Torque on Current Loop
Î Consider
a
rectangular current loop
Forces in left, right branches = 0
‹ Forces in top/bottom branches cancel
‹ No net force! (true for any shape)
‹
Î But
b
there is a net torque!
Bottom side up, top side down (RHR)
‹ Rotates around horizontal axis
‹
τ = Fd = ( iBa ) b = iBab = iBA
Îμ
B
b
a
Plane normal is ⊥ B here
= NiA ⇒ “magnetic moment” (N turns)
True for any shape!!
‹ Direction of μ given by RHR
‹
PHY2054: Chapter 19
21
General Treatment of Magnetic Moment, Torque
Îμ
= NiA is magnetic moment (with N turns)
‹ Direction
ÎTorque
of μ given by RHR
depends on angle θ between μ and B
τ = μ B sin θ
PHY2054: Chapter 19
22
Torque Example
ÎA
3-turn circular loop of radius 3 cm carries 5A current in
a B field of 2.5 T. Loop is tilted 30° to B field.
30°
2
2
Î μ = 3iπ r = 3 × 5 × 3.14 × ( 0.03 ) = 0.0339 A ⋅ m
2
Îτ
= μ B sin 30 = 0.0339 × 2.5 × 0.5 = 0.042 N ⋅ m
ÎRotation
always in direction to align μ with B field
PHY2054: Chapter 19
23
Trajectory in a Constant Magnetic Field
ÎA
charge q enters B field with velocity v perpendicular to
B. What path will q follow?
is always ⊥ velocity and ⊥ B
‹ Path will be a circle. F is the centripetal force needed to keep the
charge in its circular orbit. Let’s calculate radius R
‹ Force
x x x x x x x x x x x x x x
B
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 vx x
F
F
v
v
F
q
R
PHY2054: Chapter 19
24
Circular Motion of Positive Particle
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 x x x x x x x x x x x x x x
v
B
F
q
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 x x x x x x x x x x x x x x
2
mv
= qvB
R
mv
R=
qB
PHY2054: Chapter 19
25
Cosmic Ray Example
with energy 1 MeV move ⊥ earth B field of 0.5
Gauss or B = 5 × 10-5 T. Find radius & frequency of orbit.
ÎProtons
K=
1 mv 2
2
2K
⇒ v=
m
( )(
)
K = 106 1.6 × 10−19 =1.6 × 10−13 J
m = 1.67 × 10−27 kg
R = 2900 m
mv
2mK
R=
=
eB
eB
1
v
v
eB
f = =
=
=
T 2π R 2π ( mv / eB ) 2π m
f = 760 Hz
Frequency is independent of v!
PHY2054: Chapter 19
26
Helical Motion in B Field
ÎVelocity
of particle has 2 components
G G G
‹ v = v& + v⊥ (parallel to B and perp. to B)
‹ Only
v⊥ = v sinφ contributes to circular motion
‹ v|| = v cosφ is unchanged
ÎSo
the particle moves in a helical path
‹ v||
is the constant velocity along the B field
‹ v⊥ is the velocity around the circle
v||
v
v⊥
B
φ
mv⊥
R=
qB
PHY2054: Chapter 19
27
Helical Motion in Earth’s B Field
PHY2054: Chapter 19
28
Magnetic Field and Work
ÎMagnetic
force is always perpendicular to velocity
‹ Therefore
B field does Gno work! G
G
G
‹ Why? Because ΔK = F ⋅ Δx = F ⋅ ( v Δt ) = 0
ÎConsequences
‹ Kinetic
energy does not change
‹ Speed does not change
‹ Only direction changes
G
‹ Particle moves in a circle (if v ⊥
G
B)
PHY2054: Chapter 19
29
Magnetic Force
ÎTwo
particles of the same charge enter a magnetic field
with the same speed. Which one has the bigger mass?
‹A
‹B
‹ Both
masses are equal
‹ Cannot tell without more info
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
mv
R=
qB
x x x x x x x x x x x x
x x x x x x x x x x x x
A
Bigger mass means
bigger radius
PHY2054: Chapter 19
B
30
Mass Spectrometer
PHY2054: Chapter 19
31
Mass Spectrometer Operation
ions first enter a “velocity selector” where E ⊥ B
and values are adjusted to allow only undeflected particles
to enter mass spectrometer.
ÎPositive
‹ Balance
forces in selector ⇒ “select” v
qE = qvB
v= E/B
2
‹ Spectrometer:
Determine mass
from v and measured radius r
1
m1v
r1 =
qB
m2v
r2 =
qB
PHY2054: Chapter 19
32
Mass Spectrometer Example
ÎA
beam of deuterons travels right at v = 5 x 105 m/s
‹ What
value of B would make deuterons go undeflected through a
region where E = 100,000 V/m pointing up vertically?
eE = evB
B = E / v = 105 / 5 × 105 = 0.2T
‹ If
the electric field is suddenly turned off, what is the radius and
frequency of the circular orbit of the deuterons?
2
mv
mv
= evB ⇒ R =
=
R
eB
(
)( ) = 5.2 ×10
) ( 0.2)
3.34 × 10−27 5 × 105
(1.6 ×10
−19
−2
m
v
1
5 × 105
f = =
=
= 1.5 × 106 Hz
T 2π R ( 6.28 ) 5.2 × 10−2
(
)
PHY2054: Chapter 19
33