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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 6.00 × 106 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 equa-
tor, the Earth’s magnetic field is horizontally north. Because an electron has negative charge,~F = q~v× ~B is opposite in direction to ~v× ~B. See the illustrations in Fig. 2 which are drawn looking
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(vı × B) = −evBk. 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 par-
ticle only changes the direction of the velocity hence leaves the speed unchanged so the particle
undergoes circular motion. Therefore we can use Newtons second law in the form evB = mv2
R . 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
can now solve for the radius of the circular orbit: R = d4 + l2
d . We can now substitute this value in
the equation for the velocity and find the speed necessary for the particle to just hit the end of the
plate: v = eBm
(d4 + l2
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?
Solution: (i) Because ~FB = q~v× ~B, the charges of these particles are positive. (ii) We first find
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 v2/R. Therefore applying Newtons second law yields
qvB = mv2
R . We can solve this for the radius of the circular orbit R = mvqB . Therefore the charged
ratio q2q1
= mv/(R2B)mv(R1B) = 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,
and then enters a selector chamber, S1, in which there is an adjustable magnetic field ~B1, pointing
out of the page and a deflecting 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 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
below in terms of E ≡ | ~E|, e, x, m, B2 ≡ | ~B2|, and ∆V . (i) What magnetic field B1 in the selector
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 = 12mv
2.
The change in potential energy is ∆U = −e∆V . From conservation of energy, ∆K = −∆U , we
have that 12mv
2 = e∆V . So the speed is v =√
2e∆Vm Inside the selector the force on the charge is
given by ~F = e( ~E + ~v × ~B1). If the particle travels straight through the selector then force on the
charge is zero, therefore ~E = −~v × ~B1. Because the velocity is to the right in Fig. 5 (define this
as the +ı 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
+k direction). Then E = −vı×B1k = vB1. Thus, ~B1 = Ev k =
√m
2e∆V Ek. (ii) The force on the
charge when it enters the magnetic field ~B2 is given by ~F = evı×B2k = −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.
In particular when the particle just enters the field ~B2 the acceleration is downward ~a = − v2
x/2 .
Newtons Second Law becomes −evB2 = −m v2
x/2 . Thus, the particle hits the electronic sensor at
a distance x = 2mveB2
= 2eB2
√2em∆V from the entry slit. The mass of the particle is then m =
eB22x
2
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 electron gun with the same speed v. Since Ui = Kf ,
we have (−e)(−∆V ) = 12mev
2, yielding v =√
2e∆Vme
. For φ small, cosφ is nearly equal to 1. The
time T of passage of each electron in the chamber is given by d = vT , and so T = d√
m22e∆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 sinφ swings around according to F⊥ = ma⊥. Explicitly,
qv⊥B sin(π/2) =mv2⊥r , or equivalently eB = mev⊥
r = meω = me2πT , yielding T = me2π
eB = d√
me2e∆V .
Therefore, 2πB
√mee = d√
2∆V, which leads to B = 2π
d
√2me∆V
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 ′ = R(cosφ′ı + sinφ′) can be written as Id~s = I(d~r ′/dφ′)dφ′ = IRdφ′(− sinφ′ı + cosφ′). (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 = zk. (3) Relative position vector ~rP − ~r ′: The vector from the
current element to the field point is given by ~rP −~r ′ = −R cosφ′ı−R sinφ′+zk, and its magnitude
r = |~rP − ~r ′| =√
(−R cosφ′)2 + (−R sinφ′)2 + z2 =√R2 + z2 is the distance between the differ-
ential current element and P . Thus, the corresponding unit vector from Id~s to P can be written as
r = ~rP−~r ′
|~rP−~r ′| . (4) Simplifying the cross product: The cross product d~s×(~rP −~r ′) can be simplified as
d~s×(~rP−~r ′) = Rdφ′(− sinφ′ı+cosφ′)×(−R cosφ′ı−R sinφ′+zk) = Rdφ′(z cosφ′ı+z sinφ′+Rk).
(5) Writing down d ~B: Using the Biot-Savart law, the contribution of the current element to the
magnetic field at P is ~B = µ04π IR
∫ 2π0
z cosφ′ ı+z sinφ′ +Rk(R2+z2)3/2
dφ′. The x and the y components of ~B
can be readily shown to be zero: Bx = µ04π
IRz(R2+z2)3/2
∫ 20 π cosφ′ dφ′ = µ0
4πIRz
(R2+z2)3/2sinφ′
∣∣∣2π0
= 0 and
By = µ04π
IRz(R2+z2)3/2
∫ 20 π sinφ′ dφ′ = µ0
4πIRz
(R2+z2)3/2cosφ′
∣∣∣2π0
= 0. On the other hand the z component
is Bz = µ04π
IR2
(R2+z2)3/2
∫ 20 dφ
′ = µ0IR2
2(R2+z2)3/2. Thus, we see that along the symmetric axis, Bz is the
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 = µ0I/(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 r are
parallel or anti-parallel. For the field due to the arc segment, the magnitude of the magnetic
field due to a differential current carrying element is given in this case by d ~B = µ0I4π
d~s×rR2 =
µ04π
IRdθR2 (sin θı − cos θ) × (− cos θı − sin θ) = −µ0
4πI(sin2 θ+cos2 θ)dθ
R k = −µ04π
IdthetaR k. Therefore,
~B = −∫ π/2
0µoI4πRdθk = −µ0I
8R k into the plane of Fig. 9
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.)
Solution: The current in the ring shown in Fig. 10 is i = QT = Qω
2π . The magnetic moment is
~µ = Aik = πR2Qω2π k = QωR2
2 k. The angular momentum is ~L = Iωk = MR2ωk. So the gyromag-
netic ratio is |~µ||~L|
= QωR2/2MR2ω
= Q2M .
9. A wire ring lying in the xy-plane with its center at the origin carries a counterclockwise I.
There is a uniform magnetic field ~B = Bı in the +x-direction. The magnetic moment vector µ is
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?
Solution: The torque on a current loop in a uniform field is given by ~τ = ~µ× ~B, where µ = IA
and the vector ı is perpendicular to the plane of the loop and right-handed with respect to the
direction of current flow. The magnetic dipole moment is given by ~µ = I ~A = πIR2k. Therefore,
~τ = ~µ × ~B = πIR2k × Bı = πIR2B. Instead of using the above formula, we can calculate the
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 θı + cos θ). The force d~F
that acts on this current element is d~F = Id~s× ~B = IRdθ(− sin θı+ cos θ)×Bı = −IRB cos θdθk.
The force acting on the loop can be found by integrating the above expression, ~F =∮d~F =∫ 2π
0 (−IRB cos θ)dθk = −IRB sin θ|2π0 k = 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 θı + sin θ). The torque d~τ = ~r × d~F acting on the current element is
then d~τ = ~r× d~F = R(cos θı+ sin θ)× (−IRBdθ cos θk) = −IR2Bdθ cos θ(sin θı− cos θ). Finally,
integrate d~τ over the loop to find the total torque, ~τ =∮d~τ = −
∫ 2π0 IR2Bdθ cos θ(sin θı−cos θ) =
πIR2B. 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
918 C H A P T E R 2 9 • Magnetic Fields
10. A current-carrying conductor experiences no magneticforce when placed in a certain manner in a uniform mag-netic field. Explain.
Is it possible to orient a current loop in a uniformmagnetic field such that the loop does not tend to rotate?Explain.
12. Explain why it is not possible to determine thecharge and the mass of a charged particle separatelyby measuring accelerations produced by electric andmagnetic forces on the particle.
How can a current loop be used to determine thepresence of a magnetic field in a given region of space?
14. Charged particles from outer space, called cosmic rays,strike the Earth more frequently near the poles than nearthe equator. Why?
15. What is the net force on a compass needle in a uniformmagnetic field?
16. What type of magnetic field is required to exert a resultantforce on a magnetic dipole? What is the direction of theresultant force?
17. A proton moving horizontally enters a uniform magneticfield perpendicular to the proton’s velocity, as shown inFigure Q29.17. Describe the subsequent motion of theproton. How would an electron behave under the samecircumstances?
13.
11.
18. In the cyclotron, why do particles having different speedstake the same amount of time to complete a one-half circletrip around one dee?
19. The bubble chamber is a device used for observing tracks ofparticles that pass through the chamber, which is immersedin a magnetic field. If some of the tracks are spirals andothers are straight lines, what can you say about theparticles?
20. Can a constant magnetic field set into motion an electroninitially at rest? Explain your answer.
21. You are designing a magnetic probe that uses the Halleffect to measure magnetic fields. Assume that you arerestricted to using a given material and that you havealready made the probe as thin as possible. What, ifanything, can be done to increase the Hall voltageproduced for a given magnitude of magnetic field?
v+
! ! !
! ! !
! ! !
! ! !
! !!
Figure Q29.17
Section 29.1 Magnetic Fields and Forces
Determine the initial direction of the deflection ofcharged particles as they enter the magnetic fields asshown in Figure P29.1.
1.
(a) downward, (b) northward, (c) westward, or (d) south-eastward?
3. An electron moving along the positive x axis perpendicu-lar to a magnetic field experiences a magnetic deflectionin the negative y direction. What is the direction of themagnetic field?
4. A proton travels with a speed of 3.00 ! 106 m/s at an angleof 37.0° with the direction of a magnetic field of 0.300 Tin the " y direction. What are (a) the magnitude of themagnetic force on the proton and (b) its acceleration?
A proton moves perpendicular to a uniform magnetic fieldB at 1.00 ! 107 m/s and experiences an acceleration of2.00 ! 1013 m/s2 in the " x direction when its velocity is inthe " z direction. Determine the magnitude and directionof the field.
6. An electron is accelerated through 2 400 V from rest andthen enters a uniform 1.70-T magnetic field. What are(a) the maximum and (b) the minimum values of themagnetic force this charge can experience?
A proton moving at 4.00 ! 106 m/s through a magneticfield of 1.70 T experiences a magnetic force of magnitude8.20 ! 10#13 N. What is the angle between the proton’svelocity and the field?
7.
5.
1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide
= coached solution with hints available at http://www.pse6.com = computer useful in solving problem
= paired numerical and symbolic problems
P R O B L E M S
(a)
+
+
–
+
(c)
(b)
(d)
!!!!!!!!!!!!!!!!
45°
Bin
Bright
Bup
Bat 45°
Figure P29.1
2. Consider an electron near the Earth’s equator. In which di-rection does it tend to deflect if its velocity is directed
Figure 1: Problem 1.
vertically 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 does not depend on the value of θ.
Solution: The sphere is in translational equilibrium, thus fs −Mg 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 fsR, where R is the radius of the sphere. Hence, fsR−µB sin θ = 0. Substitut-
ing the expression for fs derived from the condition of translational equilibrium, fs = Mg sin θ into
the condition for rotational equilibrium, and cancelling out sin θ, one obtains µB = MgR. Now,
µ = NIπR2. Thus, I = MgπNBR = 0.08 kg·9.80 m/s2
5π·0.350 T·0.2 m = 0.713 A. The current must be counterclockwise
as seen from above.
Chapter 29 163
Q29.17 The proton will veer upward when it enters the field and move in a counter-clockwise semicircular
arc. An electron would turn downward and move in a clockwise semicircular arc of smaller radius
than that of the proton, due to its smaller mass.
Q29.18 Particles of higher speeds will travel in semicircular paths of proportionately larger radius. They will
take just the same time to travel farther with their higher speeds. As shown in Equation 29.15, the
time it takes to follow the path is independent of particle’s speed.
Q29.19 The spiral tracks are left by charged particles gradually losing kinetic energy. A straight path might
be left by an uncharged particle that managed to leave a trail of bubbles, or it might be the
imperceptibly curving track of a very fast charged particle.
Q29.20 No. Changing the velocity of a particle requires an accelerating force. The magnetic force is proportional
to the speed of the particle. If the particle is not moving, there can be no magnetic force on it.
Q29.21 Increase the current in the probe. If the material is a semiconductor, raising its temperature may
increase the density of mobile charge carriers in it.
SOLUTIONS TO PROBLEMS
Section 29.1 Magnetic Fields and Forces
P29.1 (a) up
(b) out of the page, since the
charge is negative.
(c) no deflection
(d) into the page
FIG. P29.1
P29.2 At the equator, the Earth’s magnetic field is
horizontally north. Because an electron has
negative charge, F v B uq is opposite in direction
to v Bu . Figures are drawn looking down.
(a) Down u North = East, so the force is
directed West .
(a) (c) (d)
FIG. P29.2
(b) North u North q sin0 0 : Zero deflection .
(c) West u North = Down, so the force is directed Up .
(d) Southeast u North = Up, so the force is Down .
Chapter 29 163
Q29.17 The proton will veer upward when it enters the field and move in a counter-clockwise semicircular
arc. An electron would turn downward and move in a clockwise semicircular arc of smaller radius
than that of the proton, due to its smaller mass.
Q29.18 Particles of higher speeds will travel in semicircular paths of proportionately larger radius. They will
take just the same time to travel farther with their higher speeds. As shown in Equation 29.15, the
time it takes to follow the path is independent of particle’s speed.
Q29.19 The spiral tracks are left by charged particles gradually losing kinetic energy. A straight path might
be left by an uncharged particle that managed to leave a trail of bubbles, or it might be the
imperceptibly curving track of a very fast charged particle.
Q29.20 No. Changing the velocity of a particle requires an accelerating force. The magnetic force is proportional
to the speed of the particle. If the particle is not moving, there can be no magnetic force on it.
Q29.21 Increase the current in the probe. If the material is a semiconductor, raising its temperature may
increase the density of mobile charge carriers in it.
SOLUTIONS TO PROBLEMS
Section 29.1 Magnetic Fields and Forces
P29.1 (a) up
(b) out of the page, since the
charge is negative.
(c) no deflection
(d) into the page
FIG. P29.1
P29.2 At the equator, the Earth’s magnetic field is
horizontally north. Because an electron has
negative charge, F v B uq is opposite in direction
to v Bu . Figures are drawn looking down.
(a) Down u North = East, so the force is
directed West .
(a) (c) (d)
FIG. P29.2
(b) North u North q sin0 0 : Zero deflection .
(c) West u North = Down, so the force is directed Up .
(d) Southeast u North = Up, so the force is Down .
Figure 2: Solution of problem 1.
Problem 3: Particle Orbits in a Uniform Magnetic Field 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, 1q and 2q . When in the magnetic field, their trajectories both curve in the same direction (see sketch), 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. (a) Are the charges of these particles positive or negative? Explain your reasoning. Solution: Because BF qv B= !
! !! , the charges of these particles are POSITIVE. (b) What is the ratio 2 1/q q ? Solution: We first find 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 2 /v R . Therefore applying Newton’s Second Law yields
2mvqvBR
= .
We can solve this for the radius of the circular orbit
mvRqB
=
Therefore the charged ratio
2 1
2 11 2
q Rmv mvR B R Bq R
! " ! "= =# $ # $% & % &
.
Problem 2: Particle Trajectory 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 the figure below. You may ignore effects of the gravitational force.
(a) Is the trajectory of the particle deflected upward or downward? (b) What is the magnitude of the velocity of the particle if it just strikes the end of the plate? Solution: Choose unit vectors as shown in the figure.
The force on the particle is given by
!F = !e(v i " Bj) = !evBk . (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
2vevB mR
= . (2)
.
Figure 3: Problem 2.
Problem 3: Particle Orbits in a Uniform Magnetic Field 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, 1q and 2q . When in the magnetic field, their trajectories both curve in the same direction (see sketch), 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. (a) Are the charges of these particles positive or negative? Explain your reasoning. Solution: Because BF qv B= !
! !! , the charges of these particles are POSITIVE. (b) What is the ratio 2 1/q q ? Solution: We first find 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 2 /v R . Therefore applying Newton’s Second Law yields
2mvqvBR
= .
We can solve this for the radius of the circular orbit
mvRqB
=
Therefore the charged ratio
2 1
2 11 2
q Rmv mvR B R Bq R
! " ! "= =# $ # $% & % &
.
Figure 4: Problem 3.
Problem 4 Mass Spectrometer Shown below 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 0V! > , and then enters a selector chamber, S1 , in
which there is an adjustable magnetic field 1B!
, pointing out of the page and a deflecting
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 field
2B!
, 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 , 2 2B ! B
!, and V! .
a) What magnetic field 1B!
in the selector chamber is needed to insure that the particle travels straight through?
Solution: 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
2(1/ 2)K mv! = . The change in potential energy is U e V! = " ! From conservation of energy, K U! = "! , we have that
2(1/ 2)mv e V= ! . So the speed is
2e Vvm!=
Inside the selector the force on the charge is given by
1( )e e= + !F E v B! ! !! .
Figure 5: Problem 4.
Problems 923
58. Review Problem. A wire having a linear mass density of1.00 g/cm is placed on a horizontal surface that has acoefficient of kinetic friction of 0.200. The wire carries acurrent of 1.50 A toward the east and slides horizontally tothe north. What are the magnitude and direction of thesmallest magnetic field that enables the wire to move inthis fashion?
59. Electrons in a beam are accelerated from rest through apotential difference !V. The beam enters an experimentalchamber through a small hole. As shown in Figure P29.59,the electron velocity vectors lie within a narrow cone ofhalf angle " oriented along the beam axis. We wish to usea uniform magnetic field directed parallel to the axis tofocus the beam, so that all of the electrons can passthrough a small exit port on the opposite side of thechamber 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 samepotential difference and the angle " is small, they allrequire the same time interval to travel the axial distance d.
long. The springs stretch 0.500 cm under the weight of thewire and the circuit has a total resistance of 12.0 #. Whena magnetic field is turned on, directed out of the page, thesprings stretch an additional 0.300 cm. What is the magni-tude of the magnetic field?
62. A hand-held electric mixer contains an electric motor.Model the motor as a single flat compact circular coil carry-ing electric current in a region where a magnetic field isproduced by an external permanent magnet. You need con-sider only one instant in the operation of the motor. (Wewill consider motors again in Chapter 31.) The coil movesbecause the magnetic field exerts torque on the coil, asdescribed in Section 29.3. Make order-of-magnitude esti-mates of the magnetic field, the torque on the coil, thecurrent in it, its area, and the number of turns in the coil, sothat they are related according to Equation 29.11. Note thatthe input power to the motor is electric, given by ! $ I !V,and the useful output power is mechanical, ! $ %&.
63. A nonconducting sphere has mass 80.0 g and radius20.0 cm. A flat compact coil of wire with 5 turns is wrappedtightly around it, with each turn concentric with thesphere. As shown in Figure P29.63, the sphere is placed onan inclined plane that slopes downward to the left, makingan angle ' with the horizontal, so that the coil is parallel tothe inclined plane. A uniform magnetic field of 0.350 Tvertically upward exists in the region of the sphere. Whatcurrent in the coil will enable the sphere to rest in equilib-rium on the inclined plane? Show that the result does notdepend on the value of '.
Exitport
d
Entranceport
!V
"
Figure P29.59
24.0 V
5.00 cm
Figure P29.61
#
B
Figure P29.63
#
B
I
g#
Figure P29.64
60. Review Problem. A proton is at rest at the plane verticalboundary of a region containing a uniform vertical mag-netic field B. An alpha particle moving horizontally makesa head-on elastic collision with the proton. Immediatelyafter the collision, both particles enter the magnetic field,moving perpendicular to the direction of the field. Theradius of the proton’s trajectory is R . Find the radius of thealpha particle’s trajectory. The mass of the alpha particle isfour times that of the proton, and its charge is twice that ofthe proton.
61. The circuit in Figure P29.61 consists of wires at the topand bottom and identical metal springs in the left andright sides. The upper portion of the circuit is fixed. Thewire at the bottom has a mass of 10.0 g and is 5.00 cm
64. A metal rod having a mass per unit length ( carries acurrent I. The rod hangs from two vertical wires in auniform vertical magnetic field as shown in Figure P29.64.The wires make an angle ' with the vertical when in equi-librium. Determine the magnitude of the magnetic field.
Figure 6: Problem 5.
Problem 5: Magnetic Field of a Ring of Current A circular ring of radius R lying in the xy plane carries a steady current I , as shown in the figure below.
What is the magnetic field at a point P on the axis of the loop, at a distance z from the center? Solution:
(a) We shall use the Biot-Savart law to find the magnetic field on the symmetry axis.
(1) Source point: In Cartesian coordinates, the differential current element located at
ˆ ˆ' (cos ' sin ' )R ! != +r i j!can be written as ˆ ˆ( '/ ') ' '( sin ' cos ' )Id I d d d IRd! ! ! ! != = " +s r i j! !
. (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
!r = zk . (3) Relative position vector
!r ! !r ' :
Problem 5: Magnetic Field of a Ring of Current A circular ring of radius R lying in the xy plane carries a steady current I , as shown in the figure below.
What is the magnetic field at a point P on the axis of the loop, at a distance z from the center? Solution:
(a) We shall use the Biot-Savart law to find the magnetic field on the symmetry axis.
(1) Source point: In Cartesian coordinates, the differential current element located at
ˆ ˆ' (cos ' sin ' )R ! != +r i j!can be written as ˆ ˆ( '/ ') ' '( sin ' cos ' )Id I d d d IRd! ! ! ! != = " +s r i j! !
. (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
!r = zk . (3) Relative position vector
!r ! !r ' :
Figure 7: Problem 6.
2
0 02 2 3/ 2 2 2 3/ 20
2sin ' ' cos ' 0
04 ( ) 4 ( )yIRz IRzB d
R z R z! !µ µ" " "
! != =# =
+ +$ . (7)
On the other hand, the z component is
22 220 0 0
2 2 3/ 2 2 2 3/ 2 2 2 3/ 20
2'4 ( ) 4 ( ) 2( )z
IRIR IRB dR z R z R z
!µ µ µ!"! !
= = =+ + +# . (8)
Thus, we see that along the symmetric axis, zB is the only non-vanishing component of the magnetic field. The conclusion can also be reached by using the symmetry arguments. The behavior of 0/zB B where 0 0 / 2B I Rµ= is the magnetic field strength at 0z = , as a function of /z R is shown in the figure below.
Figure 8: Solution of problem 6.
Problem 6: Magnetic Fields Find the magnetic field at point P due to the following current distribution.
Solution: The fields due to the straight wire segments are zero at P because d sr and r are parallel or anti-parallel. For the field due to the arc segment, the magnitude of the magnetic field due to a differential current carrying element is given in this case by
d!B =
µ0I4!
d !s " rR2 =
µ0
4!IRd#(sin# i $ cos# j)" ($cos# i $ sin# j)
R2
= $µ0
4!I(sin2# + cos2# )d#
Rk = $
µ0
4!Id#R
k.
Therefore,
!B = !
µ0I4"R
d#0
" /2
$ k = !µ0I4"R
"2
%&'
()*
k = !µ0I8R
%&'
()*
k (into the plane of the figure).
Figure 9: Problem 7.
Problem 7: Gyromagnetic Ratio A thin uniform ring of radius R and mass M carrying a charge +Q rotates about its axis with constant angular speed ! as shown in the figure below. Find the ratio of the magnitudes of its magnetic dipole moment to its angular momentum. (This is called the gyromagnetic ratio.)
Solution: The current in the ring is
I = QT
=Q!2"
.
The magnetic moment is
!µ = AI k = !R2 Q"
2!k = Q"R
2
2k .
The angular momentum is
!L = Imomnet! k = MR
2! k . So the gyromagnetic ratio is
!µ!L
=Q!R2 / 2MR2!
=Q2M
Figure 10: Problem 8.
Problems 923
58. Review Problem. A wire having a linear mass density of1.00 g/cm is placed on a horizontal surface that has acoefficient of kinetic friction of 0.200. The wire carries acurrent of 1.50 A toward the east and slides horizontally tothe north. What are the magnitude and direction of thesmallest magnetic field that enables the wire to move inthis fashion?
59. Electrons in a beam are accelerated from rest through apotential difference !V. The beam enters an experimentalchamber through a small hole. As shown in Figure P29.59,the electron velocity vectors lie within a narrow cone ofhalf angle " oriented along the beam axis. We wish to usea uniform magnetic field directed parallel to the axis tofocus the beam, so that all of the electrons can passthrough a small exit port on the opposite side of thechamber 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 samepotential difference and the angle " is small, they allrequire the same time interval to travel the axial distance d.
long. The springs stretch 0.500 cm under the weight of thewire and the circuit has a total resistance of 12.0 #. Whena magnetic field is turned on, directed out of the page, thesprings stretch an additional 0.300 cm. What is the magni-tude of the magnetic field?
62. A hand-held electric mixer contains an electric motor.Model the motor as a single flat compact circular coil carry-ing electric current in a region where a magnetic field isproduced by an external permanent magnet. You need con-sider only one instant in the operation of the motor. (Wewill consider motors again in Chapter 31.) The coil movesbecause the magnetic field exerts torque on the coil, asdescribed in Section 29.3. Make order-of-magnitude esti-mates of the magnetic field, the torque on the coil, thecurrent in it, its area, and the number of turns in the coil, sothat they are related according to Equation 29.11. Note thatthe input power to the motor is electric, given by ! $ I !V,and the useful output power is mechanical, ! $ %&.
63. A nonconducting sphere has mass 80.0 g and radius20.0 cm. A flat compact coil of wire with 5 turns is wrappedtightly around it, with each turn concentric with thesphere. As shown in Figure P29.63, the sphere is placed onan inclined plane that slopes downward to the left, makingan angle ' with the horizontal, so that the coil is parallel tothe inclined plane. A uniform magnetic field of 0.350 Tvertically upward exists in the region of the sphere. Whatcurrent in the coil will enable the sphere to rest in equilib-rium on the inclined plane? Show that the result does notdepend on the value of '.
Exitport
d
Entranceport
!V
"
Figure P29.59
24.0 V
5.00 cm
Figure P29.61
#
B
Figure P29.63
#
B
I
g#
Figure P29.64
60. Review Problem. A proton is at rest at the plane verticalboundary of a region containing a uniform vertical mag-netic field B. An alpha particle moving horizontally makesa head-on elastic collision with the proton. Immediatelyafter the collision, both particles enter the magnetic field,moving perpendicular to the direction of the field. Theradius of the proton’s trajectory is R . Find the radius of thealpha particle’s trajectory. The mass of the alpha particle isfour times that of the proton, and its charge is twice that ofthe proton.
61. The circuit in Figure P29.61 consists of wires at the topand bottom and identical metal springs in the left andright sides. The upper portion of the circuit is fixed. Thewire at the bottom has a mass of 10.0 g and is 5.00 cm
64. A metal rod having a mass per unit length ( carries acurrent I. The rod hangs from two vertical wires in auniform vertical magnetic field as shown in Figure P29.64.The wires make an angle ' with the vertical when in equi-librium. Determine the magnitude of the magnetic field.
180 Magnetic Fields
P29.62 Suppose the input power is
120 120 W V a fI : I~1 100 A A .
Suppose ZS
FHGIKJFHG
IKJ2 000
1 2200 rev min
min60 s
rad1 rev
rad s~
and the output power is 20 200 W rad s WZ W b g W ~10 1� � N m .
Suppose the area is about 3 4 cm cma f a fu , or A~10 3� m2 .
Suppose that the field is B~10 1� T .
Then, the number of turns in the coil may be found from W # NIAB :
0 1 1 10 103 1. ~ N m C s m N s C m2� � �� �Nb ge je jgiving N~103 .
*P29.63 The sphere is in translational equilibrium, thus
f Mgs � sinT 0 . (1)
The sphere is in rotational equilibrium. If torques are taken about thecenter of the sphere, the magnetic field produces a clockwise torque ofmagnitude P TBsin , and the frictional force a counterclockwise torqueof magnitude f Rs , where R is the radius of the sphere. Thus:
f R Bs � P Tsin 0 . (2)
From (1): f Mgs sinT . Substituting this in (2) and canceling out sinT ,one obtains
PB MgR . (3)
Ifs
B
MgT
TPPPP
GG
FIG. P29.63
Now P S NI R2 . Thus (3) gives IMgNBR
S S
0 08 9 80
5 0 350 0 20 713
. .
. ..
kg m s
T m A
2b ge ja fa fa f . The current must be
counterclockwise as seen from above.
P29.64 Call the length of the rod L and the tension in each wire alone T2
. Then, at equilibrium:
F T ILBx¦ � q sin sin .T 90 0 0 or T ILBsinT
F T mgy¦ � cosT 0 , or T mgcosT
tanT ILBmg
IBm L gb g or B
m L g
IgI
b g
tan tanTO
T
P29.65 F ma¦ or qvBmvr
sin .90 02
q
? the angular frequency for each ion is vr
qBm
f Z S2 and
'
'
f f fqB
m m
f f f
� �FHG
IKJ
u
u�F
HGIKJ
� u
�
�
�
12 1412 14
19
27
12 145 1
21 1 1 60 10 2 40
2 1 66 10
112 0
114 0
4 38 10 438
S S
. .
. . .
.
C T
kg u u u
s kHz
e ja fe j
Figure 11: Problem 10.
