force"
I
Biot-Savart Law
"High symmetry“ also,
only the ENCLOSED Current
Ampere’s Law AMPERIAN LOOP INTEGRAL
These are the analogous equations
Biot-Savart Law: B-field due to a current in a
wire
The magnetic field “curls” or “loops” around the wire
No dB field at P
2
since d is parallel to r
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Observations about Biot-Savart Law
d
B =
μ
o
4 π
Id
l ×
r
r
2
anddB 0 whend ||rˆ.
r ,
4 .ForagivencurrentI,dBismaximumwhend
r ).
r d sin(d ,
3.d
r
2.dB d and dB
rpointsfromd tothefield point.
× =
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Observations about Biot-Savart
d
B =
μ
o
4 π
Id
l ×
r
r
2
*Grasp element in your right hand with thumb pointing in the direction of the
current. Your fingers naturally curl around in the direction of the magnetic
field.
*To find at point P, d
B d
l ×
r
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Switch open: I = 0
Compass points north.
Switch closed: I 0
B ⊥ Id
Oersted’s Experiment
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y
x
I
d = Rd
Only if φ in radians
/2 is just the fraction of a
full circle that carries
current.
Note that as usual, the
current comes in and leaves
via un-shown wires!
Magnetic Field at the center of curvature for a
partial loop
d
B =
μ
o
I
4 π
dl
R
2
μ
o
I
4 π
Rd φ
R
2
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I
I
I
R
P
Find B at point P.
2
0
2
0
1 2 3
r
Id sin
dB
r
Id rˆ
d B
Superpositionprinciple:B B B B
θ
π
μ
×
π
μ
Line Segment Combinations
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Find B at point P.
Only the arcs contribute.
Double Arc
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Magnetic Field on Axis of Current Loop
d
B =
μ
o
4 π
Id
l ×
r
r
2
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Magnetic Field Lines of a Current Loop
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Solenoid (DEMO)
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( ) ( )
( )
( )
2
1
2 2
0 0
3 3
2 2 2 2 2 2
2
0
3
2 2 2
2
0
3
2 2 2
2 1
0
2 2 2
2 1
Result for on the axis of a current loop:
for a solenoid
2 2
no. of turns/unit length,
2
2
1
2
x x
x
x
x
x
x
B
R I R di
B dB
x R x R
N
n di nI dx
L
R nI dx
dB
x R
R nI dx
B
x R
x x
B nI
x R x
μ μ
μ
μ
μ
= → =
= = =
=
=
= −
∫
2
1 2
If - >> and >> (long solenoid),
R
x R x R
⎛ ⎞
⎜ ⎟
⎜ ⎟
⎝ ⎠
Solenoid
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B nI
x 0
=μ
B
x
at the end of a long solenoid is
exactly half of the value below,
which is for places far from either
end of a long solenoid.
This makes sense, since putting
two such long equal solenoids end
to end gets you back to full
strength, and so each of the halves
must contribute B
x
/
Remember n is the
number of turns per
meter.
Solenoid
