Examples of magnetic of magnetic fields ...
(indicated by compass needles or iron filings)
CHAPTER 26
N
N
S
S
THE MAGNETIC FIELD
• Force exerted on a charge by a magnetic field
• Motion of a charge in a magnetic field
Man made or naturally occuring ores ...
permanent magnet
• Cyclotron
• Velocity selector
I=0
I
• Torques on current loops and coils
• Magnetic moment
Direction of
the Earth’s
magnetic field
Oersted’s experiment (1819)...
electric current
I
The right hand rule:
Force on a moving charge:
!
! !
F = qv × B
!
! !
F = q v B sin θ
!
!
F
B
+
θ
!
B
θ
!
F
!
v
!
! !
F = qv × B
!
F
!
B
!
v
!
v
NOTE:
!
• Direction of F depends on sign of q.
!
v
• Charge must be moving!
!
• If θ = 0 then F = 0.
!
B
• Motion continues at constant speed ...
... Why??
!
• Direction of F given by the right-hand rule for a
positive charge ... next slide!
UNITS:
!
F⇒N
q⇒C
!
B ⇒ Tesla (T)
!
v ⇒ m/s
1 Gauss (G) = 10 −4 T
!
!
From v to B
(ccw)
!
!
From v to B
(cw)
!
F
!
The right hand rule gives the direction of F and is true
for all vector products.
!
!
!
Note that F is perpendicular to v and B.
The Earth’s magnetic field ... ~ 0.6G.
Magnets used to hold papers on a
refrigerator ... ~ 100G.
1
The magnetic fields associated with
2
sunspots ... ~ 1,000G.
The magnetic field inside atoms ...
!
B
4
5
!
B
3
~ 100,000G (10T).
Question 26.1: Electrons travel down a cathode ray tube
The strongest sustained magnetic fields
in a magnetic field that is directed vertically upward. At
produced by scientists in the lab ...
which point do they strike the screen?
~ 450,000G (45T). Pulsed fields ~ 150T.
Neutron stars ... ~ 108 T.
Magnetars ... ~ 1011 T.
Force on a straight wire in a magnetic field:
!
δℓ
Area A
!
!
B
F
+ +
I
!
θ
v
+
+
2
!
B
!
B
+ +
! !
Force on each moving charge ⇒ qv × B .
Since the charges are “held” in the wire, the total force
!
B
!
F (for +ve charge)
!
F (for −ve charge)
!
v
!
! !
F = qv × B, but q ⇒ −e.
Therefore, electrons will strike the screen at 2 .
on the element of wire length δℓ is:
!
! ! !
δF = nA δℓ qv × B ,
where n is the number of charges per unit volume and
!
A δ ℓ is the volume element. But, from Ch. 25, the
current in the wire is:
I = nqA v .
!
!
If δℓ is parallel to v , i.e., the current direction, then:
!
! !
δF = Iδℓ × B
!
So, for a straight wire of length ℓ , the total force is:
!
!!
F = I ℓ B sin θ
kˆ
!
Iℓ
!
B
!
B
Question 26.2: A straight wire segment has a current
element
!
Iℓ = (2.7)(3ˆi + 4 ˆj) A.cm
!
If the wire segment is in a magnetic field B = 1.3ˆi T,
(a) in what direction is the resulting force?
(b) What is the force on the wire?
!
Iℓ
ˆj
ˆi
! ! !
(a) We know F = I ℓ × B . Using the right hand and the
!
!
fingers in the direction from Iℓ → B , we see our thumb is
directed downwards. So, the force must be downward,
i.e., in the − kˆ direction.
! ! !
(b) F = I ℓ × B = 2.7 0.03ˆi + 0.04 ˆj × 1.3ˆi
(
(
) (
)
)
= 0.1053ˆi × ˆi + 0.1404ˆj × ˆi .
But ˆi × ˆi = 0 and ˆj × ˆi = − kˆ .
!
∴ F = −0.1404 kˆ N.
!
F
I
!
B
θ
ℓ
!
F
!
mg
Question 26.3: A straight, stiff, horizontal 25 cm long
In order to “float”, the net force on the wire must be
wire with mass of 50 g, is connected to a source of emf
zero,
by light, flexible leads. A magnetic field of 1.33 T is
horizontal and perpendicular to the wire. Find the
current necessary to “float” the wire, that is, when the
!
!
i.e., F = mg .
!!
∴F = I ℓ B sin θ = mg ,
mg
so, I = ! !
.
ℓ B sin θ
wire is released from rest it remains at rest.
With ℓ = 0.25m, B = 1.33T, θ = 90# , m = 0.05kg and
g = 9.81m/s2 .
∴I =
0.05 × 9.81
= 1.47A.
0.25 × 1.33
What’s the force if the wire isn’t straight?
Y
!
ℓ
!
ds
!
B
Break the wire XY into
!
small elements ( d s ). If
the wire carries a current I,
!
the force on an element d s
is:
A
I
B
I
!
! !
dF = Id s × B .
X
! Y ! !
∴ Total force on wire is: F = ∫ Id s × B .
X
!
! ⎧Y ! ⎫ !
Since I and B are constant: F = I ⎨ ∫ d s ⎬ × B .
⎪⎩X ⎪⎭
Y
!
But, ∫ d s is the vector sum of the elements from
X
!
→
X → Y ⇒ ℓ , i.e., the displacement XY.
! ! !
∴ F = Iℓ × B .
!
Special case: with a closed loop, ℓ = 0 so F = 0, i.e., a
closed loop will experience no net force!
!
B
I
C
I
D
Question 26.4: Four wires, A, B, C and D, shown on the
left, above, are placed in a magnetic field at the
orientations shown. If each wire carries the same
current, which one experiences the greatest force?
Motion of charges in a magnetic field ...
Cyclotron:
!
ℓ
A
I
!
ℓ
I
C
!
ℓ
B
+
!
B
I
!
B in
I
!
Dℓ
!
v
!
F
The force on the charge is
!
! !
F = qv B.
R
!
F
+
and it is always “inwards”.
Since there is no change in
!
v
speed (why??) this force is
!
2
constant and acts as a centripetal force, F = mv R , and
produces circular motion.
The force on the wires is:
! ! !
!
!!
F = Iℓ × B,
i.e.,
F = I ℓ B sin θ,
!
where, we have just shown, ℓ is the displacement of one
end of the wire relative to the other, and θ is the angle
!
!
!
!
between ℓ and B. Since I, ℓ and B are the same in each
#
case, the greatest force is for A ... where θ = 90 .
Note: the force is zero in case D.
∴qvB = m
v2
mv
⇒R =
.
R
qB
!
Note, with B into the page, the rotation is ccw for a
positive charge. The time to orbit (the cyclotron period)
is:
T=
2πR 2πm
=
.
v
qB
The cyclotron frequency, i.e., the number of orbits per
1
qB
second is:
f= =
.
T 2πm
S
Question 26.5: A proton with speed 1.00 × 106 m/s
straight lines
N
Q
enters a region with a uniform magnetic field of
(no field → no
r
magnitude 0.800 T (pointing into the page) as shown in
the figure. If the proton enters at an angle of 60! to the
RM and NS are
+
d
X
O
d
r
P
φ
R
+
is an arc of a
circle, with
2
radius r, and
"
B in
M
θ = 60!
(in the field)
d
normal, what is the exit angle?
+
force); M to N
2
its center at O.
v = 10 6 m/s
∠RMO = ∠SNO = 90!
(radius/tangent)
∠PMO = 30! = ∠MOX
(alternate angle)
By symmetry ∠NOX = 30! = 90! − φ, i.e., φ = 60! .
θ
+
"
B in
d
1
Also, sin(90 − φ) = sin 30 = 2 =
r
2
!
!
∴d = r
S
60"
Q
30"
+
N
R
Y
r
30"
O
30
X
"
d
r
P
30"
60"
R
M
!
B in
+ v = 10 6 m/s
Question 26.6: Shown here are the tracks of two protons
(R and Y) in a magnetic field that is perpendicular to the
mv
From earlier: r =
qB
=
page.
1.67 × 10−27 × 1× 106
= 0.0131m
1.6 × 10 −19 × 0.8
∴d = 0.0131 m
(13.1 mm)
(a) In which direction is the field, and
(b) which proton has the greater kinetic energy?
R
Y
!
v
Y
R
!
ds
!
F
+
!
F
!
B
(a) Using the right hand rule, the magnetic field must be
OUTWARDS.
(b) The radius of each arc is r =
mv
p
= ,
qB qB
p2
i.e., KE =
∝ r2.
2m
Proton Y has greater KE because it has the larger radius.
(c) The centripetal force due to the magnetic field,
!
! !
F = qv × B,
!
!
is perpendicular to v . But v is in the same direction as
!
!
d s (since v = d s dt ), so the work done over the
!
displacement d s is:
=0
! ! ! !
dW = F • d s = F d s cos90" = 0.
Therefore, the magnetic field does no work on either
proton. Also, the work-kinetic energy theorem
( W = ΔK) tells us that ΔK = 0, so there is no change in
the speed of the proton.
Motion of charges in a magnetic field ...
Velocity selector:
+
!
v
+
+
+
+
− − −
−
Magnetic field into the page
Electric field vertically down
Magnetic force
! !
qv B
+
!
qE
Electric force
!
v
!
B
ˆj
!
v
ˆi
!
E
−
The net force on the charge is:
!
!
! !
F = qE + (qv × B)
!
!
! !
There is no net force ( F = 0) if q E = q v B ,
!
! E !
i.e., when v =
.
B
Question 26.7: An electron enters a region where there
So, only charged particles with that particular velocity
acting on the electron the instant it enters the fields, if its
velocity is 5.00 × 105 ˆj m/s?
will pass through undeviated, no matter if charge is ±ve.
If the charge is +ve:
Faster ⇒ larger “magnetic” force
Slower ⇒ smaller “magnetic” force
is an electric field and a magnetic field that are
perpendicular to each other, as shown above. If the
strength of the electric field is 1.00 × 105 ˆi V/m and the
magnetic field strength is 1.50kˆ T, what is the net force
!
B
ˆj
Force on a coil carrying a current:
!
v
ˆi
!
E
−
!
!
F = −eE
−
!
! !
F = −ev × B
The force on the charge due to the
electric field is:
!
!
F = qE
= (−1.61 ×10 −19 )(1.00 × 105 V/m) ˆi
= −1.61 ×10 −14 N ˆi ,
1. Coil perpendicular to a magnetic field:
!
z
B
!
!
FBC
FCD
O
I nˆ C
B
!
D !
FAB
FDA
O′
A
AB = ℓ1 and BC = ℓ2
FAB = Iℓ1B
FBC = Iℓ 2B
magnetic field is:
FCD = −Iℓ1B
FDA = −Iℓ2B.
Since FAB = −FCD and FBC = −FDA
= (−1.61 ×10 −19 )(5.00 × 105 m/s ˆj ×1.50T kˆ ),
= −1.21 ×10 −13 N ˆi ,
i.e., to the left. Therefore, the net force on the charge is
!
Fnet = (−1.61 ×10 −14 − 1.21× 10 −13 )N ˆi
= −1.37 ×10
−13
x
The current (I) in each side is the same and the angles
!
!
between Iℓ and B is 90#. Therefore,
i.e., to the left. The force on the charge due to the
!
! !
F = qv × B
y
N ˆi .
there is no net force on the coil. (We could have
predicted that ... how?) Also, since the forces are all in
the same (x − y) plane as the coil (by the RH rule), there
is no net torque. The coil remains stationary in the
magnetic field.
2. Coil at an angle (θ) to a magnetic field:
nˆ
O
"
FCD
C
"
B
"
B
I
θ
B"
FAB
ℓ1
D
A
ℓ2
nˆ
θ
O′
90 − θ
B
I
A
FAB = Iℓ1Bsin(90 − θ) and FCD = Iℓ1Bsin(270 − θ).
"
"
∴ F AB = − F CD,
so there is no net force in the O − O′ direction .
"
B
nˆ
B
FBC = Iℓ 2B
θ
ℓ1 sin θ
O ℓ
1
FDA = Iℓ 2B
2. Coil at an angle (θ) to a magnetic field:
"
B
nˆ
B
FBC = Iℓ 2B
θ
ℓ1 sin θ
O ℓ
1
FDA = Iℓ 2B
The torque:
A
τ = (Iℓ1ℓ2 )Bsin θ = µBsin θ,
area of coil
where we define: µ = I(ℓ1ℓ2 ) = IA
as the MAGNETIC MOMENT of the coil. The magnetic
moment (units ⇒ A.m2 ) is a vector in the nˆ direction
given by the right hand rule, viz:
A
FBC and FDA form a couple and produce a torque that
tends to rotate the coil cw about O − O′ . The torque is:
1
1
τ = (Iℓ2B) ℓ1 sin θ + (Iℓ2B) ℓ1 sin θ
2
2
= (Iℓ1ℓ2 )Bsin θ.
"
µ
I
"
µ
I
2. Coil at an angle (θ) to a magnetic field:
In general, the magnetic moment of a coil with n turns
of wire is
!
B
µ = nIA.
So, the magnetic moment of a circular coil of radius R
and n turns is:
µ = nIπR 2 ,
!
and if it is set an angle θ to a magnetic field B, it will
experience a torque:
Question 26.8: A square coil, (10.0 × 10.00) cm2 , sits on
a flat horizontal surface in a region where there is a
magnetic field, B = 0.20 T, parallel to the plane of the
2
τ = (nIπR )Bsin θ.
coil, as shown above. The coil consists of 5 turns of
NOTE: the direction of the torque is given by the RH
wire and has a mass of 0.10 kg. What current is
rule. In vector notation:
! ! !
τ =µ×B
!
so τ = τ = µBsin θ.
required to just raise the right hand edge of the coil off
!
τ
!
µ
!
B
the surface? In what direction is the current?
"
µ
"
B
"
mg
The right hand edge of the coil will lift off the surface
when the magnitude of the magnetic torque acting on it
equals the magnitude of the gravitational torque. Note
that the gravitational force acts at the center of gravity of
the square. So, the condition for lift is
τ mag = τ grav ,
ℓ
where τ grav = mg , and τ mag = µB = nIℓ2B.
2
τ grav acts clockwise about the left hand edge, so τ mag
"
must act counter-clockwise, which means that µ is
" " "
directed upwards, since τ = µ × B . Hence the current is
counter-clockwise when the coil is viewed from above.
Thus, equating torques,
ℓ
mg
nIℓ2B = mg , i.e., I =
.
2
2nℓB
Substituting the given values:
mg
0.10 × 9.81
I=
=
= 4.91 A.
2nℓB 2 × 5 × 0.10 × 0.20
(a) The current required is 4.91 A.
(b) The direction of the current should be clockwise
when viewed from above.
y
x
!
(a) B along z
!
(b) B along x
Question 26.9: A 10-turn square coil, with edge length
coil with a uniform magnetic field of magnitude 0.30 T in
the
N
P
(a) The magnetic moment of the coil is
µ = nIA = 10 × 2.5 × (0.06 × 0.06)
6.00 cm, lies in the z = 0 plane. The wire carries a current
of 2.50 A. What is the magnitude of the torque on the
M 0.06m
!
µ
×
0.06m
I
Q
= 9.00 × 10−2 A ⋅ m2 .
From earlier, τ = µBsin θ, where θ is the angle between
!
!
µ and B ( = 0 or 180").
∴sin θ = 0, i.e., τ = 0.
(a) +z direction, and
(b) +x direction?
(b) With θ = 90" , τ = µBsin θ = 9.00 × 10 −2 × 0.3 × 1
= 2.70 × 10−2 N ⋅ m.
!
!
NOTE: The direction of τ depends on the direction of µ ,
the magnetic moment of the coil. For the direction
!
!
chosen, µ is in the −z direction and so τ is in the −y
direction.
I
0.3m
0.5m
1
2
Question 26.10: A wire loop consists of two semicircles
connected by straight segments. The inner and outer radii
(a) The magnetic moment of the coil is:
are shown in the figure. A current of 1.50 A flows in the
µ = nIA
1
where A is the shaded area = π(R12 − R 22 ).
2
1
∴µ = × 1.5 × π × 0.52 − 0.32 = 0.377A.m2 .
2
wire, which is a clockwise direction in the outer loop.
(a) What is the magnetic moment of this loop, and
(b) in which direction is it?
(
)
(b) The direction of the (net) magnetic moment is
inwards.