21.4 The Mass Spectrometer
Application: circular motion of moving
ions In a uniform magnetic field:
The mass spectrometer
mv mv
r=
=
qB eB
1
2
magnitude of
electron charge
mv 2 = eV
KE=PE
2 2
e
r 2
eBr
2
v=
→v = 2 B
m
m
2 2
r 2
e
2
1
B = eV
2 mv =
2m
 er 2  2
 B
m = 
 2V 
The mass spectrum of naturally
occurring neon, showing three
isotopes.
1
Example: A singly charged positive ion has a mass of 2.5 x 10-26 kg. After
being accelerated through a potential difference of 250 V, the ion enters a
magnetic field of 0.5 T, in a direction perpendicular to the field. Calculate
the radius of the path of the ion in the field.
2
Example: A singly charged positive ion has a mass of 2.5 x 10-26 kg. After
being accelerated through a potential difference of 250 V, the ion enters a
magnetic field of 0.5 T, in a direction perpendicular to the field. Calculate
the radius of the path of the ion in the field.
q = 1.6 x10 −19 C
m = 2.5 x10 −26 kg
∆V = 250 V
B = 0.5 T
r=?
FB = Fc
mv 2
qvB =
r
W ∆K
∆V =
=
=
q
q
v=
2∆Vq
=
m
1
2
r=
mv
qB
We need to
solve for the
velocity!
mv 2
q
2( 250)(1.6 x10 −19 )
4
=
×
5.66
10
m/s
− 26
2.5 x10
( 2.5 x10 −26 )(56,568)
r=
= 0.0177 m
−19
(1.6 x10 )(0.5)
3
21.3 The Motion of a Charged Particle in a Magnetic Field
Example: Velocity Selector
A velocity selector is a device for measuring
the velocity of a charged particle. The device
operates by applying electric and magnetic
forces to the particle in such a way that these
forces balance.
Given B and q , wow should an electric field be
applied so that the force it applies to the
particle can balance the magnetic force?
Solution: by RHR-1: the velocity is to the
right (thumb) and the magnetic field (finger) is
into the page (“x” marks the tail of an arrow),
so the magnetic force on a positive charge is
upward (palm up). Here θ = 90°  FB = qvB
We therefore need the electric force FE that
points down that has the same magnitude: for
a positive charge the electric field then needs
to point in the same direction as the desired
force, and FE = qE.
We want FE = FB qE = qvB

E = vB (and pointing downward)
4
21.5 The Force on a Current in a Magnetic Field
Magnetic force on a current
The magnetic force on the moving charges pushes the
wire to the right.
Since a current consists of moving charges then a
current is subject to the Lorentz force
In Class Demo
Magnetic force on a straight
current segment of length L
F = qvB sin θ
 ∆q 
F =  (
v∆t )B sin θ
t L
∆

I
F = ILB sin θ
Here θ is the
angle between the
direction of the
current and the
magnetic field
5
21.5 The Force on a Current in a Magnetic Field
Example: A wire carries a current of 22.0 A from west to east. Assume that at this location
the magnetic field of Earth is horizontal and directed from south to north and that it has a
magnitude of B=0.500×10−4 T.
(a) Find the magnitude and direction of the magnetic force on a 36.0 m length of wire.
(b) Calculate the gravitational force on the same length of wire if it’s made of copper and
has a cross-sectional area of 2.50×10−6 m2.
6
21.5 The Force on a Current in a Magnetic Field
Example: A wire carries a current of 22.0 A from west to east. Assume that at this location
the magnetic field of Earth is horizontal and directed from south to north and that it has a
magnitude of B=0.500×10−4 T.
(a) Find the magnitude and direction of the magnetic force on a 36.0 m length of wire.
(b) Calculate the gravitational force on the same length of wire if it’s made of copper and
has a cross-sectional area of 2.50×10−6 m2.
Solution :
(a) By RHR - 1 the force is pointing upward
FB = ILB sin θ
where θ is the angle between = the direction of the current segment and the magnetic field
Here θ = 90° → FB = ILB = ( 22.0 A )(36.0 m)(5.00 × 10 -5 T) = 3.96 × 10 −2 N (up)
(b) Density of copper is ρ = 8.94 × 10 3 kg/m3
Mass of wire : M = ρ ( volume) = ρAL
M = (8.94 × 10 3 kg/m3 )( 2.50 × 10 −6 m 2 )(36.0 m) = 0.805 kg
FG = Mg = (0.805 kg)(9.81 m/s 2 ) = 7.89 N
FG / FB = 199
7
21.7 Magnetic Fields Produced by Currents
In Class Demo
A LONG (∞), STRAIGHT WIRE
carrying current produces a
magnetic field Result to remember:
µo I
B=
2π r
Magnitude
derivation requires integral
calculus
µo = 4π × 10 −7 T ⋅ m A
permeability of
free space
Direction
Right-Hand Rule No. 2. Curl the
fingers of the right hand into the shape of
a half-circle. Point the thumb in the
direction of the conventional current,
and the tips of the fingers will point in the
direction of the magnetic field.
8
21.7 Magnetic Fields Produced by Currents
Example: What is the
Current carrying wires can exert forces on each other.
attraction/repulsion force per meter
of wire between two infinite straight
wires carrying 1.00A each, separated
by 1.00 m?
9
21.7 Magnetic Fields Produced by Currents
Example: What is the
Current carrying wires can exert forces on each other.
attraction/repulsion force per meter
of wire between two infinite straight
wires carrying 1.00A each, separated
by 1.00 m?
µ0 I 1 ( 4π × 10 −7 T ⋅ m/A)(1.00 A)
B1 =
=
= 2.00 × 10 −7 T
2π r
2π (1.00 m)
F21 = I 2 L2 B1 sin θ , θ = 90°
µII
F21
= I 2 B1 = 0 1 2
L2
2π r
= (1.00 A)( 2.00 × 10 −7 T)
N ⋅s C
N
= 2.00 × 10 −7
⋅ = 2.00 × 10 −7
C⋅m s
m
1m
This is in fact the original SI
definition of an ampere (A).
10
21.7 Magnetic Fields Produced by Currents
SOLENOID
number of turns per
unit length n = N / L
Magnetic field in the
interior of a solenoid
Is approximately uniform
In the limit of infinite
length, but finite n
B = µo nI =
µo NI
Magnitude
L
Result to remember:
derivation requires
integral calculus
11
21.7 Magnetic Fields Produced by Currents
A CIRCULAR LOOP OF WIRE

B
R.H.
R
Two different applications of RHR-2
for the same field configuration
At the center of the circular loop
only
Magnitude
B=
µo I
2R
Result to remember: derivation
requires integral calculus
12
21.7 Magnetic Fields Produced by Currents
Example: Net Magnetic Field
A long straight wire carries a current of 8.0 A and a
circular loop of wire carries a current of 2.0 A and has
a radius of 0.030 m. Find the magnitude and direction
of the magnetic field at the center of the loop.
13
21.7 Magnetic Fields Produced by Currents
Example: Net Magnetic Field
A long straight wire carries a current of
8.0 A and a circular loop of wire carries
a current of 2.0 A and has a radius of
0.030 m. Find the magnitude and
direction of the magnetic field at the
center of the loop.
µ o I1 µ o I 2 µ o  I1 I 2 

B=
−
=
− 
2π r 2 R
2 πr R 
(
4π × 10
B=
T ⋅ m A )  8 .0 A
2.0 A 

 = 1.1 × 10 −5 T
−
2
 π (0.030 m ) 0.030 m 
−7
14
21.7 Magnetic Fields Produced by Currents
The field lines around a current loop resemble those around the bar magnet.
Attraction
Repulsion
15
21.6 The Torque on a Current-Carrying Coil
Put a rectangular loop of current I and length (height) L, and width w in a uniform magnetic
field B. The loop is mounted such that it is free to rotate about a vertical axis through its
center. We will consider the forces on each segment and the resulting torque from each.
Using RHR-1: The force on segment 3 points down, and that on segment 4 points up.
F3 and F4 are also equal in magnitude and cancel one another.
The magnitudes F3 = F4 = IwBsin(90°-φ) = IwBcosφ also change with the rotation angle φ
But both F3 and F4 are directed parallel to the axis, and results in no torque.
Top view
3
3
4
φ is the angle between
the “normal” to the loop
and the magnetic field
16
21.6 The Torque on a Current-Carrying Coil
Looking at segments 1 and 2 which have the current running vertically.
By RHR-1, force F1 on segment 1 (current up) points into the page, for all values of φ. Also
by RHR-1, force F2 on segment 1 (current down) points out of the page. They cancel each
other to yield no net force on the loop.
However, F1 and F2 both tend to turn the loop in the clockwise sense (as seen in the top
view). The torques from the two forces are each
 w
τ 1, 2 = (F1, 2 )  sin φ
2
τ = τ 1 + τ 2 = Fw sin φ
Top view
F = ILB sin θ = ILB
since θ = 90°
2
1
φ is the angle between
the “normal” to the loop
and the magnetic field
17
21.6 The Torque on a Current-Carrying Coil
τ = τ 1 + τ 2 = Fw sin φ
The torque τ is maximum when the normal of the
loop is perpendicular to the magnetic field, and
zero when the normal is parallel to the field.
F = ILB sin θ = ILB
since θ = 90°
The torque tends to cause the loop normal to
become aligned to the field, just like on a bar
magnet. Current loop = magnetic dipole
Net torque = τ = ILB (w sin φ ) = IAB sin φ
Top view
A = Lw =
area of loop
magnetic
moment m

τ = NIA B sin φ
number of turns of wire
18
21.6 The Torque on a Current-Carrying Coil
Example The Torque Exerted on a Current-Carrying Coil
A coil of wire has an area of 2.0x10-4m2, consists of 100 loops or turns, and contains a
current of 0.045 A. The coil is placed in a uniform magnetic field of magnitude 0.15 T.
(a) Determine the magnetic moment of the coil.
(b) Find the maximum torque that the magnetic field can exert on the coil.
19
21.6 The Torque on a Current-Carrying Coil
Example The Torque Exerted on a Current-Carrying Coil
A coil of wire has an area of 2.0x10-4m2, consists of 100 loops or turns, and
contains a current of 0.045 A. The coil is placed in a uniform magnetic field of
magnitude 0.15 T.
(a) Determine the magnetic moment of the coil.
(b) Find the maximum torque that the magnetic field can exert on the coil.
magnetic
moment

(a) m = NIA = (100 )(0.045 A )(2.0 × 10 − 4 m 2 ) = 9.0 × 10 − 4 A ⋅ m 2
magnetic
moment

−4
2

−4
(b) τ = NIA B sin φ = 9.0 × 10 A ⋅ m (0.15 T ) sin 90 = 1.4 × 10 N ⋅ m
(
)
20
21.6 The Torque on a Current-Carrying Coil
Application: The basic components of
a dc motor.
The brushes switches the
direction of the current so
that the torque is always in
the same direction 
continuous rotation
21
21.8 Ampere’s Law
AMPERE’S LAW FOR STATIC MAGNETIC FIELDS
For any current geometry that produces a
magnetic field that does not change in time,
∑ B ∆ = µ I
||
o
net current passing
through surface
bounded by path
“positive” sense for the current
bounded is related to the direction of
the path traveled by RHR-2
Positive direction for
bound currents
22
21.8 Ampere’s Law
Example An Infinitely Long, Straight, Current-Carrying Wire
Use Ampere’s law to obtain the magnetic field.
23
21.8 Ampere’s Law
Example An Infinitely Long, Straight, Current-Carrying Wire
Use Ampere’s law to obtain the magnetic field.
1. By symmetry, the magnetic field must have cylindrical
symmetry: depends only on distance r to the wire.
2. The magnetic field should be generally perpendicular to the
current (by Right Hand Rule).
Choose a circular path
(at constant r) to match
the symmetry
The magnetic field lines
circulate around the line
by RHR-2: traverse path
in the CCW direction as
seen from the top
The field is parallel to
the path and has
constant magnitude on
the path.
∑ B ∆ = µ I
||
o
B (∑ ∆ ) = µ o I
B( 2π r ) = µo I
µo I
B=
2π r
24