
B
+
R
F= qv  B= qv B
F
v
+q
F = I B   I B sin(θ)
μ0 I =
B
||
Δl
Ampere’s Law
edge
μ0I = B 2πR
μ0 I
⇒B=
2πR
B = I n 0
4-0
Magnetic Interactions
1.) Force at a distance
Like electrostatics!
2.) Repulsive and attractive
Permanent magnets
S
N
S
break
N
S
N
repulsive
NO MAGNETIC MONOPOLES
attractive!
Magnets and Magnetic fields : Dipole Fields

Magnetic field B
S
N
{Like E}
Magnetic field lines
S
B  units Tesla =
4-1
kg
CS
N
Forces on charge due to B (magnetic field)

B
No force

B
q
v
v=0
v || B
v  B  F force
B
q
v
F
direction by right hand rule!
note:  =vector at you
= vector into page

B
v
v


v  = v sinθ
+q & -q +q
opposite
v
 F
F= qv B= qv B
= q v B sinθ
4-2
-q
B
F
v
B
-|q| flip thumb
Croft preferred RHR
direction by right hand rule!
I
+|q|
Another equivalent
RHR
B
B
v
F
4-2a
F= q v  B Units B field tesla
m
 N=C
T
s
NS
m s
kg
T=
 kg 2

 T
mC
s mC Cs
Circular motion of charge moving  to magnetic field

v2
B
F= m a = m
R
+
F
mv 2
mv
qvB=
 R=
R
R
qB
4-3
v
+q
uniform B field  v  to B  circular motion.
4-3a
General motion of charged particle in a magnetic field
|| direction
v||  constant
magnitude & direction
 direction
circular motion
speed = v 
helical-motion
4-3b
Earth’s magnetic field
deflects charge particles
from Sun
solar wind
charged particles
spiral down
magnetic field lines
near magnetic poles create aurora
* Caravaglio (2011)
4-3c
The ionized (charged) gases/plasma are following the magnetic field line protruding
from the Sun.
B
B
4-3d
Force on a straight wire with I due to B
I
B
F
Note: here
v BF
F=QvB
v

Q
= vt
or

t=
N = Am T
N
T=
Am
kg m
2
s
T=
C
m
s
kg
T
Cs
ok
F = (Q v) B = (I ) B F = (I ) B
In general I
B⊥
4-4 (4.5at end)
v
Q
but I =
t
Q Qv

so I =
v
and so I  Q v
units
θ
B
F = I B   I B sin(θ)
Flowing charge (current) creates magnetic field.
!!!!!!!!!!!!!!!!
2’nd right hand rule: RHR for B direction created by current
4-6
Magnitude of field (Biot & Savart Law)
{Like point charge electric field law but for current}
{NOT REQUIRED FOR THIS COURSE : BUT GOOD TO KNOW}
μ0 I Δ
ΔB =
sin(θ)
2
4π r
To calculate tot field add (*as vectors) up all B from each
Δ
[B&S Law- magnetic field equivalent of electric field point charge]
IΔ
ΔB = k' 2 sin(θ)
r
4-6a
(using Biot & Savart Law)
{NOT REQUIRED FOR THIS COURSE :
BUT GOOD TO KNOW}
4-7
Ampere’s Law
B||=B cos ()
here
μ0 I =
B
||
edge
Δl
μ0 I =

edge
4-8
B|| dl
recall Ampere’s Law
 I=
0
I= current
through surface
I 
B
 [B l ]
||
edge
B along l
Add up around surface edge.
 B field around an  wire with I
B along circle & B always constant on circle.
μ0 I = B 2πR
μ0 I
⇒B=
2πR
example
Suppose I = 1A : R = 1m : B=?
Units check
Recall: F=qVB F=I B
F
N
B=
 T=
I
Am
Ampere’s Law
or
For a wire
μ0
k'=
4π
2k'I
B=
R
N
] 1A
2
A
B=
2 1m
NA
N
 2(10)-7 2
B= 2(10)-7
A m
Am
[4 (10)-7
B= 2(10)-7 T very small!
4-9
 0 I1
B1 
 2.0 (107 ) T
2r1
 0I 2
B2 
 4.0 (107 ) T
2r2
B  B1  B 2
B  B1  B2  2.0 (107 ) T
A. Brahmia
4-9a
Force between 2 parallel wires L=1 m, at 0.1m, I = 50 A
.01 m  .05 N
.001 m  .5 N
4-10
4-10
Magnetic field of long solenoid
1.) B field outside goes to 0
4-11
Magnetic field of long solenoid
2.) B field inside


0= +I n L - I n L
Bout=0
2.) B field inside
4-11a
B
Superconducting solenoid
B = I n 0
R=0.025m
Q: What is outward
force on inside of such a
solenoid?
R=0.025m
turns
-7 N
 100A [100,000
] [ 4 (10) ] 2
m
A
2
5
-7 NA
=(10) 10 4 (10)
= 4 [3.14] T
2
mA
l= 2  R= 2  0.025m
F = I l B= (100A) (.025m 2π) 12T
ANm
= (100) (.025) (6.28)12
Am
=(2.5)(6.28) 12N
=150N
~33.7 lbs.
4-12
B = 12T
Magnetic Field Review

F= qv  B or F=qvB sin θ
B

v
F= I
IΔl
ΔB = k' 2 sin θ
r
Long wire:
μ0I
2I
B= k'
or B=
R
2πR
Center of Loop (circular):
R
Bcenter =
B
Right Hand
Rule
I
B or F= I B sinθ
Not required:
R
μ I
2πk'I
= 0
R
2R
Solenoid:
B = μ 0 nI = 4π k' nI
4-13
μ0
k'=
4π

Torque, , on current loop (assume square) in B field
 = magnetic moment (like electric dipole)
θ
loop plane)
θ
B
τ = μ B sin(θ)
=I L B
 = I L2 B F
 = I A B sin()
θ
F
θ
=I L B
4-13 (old4-5)