Prof. Anchordoqui
Problems set # 7
Physics 169
March 31, 2015
1. (i) Determine the initial direction of the deflection of charged particles as they enter the
magnetic fields as shown in Fig. 1. (ii) At the Equator near Earths surface, the magnetic field is
approximately 50.0 µT northward and the electric field is about 100 N/C downward in fair weather.
Find the gravitational, electric, and magnetic forces on an electron with an instantaneous velocity of 6.00x106 m/s directed to the east in this environment. (iii) Consider an electron near the
Earths equator. In which direction does it tend to deflect if its velocity is directed: (a) downward,
(b) northward, (c) westward, or (d) south-eastward?
Solution: (i) (a) up, (b) out of the page, since the chage is negative, (c) no deflection, (d) into the
page; see Fig. 2. (ii) Gravitational force: Fg = mg = 9.11 × 10−31 kg · 9.80 m/s2 = 8.93 × 10−30 N,
downward. Electric force: Fe = qE = (−1.60 × 10−19 C) · (−100 N/C) = 1.60 × 10−17 N, upward.
Magnetic force: Fm = qvB sin θ = (−1.60 × 10−19 C) · 6.00 × 106 m/s · 50.0 × 10−6 T · sin(π/2) =
4.80 × 10−17 N in direction opposite right hand rule prediction, i.e., downward. (iii) At the equator, the Earth’s magnetic field is horizontally north. Because an electron has negative charge,
~ is opposite in direction to ~v × B.
~ See the illustrations in Fig. 2 which are drawn looking
F~ = q~v × B
down. (a) Down × North = East, so the force is directed West; (b) North × North = 0; (c) West
× North = Down, so the force is directed Up; (d) Southeast × North = Up, so the force is Down.
2. A particle of charge −e is moving with an initial velocity v when it enters midway between
two plates where there exists a uniform magnetic field pointing into the page, as shown in Fig. 3.
You may ignore effects of the gravitational force. (i) Is the trajectory of the particle deflected
upward or downward? (ii) What is the magnitude of the velocity of the particle if it just strikes
the end of the plate?
Solution: (i) Choose unit vectors as shown in Fig. 3. The force on the particle is given by
~
F = −e(vı̂ × B̂) = −evB k̂. The direction of the force is downward. (ii) Remember that when a
charged particle moves through a uniform magnetic field, the magnetic force on the charged particle only changes the direction of the velocity hence leaves the speed unchanged so the particle
2
undergoes circular motion. Therefore we can use Newtons second law in the form evB = m vR . The
speed of the particle is then v = eBR/m. In order to determine the radius of the orbit we note that
the particle just hits the end of the plate. From the figure above, by the Pythagorean theorem, we
have that R62 = (R − d/2)2 + l2 Expanding this equation yields R2 = R2 − Rd + d2 /4 + l2 . We
2
can now solve for the radius of the circular orbit: R = d4 + ld . We can now substitute this value in
the equation for
and find the speed necessary for the particle to just hit the end of the
the 2velocity
d
l
plate: v = eB
+
.
m
4
d
3. The entire x − y plane to the right of the origin O is filled with a uniform magnetic field of
magnitude B pointing out of the page, as shown in Fig. 4. Two charged particles travel along the
negative x axis in the positive x direction, each with velocity ~v , and enter the magnetic field at the
origin O. The two particles have the same mass m, but have different charges, q1 and q2 . When
propagate thorugh the magnetic field, their trajectories both curve in the same direction (see sketch
in Fig. 4), but describe semi-circles with different radii. The radius of the semi-circle traced out by
particle 2 is exactly twice as big as the radius of the semi-circle traced out by particle 1. (i) Are the
charges of these particles positive or negative? Explain your reasoning. (ii) What is the ratio q2 /q1 ?
~ the charges of these particles are positive. (ii) We first find
Solution: (i) Because F~B = q~v × B,
an expression for the radius R of the semi-circle traced out by a particle with charge q in terms of
q, v ≡ |~v |, B, and m. The magnitude of the force on the charged particle is qvB and the magnitude
of the acceleration for the circular orbit is v 2 /R. Therefore applying Newtons second law yields
2
mv
qvB = mv
R . We can solve this for the radius of the circular orbit R = qB . Therefore the charged
ratio
q2
q1
=
mv/(R2 B)
mv(R1 B)
=
R1
R2 .
4. Shown in Fig. 5 are the essentials of a commercial mass spectrometer. This device is used to
measure the composition of gas samples, by measuring the abundance of species of different masses.
An ion of mass m and charge q = +e is produced in source S, a chamber in which a gas discharge is
taking place. The initially stationary ion leaves S, is accelerated by a potential difference ∆V > 0,
~ 1 , pointing
and then enters a selector chamber, S1 , in which there is an adjustable magnetic field B
~ pointing from positive to negative plate. Only
out of the page and a deflecting electric field E,
particles of a uniform velocity ~v leave the selector. The emerging particles at S2 , enter a second
magnetic field B2 , also pointing out of the page. The particle then moves in a semicircle, striking
an electronic sensor at a distance x from the entry slit. Express your answers to the questions
~ e, x, m, B2 ≡ |B
~ 2 |, and ∆V . (i) What magnetic field B1 in the selector
below in terms of E ≡ |E|,
chamber is needed to insure that the particle travels straight through? (ii) Find an expression for
the mass of the particle after it has hit the electronic sensor at a distance x from the entry slit.
Solution: (i) We first find an expression for the speed of the particle after it is accelerated by the
potential difference ∆V , in terms of m, e, and ∆V . The change in kinetic energy is ∆K = 12 mv 2 .
The change in potential energy is ∆U = −e∆V
q . From conservation of energy, ∆K = −∆U , we
have that 12 mv 2 = e∆V . So the speed is v = 2e∆V
Inside the selector the force on the charge is
m
~
~
~
given by F = e(E + ~v × B1 ). If the particle travels straight through the selector then force on the
~ = −~v × B
~ 1 . Because the velocity is to the right in Fig. 5 (define this
charge is zero, therefore E
as the +ı̂ direction), the electric field points up (define this as the +̂ direction) from the positive
plate to the negative plate, and the magnetic field is pointing out of the page (define this as the
p m
~ 1 = E k̂ =
+k̂ direction). Then E̂ = −vı̂ × B1 k̂ = vB1 ̂. Thus, B
v
2e∆V E k̂. (ii) The force on the
~
~
charge when it enters the magnetic field B2 is given by F = evı̂ × B2 k̂ = −evB2 ̂. This force points
downward and forces the charge to start circular motion. You can verify this because the magnetic
field only changes the direction of the velocity of the particle and not its magnitude which is the
condition for circular motion. Recall that in circular motion the acceleration is towards the center.
~ 2 the acceleration is downward ~a = − v2 ̂.
In particular when the particle just enters the field B
x/2
2
v
Newtons Second Law becomes −evB2 = −m x/2
. Thus, the particle hits the electronic sensor at
√
eB22 x2
2
a distance x = 2mv
eB2 = eB2 2em∆V from the entry slit. The mass of the particle is then m = 8∆V .
5. Electrons in a beam are accelerated from rest through a potential difference V . The beam
enters an experimental chamber through a small hole. As shown in Fig. 6, the electron velocity
vectors lie within a narrow cone of half angle φ oriented along the beam axis. We wish to use a
uniform magnetic field directed parallel to the axis to focus the beam, so that all of the electrons
can pass through a small exit port on the opposite side of the chamber after they travel the length
d of the chamber. What is the required magnitude of the magnetic field? [Hint: Because every
electron passes through the same potential difference and the angle φ is small, they all require the
same time interval to travel the axial distance d.
Solution The electrons are all fired from the
qelectron gun with the same speed v. Since Ui = Kf ,
we have (−e)(−∆V ) = 21 me v 2 , yielding v =
2e∆V
me
. For φ small, cos φ is nearly equal to 1. The
p m2
time T of passage of each electron in the chamber is given by d = vT , and so T = d 2e∆V
.
Each electron moves in a different helix, around a different axis. If each completes just one
revolution within the chamber, it will be in the right place to pass through the exit port. Its
transverse velocity component v = v sin φ swings around according to F⊥ = ma⊥ . Explicitly,
p me
mv 2
e 2π
qv⊥ B sin(π/2) = r ⊥ , or equivalently eB = merv⊥ = me ω = me 2π
, yielding T = meB
= d 2e∆V
.
T
q
p
me
2me ∆V
2π
√ d
Therefore, 2π
.
B
e = 2∆V , which leads to B = d
e
6. A circular ring of radius R lying in the xy plane carries a steady current I, as shown in the
Fig. 7. What is the magnetic field at a point P on the axis of the loop, at a distance z from the
center?
Solution: We use the Biot-Savart law to find the magnetic field on the symmetry axis, see
Fig. 7. (1) Source point: In Cartesian coordinates, the differential current element located at
~r 0 = R(cos φ0 ı̂ + sin φ0 ̂) can be written as Id~s = I(d~r 0 /dφ0 )dφ0 = IRdφ0 (− sin φ0 ı̂ + cos φ0 ̂). (2)
Field point: Since the field point P is on the axis of the loop at a distance z from the center,
its position vector is given by ~rP = z k̂. (3) Relative position vector ~rP − ~r 0 : The vector from the
current element to the field point is given by ~rP −~r 0 = −R cos φ0 ı̂−R sin φ0 ̂+z k̂, and its magnitude
p
√
r = |~rP − ~r 0 | = (−R cos φ0 )2 + (−R sin φ0 )2 + z 2 = R2 + z 2 is the distance between the differential current element and P . Thus, the corresponding unit vector from Id~s to P can be written as
r0
r̂ = |~~rrPP −~
s × (~rP −~r 0 ) can be simplified as
−~
r 0 | . (4) Simplifying the cross product: The cross product d~
d~s×(~rP −~r 0 ) = Rdφ0 (− sin φ0 ı̂+cos φ0 ̂)×(−R cos φ0 ı̂−R sin φ0 ̂+z k̂) = Rdφ0 (z cos φ0 ı̂+z sin φ0 ̂+Rk̂).
~ Using the Biot-Savart law, the contribution of the current element to the
(5) Writing down dB:
R
sin φ0 ̂+Rk̂
~
~ = µ0 IR 2π z cos φ0 ı̂+z
dφ0 . The x and the y components of B
magnetic field at P is B
4π
0
(R2 +z 2 )3/2
2π
R2
µ0
IRz
IRz
0 dφ0 = µ0
0
can be readily shown to be zero: Bx = 4π
π
cos
φ
sin
φ
= 0 and
3/2
3/2
2
2
2
2
4π
0
(R +z )
(R +z )
0
R2
2π
µ0
µ0
IRz
IRz
By = 4π
π sin φ0 dφ0 = 4π
cos φ0 = 0. On the other hand the z component
(R2 +z 2 )3/2 0
(R2 +z 2 )3/2
0
R2 0
µ0
µ0 IR2
IR2
is Bz = 4π
dφ
=
.
Thus,
we
see
that along the symmetric axis, Bz is the
(R2 +z 2 )3/2 0
2(R2 +z 2 )3/2
only non-vanishing component of the magnetic field. The conclusion can also be reached by using
the symmetry arguments. The behavior of Bz /B0 where B0 = µ0 I/(2R) is the magnetic field
strength at z = 0, as a function of z/R is shown in Fig. 8.
7. Find the magnetic field at point P due to the current distribution shown in Fig. 9.
Solution: The fields due to the straight wire segments are zero at P because d~s and r̂ are
parallel or anti-parallel. For the field due to the arc segment, the magnitude of the magnetic
~ = µ0 I d~s×r̂
field due to a differential current carrying element is given in this case by dB
=
4π R2
µ0 IRdθ
µ0 I(sin2 θ+cos2 θ)dθ
k̂
4π R2 (sin θı̂ − cos θ̂) × (− cos θı̂ − sin θ̂) = − 4π
R
R π/2 µo I
µ0 I
~
B = − 0 4πR dθk̂ = − 8R k̂ into the plane of Fig. 9
µ0 Idtheta
= − 4π
k̂.
R
Therefore,
8. A thin uniform ring of radius R and mass M carrying a charge +Q rotates about its axis
with constant angular speed ω. Find the ratio of the magnitudes of its magnetic dipole moment to
its angular momentum. (This is called the gyromagnetic ratio.)
Qω
Solution: The current in the ring shown in Fig. 10 is i = Q
T = 2π . The magnetic moment is
2
QωR
2
~
µ
~ = Aik̂ = πR2 Qω
2π k̂ =
2 k̂. The angular momentum is L = Iω k̂ = M R ω k̂. So the gyromag-
netic ratio is
|~
µ|
~
|L|
=
QωR2 /2
M R2 ω
=
Q
2M .
9. A wire ring lying in the xy-plane with its center at the origin carries a counterclockwise I.
~ = Bı̂ in the +x-direction. The magnetic moment vector µ is
There is a uniform magnetic field B
perpendicular to the plane of the loop and has magnitude µ = IA and the direction is given by
right-hand-rule with respect to the direction of the current. What is the torque on the loop?
~ where µ = IA
~ × B,
Solution: The torque on a current loop in a uniform field is given by ~τ = µ
and the vector ı̂ is perpendicular to the plane of the loop and right-handed with respect to the
~ = πIR2 k̂. Therefore,
direction of current flow. The magnetic dipole moment is given by µ
~ = IA
2
2
~ = πIR k̂ × Bı̂ = πIR B̂. Instead of using the above formula, we can calculate the
~τ = µ
~ ×B
torque directly as follows. Choose a small section of the loop of length ds = Rdθ. Then the vector
describing the current-carrying element is given by Id~s = IRdθ(− sin θı̂ + cos θ̂). The force dF~
~ = IRdθ(− sin θı̂ + cos θ̂) × Bı̂ = −IRB cos θdθk̂.
that acts on this current element is dF~ = Id~s × B
H
The force acting on the loop can be found by integrating the above expression, F~ = dF~ =
R 2π
2π
0 (−IRB cos θ)dθ k̂ = −IRB sin θ|0 k̂ = 0. We expect this because the magnetic field is uniform
and the force on a current loop in a uniform magnetic field is zero. Therefore we can choose any
point to calculate the torque about. Let ~r be the vector from the center of the loop to the element
Id~s. That is, ~r = R(cos θı̂ + sin θ̂). The torque d~τ = ~r × dF~ acting on the current element is
then d~τ = ~r × dF~ = R(cos θı̂ + sin θ̂) × (−IRBdθ cos θk̂) = −IR2 Bdθ cos θ(sin θı̂ − cos θ̂). Finally,
H
R 2π
integrate d~τ over the loop to find the total torque, ~τ = d~τ = − 0 IR2 Bdθ cos θ(sin θı̂ − cos θ̂) =
πIR2 B̂. This agrees with our result above.
10. A nonconducting sphere has mass 80.0 g and radius 20.0 cm. A flat compact coil of wire with
5 turns is wrapped tightly around it, with each turn concentric with the sphere. As shown in Fig. 11,
the sphere is placed on an inclined plane that slopes downward to the left, making an angle θ with
the horizontal, so that the coil is parallel to the inclined plane. A uniform magnetic field of 0.350 T
1.
Determine the initial direction of the deflection of
charged particles as they enter the magnetic fields as
shown in Figure P29.1.
(a)
+
(c)
(b)
Bin
×
×
×
×
×
×
×
×
×
×
×
×
Bright
Bup
×
×
×
×
–
(d)
+
Bat 45°
45°
+
Figure P29.1
Figure 1: Problem 1.
vertically
upward exists
the region ofnear
the sphere.
current equator.
in the coil willIn
enable
the sphere
2.
Consider
an inelectron
theWhat
Earth’s
which
dito rest in equilibrium on the inclined plane? Show that the result does not depend on the value of θ.
rection does it tend to deflect if its velocity is directed
Solution: The sphere is in translational equilibrium, thus fs − M g sin θ = 0, see Fig. 11. The
sphere is in rotational equilibrium. If torques are taken about the center of the sphere, the magnetic
field produces a clockwise torque of magnitude µB sin θ, and the frictional force a counterclockwise
torque of magnitude fs R, where R is the radius of the sphere. Hence, fs R−µB sin θ = 0. Substituting the expression for fs derived from the condition of translational equilibrium, fs = M g sin θ into
the condition for rotational equilibrium, and cancelling out sin θ, one obtains µB = M gR. Now,
0.08 kg·9.80 m/s2
Mg
µ = N IπR2 . Thus, I = πN
BR = 5π·0.350 T·0.2 m = 0.713 A. The current must be counterclockwise
as seen from above.
e
3. A
l
i
m
4. A
o
i
m
5. A
B
2
t
o
6. A
t
(
m
7. A
fi
8
v
(c)
no deflection
(d)
into the page
ng the velocity of a particle requires an accelerating force. The magnetic force is proportional
d of the particle. If the particle is not moving, there can be no magnetic force on it.
e current in the probe. If the material is a semiconductor, raising its temperature may
e density of mobile charge carriers in it.
Problem 2: Particle Trajectory
PROBLEMS
!
A particle of charge !e is moving with an initial velocity v when it enters midway
between two plates where there exists a uniform magnetic field pointing into the page, as
FIG.
shown in the figure below. You
mayP29.1
ignore effects of the gravitational force.
netic Fields and Forces
of the page, since the
roblem
3: Particle Orbits in a Uniform Magnetic Field
arge is negative.
29.2
At the equator, the Earth’s magnetic field is
deflection horizontally north. Because an electron has
he entire x-y plane to the right of the origin O is filled with a uniform magnetic field of
charge,
F page,
qv Basisshown.
opposite
direction
o the page negative
B pointing
agnitude
out of the
Twoincharged
particles travel along the
!
v
B
to
.
Figures
are
drawn
looking
down.
gative x axis in the positive x direction, each with velocity v , and enter the magnetic
(c)
(d)
eld at the origin O. The two particles have the same mass m , but have different (a)
(a)
Down
North
=
East,
so
the
force
is
arges, q1 and q2 . When in the magnetic
field, their trajectories
both curve
the same
(a) Is the trajectory
of the in
particle
deflected upward or downward?
FIG. P29.1
FIG. P29.2
West
.
directed
rection (see sketch), but describe semi-circles with
different
radii.
The
radius
of
the
Figure 2: Solution of problem 1.
tor, the Earth’s magnetic field is
(b)
What
is
the
magnitude
of
the
velocity
of the particle if it just strikes the end of the
mi-circle
traced
out by
y north. Because
an electron
has particle 2 is exactly twice as big as the radius of the semi-circle
plate?
arge, Fout
qvby
B is
opposite in
aced
particle
1.direction
(b)looking down.
North North sin 0 0 : Zero deflection .
gures are drawn
(a)
wn North = East, so the force is
ected West .
rth
st
(c)
North
West
(c)
Solution:
(d) Choose unit vectors as shown in the figure.
FIG. P29.2
North = Down,
so the force is directed Up .
Problem
3: Particle
Orbits in a Uniform Magnetic Field
0 : Zero deflection
.
sin 0
(d)
Southeast
North = Up, so the force is Down .
North = Down, so the force is directed Up .
The entire x-y plane to the right of the origin O is filled with a uniform magnetic field of
magnitude B pointing out of the page, as shown. Two charged particles travel along the
!
negative x axis in the positive x direction, each with velocity v , and enter the magnetic
field at the origin O. The two particles have the same mass m , but have different
charges, q1 and q2 . When in the magnetic field, their trajectories both curve in the same
direction
sketch),
butor describe
semi-circles
with different radii. The radius of the
) Are the charges
of these(see
particles
positive
negative?
Explain
your
The
force
on
the
particle
is
by
3: Problem
2.given
semi-circle traced out by particle 2Figure
is exactly
twice
as big
as the radius of the semi-circle
asoning.
!
traced
F = !e(v î " Bĵ) = !evBk̂ .
! out by
! particle 1.
utheast
North = Up, so the force is Down .
!
olution: Because FB = qv ! B , the charges of these particles are POSITIVE.
(1)
The direction of the force is downward. Remember that when a charged particle moves
through a uniform magnetic field, the magnetic force on the charged particle only
changes the direction of the velocity hence leaves the speed unchanged so the particle
undergoes circular motion. Therefore we can use Newton’s second law in the form
) What is the ratio q2 / q1 ?
olution: We first find an expression for the radius R of the semi-circle traced out by a
2
!
rticle with charge q in terms of q , v ! v , B , and m . The magnitude of the force onevB = m v .
R
e charged particle is qvB and the magnitude of the acceleration
for the circular orbit is
.
/ R . Therefore applying Newton’s Second Law yields
mv 2
qvB =
.
R
e can solve this for the radius of the circular orbit
R=
mv
qB
herefore the charged
(a) Areratio
the charges of these particles
positive
or negative?
Figure
4: Problem
3.
!
"
!
"
q
R
mv
mv
2
1
reasoning.
.
=#
$ #
$=
q1 % R2 B & % R1 B & R2
Explain your
!
! !
Solution: Because FB = qv ! B , the charges of these particles are POSITIVE.
(2)
electric field E , pointing from positive to negative plate. Only particles of a uniform
!
velocity v leave the selector. The emerging particles at S2 , enter a second magnetic
!
B 2 , also pointing out of the page. The particle then moves in a semicircle, striking an
electronic sensor at a distance x from the entry slit. Express your answers to the
!
!
questions below in terms of E ! E , e , x , m , B2 ! B 2 , and !V .
58. Review Problem. A wire having a linear mass density of
1.00 g/cm is placed on a horizontal surface that has a
coefficient of kinetic friction of 0.200. The wire carries a
current of 1.50 A toward the east and slides horizontally to
the north. What are the magnitude and direction of the
smallest magnetic field that enables the wire to move in
this fashion?
long. The springs
wire and the circ
a magnetic field i
springs stretch an
tude of the magn
62. A hand-held ele
Model the motor
ing electric curre
59. Electrons in a beam are accelerated from rest through a
produced by an e
potential difference !V. The beam enters an experimental
sider only one in
chamber through a small hole. As shown in Figure P29.59,
will consider mot
the electron velocity vectors lie within a narrow cone of
because the mag
half angle " oriented along the beam axis. We wish to use
described in Sec
a uniform magnetic field directed parallel to the axis to
mates of the ma
focus the beam, so that all of the electrons can pass
!
Figure
5:
Problem
4.
current in it, its ar
through a small exit port on the opposite side of the
a) What magnetic
field
in
the
selector
chamber
is
needed
to
insure tha
B
that they are relat
chamber after 1they travel the length d of the chamber.
the input power t
What
is the through?
required magnitude of the magnetic field?
particle travels
straight
and the useful ou
Hint: Because every electron passes through the same
potential difference and the angle " is small, they all
63. A nonconductin
Solution: We first find
an expression
for thetospeed
ofaxial
the distance
particled. after it20.0
is accelerate
require
the same time interval
travel the
cm. A flat co
around
the potential difference !V , in terms of m , e , and !V . The change intightly
kinetic
energi
sphere. As shown
o
!K = (1/ 2)mv 2 . The change in potential energyd is !U = "e!V From conservation
an inclined plane
an angle ' with th
energy, !K = "!U , we have that
the inclined plan
(1/ 2)mvφ 2 = e!V .
vertically upward
Exit
current in the co
So the speed is
port
rium on the incli
2e!V
depend on the va
v
=
Entrance
∆V
m
port
Figure P29.59
Figure 6: Problem 5.
Inside the selector the force on the charge is given by
60. Review Problem. A proton is at rest at the plane vertical
! containing
! !a uniform
!
boundary of a region
vertical magFe particle
= e(E +moving
v ! Bhorizontally
netic field B. An alpha
makes
1) .
a head-on elastic collision with the proton. Immediately
after the collision, both particles enter the magnetic field,
moving perpendicular to the direction of the field. The
2!
2!
µ0 IRz
µ0 IRz
What is
the magnetic
loop, at
By =
sin
" ' d" field
' = #at a point P on2 the axis2 of3/the2 cos
"a'distance=z fr0
2
2 3/ 2 center?
$
0
0
4! ( R + z )
4! ( R + z )
5: Magnetic Field of a Ring of Current
Solution:
ring of radius R lying in the xy plane carries a steady current I , as shown in
(a) We shall use the Biot-Savart law to find the magnetic field on the symm
below.
hand, the z component is
µ0
IR 2
Bz =
4! ( R 2 + z 2 )3/ 2
#
2!
0
µ0 2! IR 2
µ0 IR 2
d" ' =
=
.
2
2 3/ 2
2
2 3/ 2
4! ( R + z )
2( R + z )
e that along the symmetric axis, Bz is the only non-vanishing com
(1) Source
chefield.
The conclusion
canpoint:
also
be
reached
by using the symmetr
magnetic field at a point P on the axis of theFigure
loop, 7:at Problem
a distance
6. z from the
= µ0 Icoordinates,
/ 2 R isthethe
r of Bz / B0 where BIn0 Cartesian
magnetic
field
strength
at
differential
current element
located
at
ˆi + sin ! ' ˆj) can be written as Id !s = I (dr! '/ d! ')d! ' = IRd! '(" sin ! ' ˆi +
z / R is shown in the r!figure
' = R(cos ! 'below.
e shall use the Biot-Savart law to find the magnetic field on the symmetry axis.
(2) Field point:
Since the field point P is on the axis of the loop at a distance z from the center,
!
position vector is given by r = zk̂ .
! !
(3) Relative position vector r ! r ' :
e point:
an coordinates, the differential current element located at
!
!
! ' ˆi + sin ! ' ˆj) can be written as Id s = I (dr '/ d! ')d! ' = IRd! '(" sin ! ' ˆi + cos ! ' ˆj) .
point:
Figure 8: Solution of problem 6.
field point P is on the axis of the loop at a distance z from the center, its
!
ector is given by r = zk̂ .
! !
ve position vector r ! r ' :
Magnetic Fields
gnetic field at point P due to the following current distribution.
netic Ratio
radius R and mass M carrying a charge +Q
peed ! as shown in the figure below. Find the
netic dipole moment to its angular momentum.
Figure 9: Problem 7.
r
ue to the straight wire segments are zero at P because d s and
lel. For the field due to the arc segment, the magnitude of the m
erential current carrying element is given in this case by
! µ0 I d !s " r̂ µ0 IRd# (sin # î $ cos# ĵ) " ($ cos# î $ sin # ĵ)
dB =
=
2
4! R
4!
R2
µ0 I(sin 2 # + cos 2 # )d#
µ0 Id#
=$
k̂ = $
k̂
4!
R
4! R
in the ring is
$
" /2
0
Figure 10: Problem 8.
%Qµ!
µ0 I
µ0 I % " (Q
I(
0
d# k̂ = !
k̂ ==! '
k̂ (into the plane of the
.
I
=
'
*
*
4" R
4" R & 2 )T
&28R
")
FG
H
IJ FG
KH
IJ
K
2 tude
000ofrev
min field?
~ 200 rad s
the magnetic
60
s
1
rev
62. A hand-held electric mixer contains an electric motor.
he
in
b
f
g
Model the motor as a single flat compact circular coil carry1
a region
a magnetic
field .is
a20 W ing electric current
N m
~ 10
200 inrad
s where
produced by an external permanent magnet. You need conal
sider only one instant in the operation of the motor. (We
9,
3 moves
2
will consider motors again in Chapter 31.) The coil
of
3
cm
4
cm
A
~
10
m
,
or
.
because the magnetic field exerts torque on the coil, as
se
described in Section 29.3. Make order-of-magnitude estito
mates
ss
1 of the magnetic field, the torque on the coil, the
currentT
in it,. its area, and the number of turns in the coil, so
he B ~ 10
that they are related according to Equation 29.11. Note that
er.
input power to the motor is electric, given by ! $ I !V,
d?
NIAB :
the
coilthe
may
be found from
and the useful output power is mechanical, ! $ %&.
me
all
63. A nonconducting sphere has
3 mass
2 80.01g and radius
m cm.
~N
1
C
s
10
m
105 turnsN
sC m
d.0.1 N 20.0
A flat compact coil of wire with
is wrapped
a
f a
b
ge
je
tightly around it, with each turn concentric with the
N ~ sphere.
10 3 .As shown in Figure P29.63, the sphere is placed on
an inclined plane that slopes downward to the left, making
an angle ' with the horizontal, so that the coil is parallel to
the inclined plane. A uniform magnetic field of 0.350 T
equilibrium,
vertically thus
upward exists in the region of the sphere. What
current in the coil will enable the sphere to rest in equilibrium on the inclined plane? Show that the result
(1)does not
depend on the value of '.
j
B
uilibrium. If torques are taken about the
B
netic field produces a clockwise torque
of
rictional force a counterclockwise torque
al
gthe
radius of the sphere. Thus:
es
ely
d,
he
tuting
he
is
of
op
nd
he
es
m
I
I
(2)
Mg
this in (2)θ and canceling out sin ,
Figure P29.63
Figure 11: Problem 10.
b
ge
(3)
64. A metal rod having a mass per unit length ( carries a
current I. The0.rod
08 hangs
kg 9from
.80 two
m vertical
s 2 wires in a
Mg
uniform vertical magnetic field as shown in Figure P29.64.
0.713
The wires make an angle ' with the vertical when in equi5
0
.
350
T
0
.
2
m
NBR
librium. Determine the magnitude of the magnetic field.
m above.
a fa
fa
d the tension in each wireθ alone
0
fs
or
or
or
T sin
B
ILB
mg
I
bm Lgg tan
Figure P29.64
A . The current must be
T
. Then, at equilibrium:
2
g
T cos
B
θ
j
f
FIG. P29.63
g
tan