Prof. Anchordoqui
Problems set # 8
Physics 169
April 14, 2015
1. The unit of magnetic flux is named for Wilhelm Weber. The practical-size unit of magnetic
field is named for Johann Karl Friedrich Gauss. Both were scientists at Gottingen, Germany. Along
with their individual accomplishments, together they built a telegraph in 1833. It consisted of a
battery and switch, at one end of a transmission line 3 km long, operating an electromagnet at the
other end. (André Ampérè suggested electrical signaling in 1821; Samuel Morse built a telegraph
line between Baltimore and Washington in 1844.) Suppose that Weber and Gausss transmission
line was as diagrammed in Fig. 1. Two long, parallel wires, each having a mass per unit length of
40.0 g/m, are supported in a horizontal plane by strings 6.00 cm long. When both wires carry the
same current I, the wires repel each other so that the angle θ between the supporting strings is
16.0◦ . (i) Are the currents in the same direction or in opposite directions? (ii) Find the magnitude
of the current.
Solution The separation between the wires is a a = 2 · 6.00cm · sin 8.00◦ = 1.67 cm. (i) Because
the wires repel, the currents are in opposite directions. (ii) Because the magnetic force acts horiµ0 I 2 `
◦
= tan 8.00◦ , yielding I 2 = mg2πa
zontally, FFBg = 2πamg
µ0 tan 8.00 and so I = 67.8 A.
2. Figure 2 is a cross-sectional view of a coaxial cable. The center conductor is surrounded by
a rubber layer, which is surrounded by an outer conductor, which is surrounded by another rubber
layer. In a particular application, the current in the inner conductor is 1.00 A out of the page and
the current in the outer conductor is 3.00 A into the page. Determine the magnitude and direction
of the magnetic field at points a and b.
µ 0 Ia
Solution From Ampere’s law, the magnetic field at point a is given by Ba = 2πr
, where Ia is
a
the net current through the area of the circle of radius ra . In this case, Ia = 1.00 A out of the page
−7 T·m/A·1 A
(the current in the inner conductor), so Ba = 4π×10
= 200 µT toward top of the page.
2π·1.00×10−3 m
µ0 Ib
Similarly at point b: Bb = 2πrb , where Ib is the net current through the area of the circle having
radius rb . Taking out of the page as positive, Ib = 1.00 A − 3.00 A = −2.00 A, or Ib = 2.00 A into
−7 T·m/A·2 A
the page. Therefore, Bb = 4π×10
= 133 µT toward the bottom of the page.
2π·3.00×10−3 m
3. A long cylindrical conductor of radius R carries a current I as shown in Fig. 3. The current
density J, however, is not uniform over the cross section of the conductor but is a function of the
radius according to J = br, where b is a constant. Find an expression for the magnetic field B
(i) at a distance r1 < R and (ii) at a distance r2 > R, measured from the axis.
H
H
~ · d~s = µ0 I. For current density J,
~ this becomes B
~ · d~s =
Solution Use Ampérè’s law, B
R
R r1
2
µ
br
0 1
~ (i) For r1 < R, this gives B2πr1 = µ0
µ0 J~ · dA.
(for r1 < R or
3
0 br2πrdr and B R=
3
R
inside the cylinder. (ii) When r2 > R, Ampérè’s law yields 2πr2 )B = µ0 0 br2πrdr = 2πµ30 bR or
B=
µ0 bR3
3r2
(for r2 > Ror outside the cylinder).
4. A toroid with a mean radius of 20.0 cm and 630 turns is filled with powdered steel whose
magnetic susceptibility χ is 100. The current in the windings is 3.00 A. Find B (assumed uniform)
inside the toroid.
Solution: Assuming a uniform B inside the toroid is equivalent to assuming r R (see Fig. 4);
630·3.00
then B0 = µ0 H ≈ µ0 N I as for a tightly wound solenoid. This leads to B0 = µ0 2π·0.200
= 0.00189 T.
With the steel, B = (1 + χ)µ0 H = 101 · 0.00189 T = 0.191 T.
5. At saturation, when nearly all of the atoms have their magnetic moments aligned, the magnetic field in a sample of iron can be 2.00 T. If each electron contributes a magnetic moment of
9.27×10−24 A · m2 (one Bohr magneton), how many electrons per atom contribute to the saturated
field of iron? Iron contains approximately 8.50 × 1028 atoms/m3 .
Solution Consider a torus made of unmagnetized iron. If the current in the primary coil is
increased from zero to some value I, the magnitude of the magnetic field strength H increases
linearly with I according to H = nI. Furthermore, the magnitude of the total field B also increases
with increasing current, as shown by the curve from point O to point a in Fig. 5. At point O, the
domains in the iron are randomly oriented, corresponding to M = 0. As the increasing current
in the primary coil causes the external field to increase, the aligned domains grow in size until
nearly all magnetic moments are aligned at point a. At this point the iron core is approaching
saturation, which is the condition in which all magnetic moments in the iron are aligned. Next,
suppose that the current is reduced to zero, and the external field is consequently eliminated. The
B-versus-H curve, called a magnetization curve, now follows the path ab in Fig. 5. Note that
at point b, B is not zero even though the external field H is zero. The reason is that the iron
is now magnetized due to the alignment of a large number of its magnetic moments. At this
point, the iron is said to have a remanent magnetization. Shortly after removing the external
field we have 2.00 T = µ0 M . In addition, M = xnµB , where n is the number of atoms per volume and x is the number of electrons per atom contributing. Therefore B = µ0 µB xn, yielding
2.00 T
x = µ0 B
µB n = 8.50×10−28 m−3 ·9.27×10−24 N·m/T·4π×10−7 T·m/A = 2.02.
6. The magnetic moment of the Earth is approximately 8.00 × 1022 A · m2 . (i) If this were
caused by the complete magnetization of a huge iron deposit, how many unpaired electrons would
this correspond to? (ii) At two unpaired electrons per iron atom, how many kilograms of iron
would this correspond to? [Hint: Iron has a density of 7, 900 kg/m3 , and approximately 8.50 × 1028
iron atoms/m3 .]
22
2
8.00×10 A·m
45
Solution (i) Number of unpaired electrons = 9.27×10
−24 A·m2 = 8.63 × 10 . Each iron atom
has two unpaired electrons, so the number of iron atoms required is 21 8.63 × 1045 . (ii) Mass =
4.31×1045 atoms·7,900 kg/m3
8.50×1028 atoms/m3
= 4.01 × 1020 kg.
7. Find the magnetic field on the axis at a distance a above a disk of radius R with charge
density σ rotating at an angular speed (clockwise when viewed from the point P ).
Solution Consider a thin ring between r and r + dr as shown in Fig. 6. The charge per length
on the ring is q = σdr and each point on it is moving at a speed v = ω~r, so the current is
I = vq = σωrdr. By symmetry, the magnetic field is in the z direction. Using the result from
dr)
r2
problem 6 of the previous homework, the contribution from the ring is dBz = µ0 (σωr
.
2
(r2 +z 2 )3/2
R
3
R
The magnetic field for the entire disk is Bz = µ02σω 0 (r2 +zr 2 )3/2 dr. Let u = r2 , so du = 2rdr. This
R2
R R2
2(u+2z 2 ) =
substitution (dont forget to change limits) gives Bz = µ04σω 0 (u+zu2 )3/2 du = µ04σω (u+z
2 )1/2 0
h 2 2
i
µ0 σω
R +2z
− 2z .
2
(R2 +z 2 )1/2
8. A right-circular solenoid of finite length L and radius a has N turns per unit length and carries a current I. Show that the magnetic induction on the cylinder axis in the limit N L → inf ty
is Bz = µ02N I (cos θ1 + cos θ2 ), where the angles are defined in Fig. 7.
Solution We start by computing the magnetic field on axis for a single loop of wire carrying a
current I shown in Fig. 8. This may be done by an elementary application of the Biot-Savart law.
R [d~`×R]
R d`R sin α
~ z
0I
0I
0I
By symmetry, only the z component contributes Bz = µ4π
= µ4π
= µ4π
2πa Ra3 =
R3
R3
µ0 Ia2
.
2R3
2
0 Ia
Substituting in R2 = a2 + z 2 yields Bz = 2(aµ2 +z
2 )3/2 . We now use linear superposition
to obtain the field of the solenoid. Defining z1 and z2 as in Fig. 8 (where z1 + z2 = L) we
2 Rz
dz
have Bz = µ0 N2Ia −z2 1 (a2 +z
2 )3/2 . A simple trig substitution z = a tan α converts this integral to
+ tan−1 (z2 /a)
R + tan−1 (z /a)
Bz = µ02N I − tan−1 (z12/a) cos αdα = µ02N I sin α
. A bit of geometry then demonstrates
−1
− tan
that this is equivalent to Bz =
µ0 N I
2 (cos θ1
(z1 /a)
+ cos θ2 ).
9. Two circular coils of radius a, each with N turns, are perpendicular to a common axis. Each
coil carries a steady current I in the same direction, as shown in Fig. 9 (i) Find the total magnetic
field on the axis Bztot as a function of z, using the midway point as the origin (z = 0). (ii) Show that
dBztot
dz = 0 at the midpoint. This means that, functionally, at least at that one location, the field is
2
tot
Bz
not changing. (iii) If the separation b is picked correctly, then d dz
= 0 at the midpoint. This
2
configuration is known as Helmholtz coils. Determine the appropriate value of b. (iv) Show that
the magnetic field on the axis near the origin, as an expansion in powers of z (up to z 4 inclusive)
is given by
µ0 N Ia2
3(b2 − a2 )z 2 15(b4 − 6b2 a2 + 2a4 )z 4
Bztot =
1
+
+
+
·
·
·
,
d3
2d4
16d8
where d2 = a2 + b2 /4. (v) Show that the magnetic field on the z axis for large |z| is given by the
expansion in inverse odd powers of |z| obtained from the small z expansion of part (iv) by the
formal substitution d → |z|. (vi) What is the maximum permitted value of |z|/a if the field is to
be uniform to one part in 104 (or one part in 102 ) in the Helmholtz configuration?
Solution (i) The magnetic field due to a circular hoop of radius a carrying a current I, which
2
0 Ia
lies in the x-y plane with its center at the origin, is Bz = 2(z 2µ+a
2 )3/2 . If the coils on the system
have N turns, then the field from each coil is just N times larger. The magnetic field from both
coils is then given by
Bztot = Bz1 + Bz2
(
)
µ0 N Ia2
1
1
=
+
2
[a2 + (b/2 + z)2 ]3/2 [a2 + (b/2 − z)2 ]3/2
(1)
(ii) The derivative is
dBztot
3µ0 N Ia2
=
dz
2
It is easy to see that
again gives
d2 Bztot
dz 2
=
dBztot dz z=0
3µ0 N Ia2
2
n
(
b/2 − z
[a2 + (b/2 − z)2 ]5/2
−
b/2 + z
[a2 + (b/2 + z)2 ]5/2
)
.
= 0. This is true regardless of the separation b. (iii) Differentiating
o
5(b/2+z)2
5(b/2−z)2
1
1
−
+
−
.
7/2
5/2
7/2
5/2
2
2
2
2
2
2
2
2
[a +(b/2+z) ]
[a +(b/2+z) ]
[a +(b/2−z) ]
[a +(b/2−z) ]
At the midpoint, the second derivative becomes
)
(
d2 Bztot 3µ0 N Ia2
10b2 /4
2
=
−
dz 2 z=0
2
[a2 + b2 /4]7/2 [a2 + (b/2)2 ]5/2
#
"
2
2
2
2 a + b /4 − 5b /4
= 3µ0 N Ia
(a2 + b2 /4)7/2
"
#
a2 − b2
2
= 3µ0 N Ia
.
(a2 + b2 /4)7/2
2 B tot z
Therefore, the condition d dz
= 0 is met if b = a. (iv) All we must do now is to Taylor expand
2 z=0
the terms to order z 4 . Noting that we are seeking an expansion in powers of z/d2 , we may rewrite
(1) as
Bztot =
=
=
µ0 N Ia2
2
µ0 N Ia2
2d3
µ0 N Ia2
2d3
h
i
(d2 − bz + z 2 )−3/2 + (d2 + bz + z 2 )−3/2
h
i
(1 − bζ + d2 ζ 2 )−3/2 + (1 + bζ + d2 ζ 2 )−3/2
n
−3/2 −3/2 o
1 − bζ + (a2 + b2 /4)ζ 2
+ 1 + bζ + (a2 + b2 /4)ζ 2
,
(2)
where we have introduced ζ = z/d2 . Expanding this in powers of ζ yields
3 2
15 4
µ0 N Ia2
2 2
2 2
4 4
tot
Bz =
1 + (b − a )ζ + (b − 6b a + 2a )ζ + · · · ,
2d3
2
16
which is the desired result. The magnetic field along the axis changes very slowly when z is very
small. (v) For large |z| we Taylor expand Bztot in inverse powers of z, that is
Bztot =
o
µ0 N Ia2 n
−1
2
2
−2 −3/2
−1
2
2
−2 −3/2
1
−
bz
+
(a
+
b
/4)z
+
1
+
bz
+
(a
+
b
/4)z
.
2|z|3
Comparing this with the last line of (2) shows that the Taylor series is formally equivalent under
the substitution ζ → z −1 , which may be accomplished by taking d → |z|. (vi) For b = a the field
is of the form
µ0 N Ia2
45 a4 z 4
4µ0 N Ia2
144 z 4
tot
Bz =
1−
+ ··· =
1−
+ ··· .
2d3
16 d8
125 a
53/2 a3
tot
144 z 4
z
Taking the (|z|/a)4 term as a small correction, the field non-uniformity is δB
Bztot ≈ 125 a . For
uniformity to one part in 104 , we find |z|/a < 0.097, while for uniformity to one part in 102 , we
instead obtain |z|/a < 0.305. These numbers are actually pretty good because of the fourth power.
For example, the first value indicates we can move ≈ ±10% of the distance between the coils while
maintaining field uniformity at the level of 0.01%. Helmholtz coils are very useful in the lab for
canceling out the Earth’s magnetic field.
10. Let us treat the motion of an electron (charge −e, mass m) in a hydrogen atom classically.
Suppose that an electron follows a circular orbit of radius r around a proton. What is the angular
~ perpendicular to the
frequency ω0 of the orbital motion? Suppose now that a small magnetic field B
plane of the orbit is switched on. Assuming that the radius of the orbit does not change, calculate
the shift in the angular frequency ∆ω = ωf − ω0 of the orbital motion in terms of the magnitude
B of the magnetic field, charge −e, mass m, and radius r. This is known as the “Zeeman effect.”
Solution We first apply Newton’s second law to find the angular frequency ω) of the orbital
motion. Coulombs law describes the force acting on the electron due to the electric interaction
1 e2
between the proton and the electron F~elec = − 4π
2 r̂, where r̂ is a unit vector in the plane of
0 r
the circular orbit pointing radiallly outward. Because we are assuming the motion is circular the
acceleration is ~a = −rω02 r̂. Newtons second law is then
−
1 e2
r̂ = −mrω02 r̂.
4π0 r2
(3)
We can now solve for the angular frequency
ω0 =
s
e2
.
4π0 mr3
(4)
~ perpendicular to the plane of the orbit is switched on there is
When a small magnetic field B
~ Let’s first assume that the magnetic
magnetic force acting on the electron F~mag = −e~v × B.
force points radially inward and that the radius of the orbit does not change. This will cause
the electron to speed up hence increasing the angular frequency to ωf . Choose coordinates as
shown in Fig. 10. Assume that the magnetic field points in the positive z-direction. In order to have a radially inward force, we require that the velocity of the electron is ~v = rωf θ̂.
~ = −erωf θ̂ × B k̂ = −erωf Br̂. Therefore
Then the magnetic force is given by F~mag = −e~v × B
e2
Newton’s second law is now − 4π0 r2 r̂ − erωf Br̂ = −mrωf2 r̂. Using (3) we can rewrite this as
−mrω02 r̂ − erωf Br̂ = −mrωf2 r̂ or 0 = ωf2 − eωmf B − ω0 . We can solve this quadratic equation to
q
2 2
eB
obtain ωf = 2m ± e4mB2 + ω02 . We choose the positive square root to keep ωf > 0. We now
rents in the same direction or in opposite dire
Find the magnitude of the current.
y
6.00 cm
θ
16.0°
z
x
Figure P30.20
Figure 1: Problem 1.
q
2
2 2
eB
substitute (4) into the above equation yielding ωf = 2m
+ e4mB2 + 4πe0 mr3 . Then the change
q
q
2 2
2
2
eB
in angular frequency is ∆Ω = ωf − ω0 = 2m
+ e4mB2 + 4πe0 mr3 − 4πe0 mr3 . Suppose we reverse the direction of the magnetic field as shown in Fig. 10. Then the magnetic force points
~ = −erωf θ̂ × −B k̂ = +erωf Br̂, resulting in a smaller angular freoutward, F~mag = −e~v × B
quency ωf . Repeating the analysis we just finished we have that −mrω02 r̂ + erωf Br̂ = −mrωf2 r̂
or 0 = ωf2 + eωmf B − ω0 . We can solve this quadratic equation for ωf choosing the positive square
q
2 2
eB
root to keep ωf > 0, that is ωf = − 2m
± e4mB2 + ω02 . The change in angular frequency is now
q
q
2 2
2
2
eB
∆Ω = ωf − ω0 = − 2m
+ e4mB2 + 4πe0 mr3 − 4πe0 mr3 .
n 30.3 Ampère’s Law
Four long, parallel conductors carry equal curr
5.00 A. Figure P30.21 is an end view of the cond
e current direction is into the page at points A
dicated by the crosses) and out of the page at C
dicated by the dots). Calculate the magnitud
ection of the magnetic field at point P, located
ter of the square of edge length 0.200 m.
e.
r,
s
r
A
s
d
e
s
o
y
s
y
e
p
surrounded by another rubber layer. In a particular
application, the current in the inner conductor is 1.00 A
28. N
out of the page and the current in the outer conductor is
b
3.00 A into the page. Determine the magnitude and
su
ofdirection
the can of
and
I
the
upward
current,
uniformly
distributed
the magnetic field at points a and b.
m
over its curved wall. Determine the magnetic field (a) just
ca
inside the wall and (b) just outside. (c) Determine the
e
pressure on the wall.
×
28. Niobium metal becomes
a superconductor
when cooled
×
×
3.00 A
below 9 K. Its superconductivity is destroyed when the
surface magnetic field exceeds
a 0.100 T.b Determine the
× 1.00
×
maximum current
a A2.00-mm-diameter
niobium wire can
carry and remain superconducting, in the absence of any
external magnetic×field.
×
.
.
29. A
as
is
is
co
(a
m
×
29. A long cylindrical conductor
of radius R carries a current I
as shown in Figure P30.29. The current density J, however,
1 mm
1 mm 1of
mm
is not uniform over the cross
section
the conductor but
is a function of the radius according to J ! br, where b is a
Figure
P30.232.
Figure 2: Problemfor
constant. Find an expression
the magnetic field B
(a) at a distance r 1 " R and (b) at a distance r 2 # R,
fromfield
the axis.
24.measured
The magnetic
40.0 cm away from a long straight wire
carrying current 2.00 A is 1.00 &T. (a) At what distance is
it 0.100 &T? (b) What If? At one instant, the two
I
conductors in a long household
extension cord carry
equal 2.00-A currents in opposite directions. The two wires
are 3.00 mm apart. Find the magnetic field 40.0 cm away
r2
from the middle
of the straight cord, in the plane of the
r 1 what distance is it one tenth as large?
two wires. (c) At
R
(d) The center wire in a coaxial cable carries current
2.00 A in one direction and the sheath around it carries
current 2.00 A in the
opposite
direction. What magnetic
Figure
P30.29
Figure 3: Problem 3.
field does the cable create at points outside?
25.In Figure
A packed
bundle
100 long,
insulated
wires
30.
P30.30,
bothofcurrents
in straight,
the infinitely
long
wires
forms
a cylinder
of radius
R ! 0.500
(a) Ifthe
each
wire
are
in the
negative
x direction.
(a)cm.
Sketch
magnetic
carries
2.00 A,inwhat
the magnitude
and direction
field
pattern
the are
yz plane.
(b) At what
distance of
d the
along
30. In
ar
fi
th
fa
0 m 1.30 T
7
201
Chapter 30
f
277 mA
ja f
Wb A m 470
lent to assuming
solenoid.
g B
5.00 fe9.27
89 T
0.191 T
10
24
Figure 4:P30.41
Problem 4.
FIG.
j
2
950
J T
C HAPTE R 3 0 •KSources
of the Magnetic Field
J
2.97 10 4
T 2 m3
B
a
b
o M xn B . Then B
0 B xn where n is the
r of electrons per atom contributing.
c
2.00 T
O
2.f02 .
24
7
10
N mT 4
10 T m A
je
H
j
e
e see that the applied field is described by
d
B0 C
C
definition of susceptibility
0 H , and the
Figure
30.31
Magnetization curve
T T
Figure 5: Problem 5.
4
for a ferromagnetic material.
K A
the magnitude
the curve from
are randomly
primary coil ca
until nearly all
approaching sa
iron are aligned
Next, supp
consequently e
follows the path
external field B
alignment of a
the iron is said
If the curre
magnetic field
unmagnetized
Find the magnetic field on the axis at a distance a above a disk of radius R with
charge density rotating at an angular speed (clockwise when viewed from the
point P).
Consider a thin ring between r and r + dr as shown below. The charge per length on
dr and each point on it is moving at a speed v
r , so the current is
the ring is
I
v
r dr .
P
where the anglesz are defined in the figure.
θ1
dr
θ2
r
We start by computing the magnetic field on axis for a single loop
a current I. This may be done by an elementary application of
law.
By symmetry, the magnetic field is
in the
z direction.
Figure
6: Problem
7.dl Using the result from Ex. 5.6
(Eq. 5.38), the contribution from the ring is
R
2
r
dr
where the angles are defined
in
the
figure.
r
0
B
dBz
a32 . α
2
2
2
the angles are defined in the figure.
r z
z
1 disk
2
The magnetic
θ field for the entire
θ is
θ
θ
1
2
R
3
ronly
dr the z component contributes
0
B By symmetry,
.
z
2
r2
zZ2
32
Z
~ ]z
µ0 I
[d~` ⇥ R
µ0 I
d` R sin ↵
µ0 I
a
B
=
=
=
2⇡a
Figure
7:
Problem
8.
z
startWe
by computing
the
magnetic
field
on
axis
for
a
single
loop
of
wire
carrying
3
3
byducomputing
the magnetic
on
for
a single
lo3
4⇡
Rfield
4⇡ axis R
4⇡
R
2r dr . This substitution (don’t forget to change limits) gives
Letstart
u r 2 , so
0
e
current
This may
donemay
by anbe
elementary
application
of the Biot-Savart
a I.
current
I. be
This
done
application
2
2
2
Substituting
in Rby
= aan
+ zelementary
yields
w.
law.
dl
dl
R
a
Bz =
α
Z
2
We now use linear superposition to obtain the field of the solen
and z2Bas follows
R
az
y symmetry, only the z
µ0 Ia2
2(a2 + z 2 )3/2
−z 1
α
θ1
z2
B
θ2
z
(where
z1 + z2 = L) we have
component
contributes
Figure 8: Solution of problem 8.
Z
µ0 Ia2
B
=
µ0 Iz
2a
Z
z2
N dz
2
~ ]z
2 + z 2 )3/2
µ0 I
[d~` ⇥ R
µ0 I
d` R sin ↵
µ0 Ia
z (a
Bz =
=
=
2⇡a 3 =
3
By symmetry,
z
component
contributes
4⇡
R3 only the
4⇡
R
4⇡
R
2R3
A simple trig substitution z = a tan ↵ converts this integral to
1
point midway between the coils. This means the magnetic
field in the region midway between the coils is uniform
Coils in this configuration are called Helmholtz coils.
I
I
a
R
R
a
>
z
R
b
Figure Problems
9: Problem 9.
Figure P30.55
55 and 56.
56. Two identical, flat, circular coils of wire each have
100 turns and a radius of 0.500 m. The coils are arranged
Suppose we reverse the direction of the magnetic field as in the figure below.
as a set of Helmholtz coils (see Fig. P30.55), parallel and
with separation 0.500 m. Each coil carries a current o
10.0 A. Determine the magnitude of the magnetic field a
a point on the common axis of the coils and halfway
Then!the10:
magnetic
force10.
points outward,
Figure
Problem
between them.
v = r! "ˆ
!
!
Assume that the magnetic field points in the positive z-direction. In order to have a
radially inward force, we require that the velocity of the electron is
f . Then the
!
Fmag = !ev " B = !er# f $̂ " !Bk̂ = +er# f B r̂ ,
magnetic force is given by
!
! !
57. We have
seen
F = !e
v " B = !er#that
$̂ " Bk̂ = !era
# B r̂long solenoid produces a uniform
resulting in a smaller angular frequency ! .
magnetic field directed
along the axis of a cylindrica
Repeating the analysis we just finished we have that
"
"to produce a !mr
region. However,
uniform
field
" r̂ + er" Br̂ = !mr"magnetic
r̂
directed parallel to a ordiameter of a 0 cylindrical
region, one
e! B
=! +
"! .
mag
f
f
.
f
Therefore Newton’s Second Law is now
!
ke2
r̂ ! er
r2
f
Br̂ = !mr
2
f
r̂ .
2
2
0
f
Using Eq. (1) we can rewrite this as
2
f
f
f
2
0