MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Exam 2 Practice Problems Part 2 Solutions
Problem 1: Short Questions
(a) Can a constant magnetic field set into motion an electron, which is initially at rest?
Explain your answer.
Solution: 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.
(b) Is it possible for a constant magnetic field to alter the speed of a charged particle?
What is the role of a magnetic field in a cyclotron?
!
! !
Solution: No, it is not possible. Because FB = q( v ! B) , the acceleration produced by a
!
! !
magnetic field on a particle of mass m is a B = q( v ! B) / m . For the acceleration to
change the speed, a component of the acceleration must be in the direction of the
velocity. The cross product tells us that the acceleration must be perpendicular to the
velocity, and thus can only change the direction of the velocity.
The magnetic field in a cyclotron essentially keeps the charged particle in the electric
field for a longer period of time, and thus experiencing a larger change in speed from the
electric field, by forcing it in a spiral path. Without the magnetic field, the particle would
have to move in a straight line through an electric field over a distance that is very large
compared to the size of the cyclotron.
(c) How can a current loop be used to determine the presence of a magnetic field in a
given region of space?
Solution: If the current loop feels a torque, it must be caused by a magnetic field. If the
current loop feels no torque, try a different orientation—the torque is zero if the field is
along the axis of the loop.
(d) If a charged particle is moving in a straight line through some region of space, can
you conclude that the magnetic field in that region is zero? Why or why not?
Solution: Not necessarily. If the magnetic field is parallel or anti-parallel to the velocity
of the charged particle, then the particle will experience !no magnetic
! ! force.
! There may
also be an electric force acting on the particle such that Fq = q(E + v q ! B) = 0 .
(e) List some similarities and differences between electric and magnetic forces.
Solution:
Similarities:
1. Both can accelerate a charged particle moving through the field.
2. Both exert forces directly proportional to the charge of the particle feeling the force.
Differences:
1. The direction of the electric force is parallel or anti-parallel to the direction of the
electric field, but the direction of the magnetic force is perpendicular to the magnetic
field and to the velocity of the charged particle.
2. Electric forces can accelerate a charged particle from rest or stop a moving particle,
but magnetic forces cannot.
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 î " Bĵ) = !evBk̂ .
(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
evB = m
.
v2
.
R
(2)
The speed of the particle is then
v=
eBR
.
m
(3)
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
R 2 = ( R ! d / 2) 2 + l 2 .
(4)
R 2 = R 2 ! Rd + d 2 / 4 + l 2 .
(5)
Expanding Eq. (4) yields
Solve Eq. (5) for the radius of the circular orbit:
R=
d l2
+ .
4 d
(6)
We can now substitute the Eq. (6) into Eq. (3) and find the speed necessary for the
particle to just hit the end of the plate:
v=
eB ! d l 2 "
# + $.
m%4 d &
(7)
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, q1 and q2 . 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 FB = qv ! B , the charges of these particles are POSITIVE.
(b) What is the ratio q2 / q1 ?
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
v 2 / R . Therefore applying Newton’s Second Law yields
mv 2
qvB =
.
R
We can solve this for the radius of the circular orbit
R=
mv
qB
Therefore the charged ratio
q2 ! mv " ! mv " R1
.
=#
$ #
$=
q1 % R2 B & % R1 B & R2
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 !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
!
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 .
!
a) What magnetic field B1 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
!K = (1/ 2)mv 2 . The change in potential energy is !U = "e!V From conservation of
energy, !K = "!U , we have that
(1/ 2)mv 2 = e!V .
So the speed is
2e!V
v=
m
Inside the selector the force on the charge is given by
!
! ! !
Fe = 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 the figure above (define this as the +iˆ direction),
the electric field points up (define this as the +jˆ direction) from the positive plate to the
negative plate, and the magnetic field is pointing out of the page (define this as the +kˆ
direction). Then
Eˆj = !vˆi " B1kˆ = vB1ˆj .
So
!
E
m
B1 = kˆ =
E kˆ
v
2e!V
b) 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: The force on the charge when it enters the magnetic field B 2 is given by
!
Fe = evˆi ! B2kˆ = "evB2 ˆj .
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. When in circular motion
the acceleration is towards the center. In particular when the particle just enters the field
!
B 2 , the acceleration is downward
!
v2 ˆ
a=!
j.
x/2
Newton’s Second Law becomes
!evB2 ˆj = !m
v2 ˆ
j.
x/2
Thus the particle hits the electronic sensor at a distance
x=
2mv
2
=
2e!Vm
eB2 eB2
from the entry slit. The mass of the particle is then
m=
eB2 2 x 2
.
8!V
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
!
!
!
r ' = R(cos ! ' ˆi + sin ! ' ˆj) can be written as Id s = I (dr '/ d! ')d! ' = IRd! '(" sin ! ' ˆi + cos ! ' ˆj) .
(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 = zk̂ .
! !
(3) Relative position vector r ! r ' :
The vector form the current element ot the field point is given by
! !
r ! r ' = ! Rcos " ' î ! Rsin " ' ĵ + z k̂ ,
(1)
and its magnitude
(
)
2
! !
r = r ! r ' = (! Rcos " ')2 + ! Rsin " ' + z 2 =
R2 + z 2
(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
! !
r ! r'
r̂ = ! ! .
|r ! r' |
(4) Simplifying the cross product:
! ! !
The cross product d s ! (r " r ') can be simplified as
! ! !
d s ! (r " r ') = R d# '(" sin # ' î + cos# ' ĵ) ! (" Rcos# ' î " Rsin # ' ĵ + z k̂)
= R d# '(z cos# ' î + z sin # ' ĵ + R k̂).
(3)
!
(5) Writing down dB :
Using the Biot-Savart law, the contribution of the current element to the magnetic field at
P is
! µ0 I d !s " r̂ µ0 I d !s " (r! # r! ')
dB =
=
! !
4! r 2
4! | r # r ' |3
.
(4)
µ0 IR z cos $ ' î + z sin $ ' ĵ + R k̂
=
d$ '
4!
(R 2 + z 2 )3/ 2
(6) Carrying out the integration:
Using the result obtained above, the magnetic field at P is
! µ IR 2! z cos " ' î + z sin " ' ĵ + R k̂)
B= 0 # (
d" ' .
4! 0
(R 2 + z 2 )3/ 2
(5)
!
The x and the y components of B can be readily shown to be zero:
Bx =
µ0 IRz
4! ( R 2 + z 2 )3/ 2
#
2!
0
cos " ' d" ' =
2!
µ0 IRz
sin " '
=0.
2
2 3/ 2
0
4! ( R + z )
(6)
By =
µ0 IRz
4! ( R 2 + z 2 )3/ 2
$
2!
0
sin " ' d" ' = #
2!
µ0 IRz
cos " '
=0.
2
2 3/ 2
0
4! ( R + z )
(7)
On the other hand, the z component is
Bz =
µ0
IR 2
4! ( R 2 + z 2 )3/ 2
#
2!
0
d" ' =
µ0 2! IR 2
µ0 IR 2
=
.
4! ( R 2 + z 2 )3/ 2 2( R 2 + z 2 )3/ 2
(8)
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 = µ0 I / 2 R is the magnetic field strength at z = 0 , as a
function of z / R is shown in the figure below.
Problem 6: Magnetic Fields
Find the magnetic field at point P due to the following current distribution.
Solution:
r
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 field
due to a differential current carrying element is given in this case by
Therefore,
! µ0 I d !s " r̂ µ0 IRd# (sin # î $ cos# ĵ) " ($ cos# î $ sin # ĵ)
dB =
=
4! R 2
4!
R2
.
µ0 I(sin 2 # + cos 2 # )d#
µ0 Id#
=$
k̂ = $
k̂
4!
R
4! R
!
" /2 µ I
%µ I(
µ I %"(
0
B = !$
d# k̂ = ! 0 ' * k̂ = ! ' 0 * k̂ (into the plane of the figure).
0
4" R
4" R & 2 )
& 8R )
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=
Q Q!
.
=
T
2"
The magnetic moment is
!
Q"
Q" R 2
µ = AI k̂ = ! R 2
k̂ =
k̂ .
2!
2
The angular momentum is
!
L = I momnet! k̂ = MR 2! k̂ .
So the gyromagnetic ratio is
!
µ Q! R 2 / 2
Q
! =
=
2
MR !
2M
L
Problem 8: Torque on Circular Current Loop
A wire ring lying in the xy-plane with its center at the origin carries a counterclockwise
!
current I. There !is a uniform magnetic field B = Bˆi 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
! ! !
! = µ " B,
!
where µ = IA and the vector ì 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 = IA = I(!R 2k̂) = !IR 2k̂ .
Therefore,
! ! !
! = µ " B = (#IR 2k̂) " (B î) = #IR 2 B ĵ .
Instead of using the above formula, we can calculate the torque directly as follows.
Choose a small section of the loop of length ds = Rd! . Then the vector describing the
current-carrying element is given by
!
Id s = IRd! (" sin ! ˆi + cos ! ˆj)
!
The force dF that acts on this current element is
!
! !
dF = Id s " B
= I Rd! (# sin ! ˆi + cos ! ˆj) " ( Bˆi )
= # IRB cos ! d! kˆ
The force acting on the loop can be found by integrating the above expression.
!
!
F="
d
F
! =
!
2$
0
(" IRBcos# )d# k̂
2$
= " IRB %&sin # '(0 k̂ = 0
We expect this because the magnetic field is uniform and the force ona 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 Ids . That is,
!
! !
!
r = R(cos! î + sin ! ĵ) . The torque d ! = r " dF acting on the current element is then
!
! !
d ! = r " dF
(
) (
= R cos# î + sin # ĵ " $ IRBd# cos# k̂
)
= $ IR 2 Bd# cos# (sin # î $ cos# ĵ)
!
!
Integrate dô over the loop to find the total torque ô .
!
!
! = "d!
(
2%
)
= " # IR 2 B d$ cos$ sin $ î # cos$ ĵ
0
2%
= # IR 2 B " (sin $ cos$ î # cos 2 $ ĵ)d$
0
= % IR B ĵ
2
This agrees with our result above.
Problem 9: Challenge Zeeman Effect
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 frequency ! 0 of the orbital motion? Suppose now that a small
!
magnetic field B perpendicular to the plane of the orbit is switched on. Assuming that
the radius of the orbit does not change, calculate the shift in the angular frequency
!" = " f # " 0 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 ! 0 of the
orbital motion. Coulomb’s Law describes the force acting on the electron due to the
electric interaction between the proton and the electron
!
ke2
Felec = ! 2 r̂
r
where r̂ is a unit vector in the plane of the circular orbit pointing radiallly outward.
Because we are assuming the motion is circular and the acceleration is
!
a = !r" 0 2 r̂ .
Newton’s Second Law is then
ke2
! 2 r̂ = !mr" 0 2 r̂
r
(1)
We can now solve for the angular frequency ! 0
!0 =
ke2
.
mr 3
(2)
!
When a small magnetic field B perpendicular to the plane of the orbit is switched on
there is magnetic force acting on the electron
!
! !
Fmag = !ev " B .
Let’s first assume that the magnetic 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 ! f . Choose coordinates as shown in the figure below.
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 = r! f "ˆ . Then the
magnetic force is given by
!
! !
Fmag = !ev " B = !er# f $̂ " Bk̂ = !er# f B r̂ .
Therefore Newton’s Second Law is now
!
ke2
r̂ ! er" f Br̂ = !mr" f 2 r̂ .
r2
Using Eq. (1) we can rewrite this as
!mr" 0 2 r̂ ! er" f Br̂ = !mr" f 2 r̂
or
0 = !f2 "
e! f B
m
" ! 02 .
We can solve this quadratic equation for ! f
!f =
eB
e2 B 2
±
+ ! 02 .
2m
4m 2
We choose the positive square root to keep ! f > 0 . We now substitute Eq. (2) into the
above equation yielding
eB
e2 B 2 ke2
!f =
+
+
.
2m
4m 2 mr 3
(3)
Then the change in angular frequency is
!" = " f # " 0 =
eB
e2 B 2 ke2
ke2
+
+
#
.
2m
4m 2 mr 3
mr 3
(4)
Suppose we reverse the direction of the magnetic field as in the figure below.
Then the magnetic force points outward,
!
! !
Fmag = !ev " B = !er# f $̂ " !Bk̂ = +er# f B r̂ ,
resulting in a smaller angular frequency ! f .
Repeating the analysis we just finished we have that
!mr" 0 2 r̂ + er" f Br̂ = !mr" f 2 r̂
or
0 = !f2 +
e! f B
m
" ! 02 .
We can solve this quadratic equation for ! f choosing the positive square root to keep
!f > 0:
!f = "
eB
e2 B 2
±
+ ! 02 .
2m
4m 2
The change in angular frequency is now
eB
e2 B 2 ke2
ke2
!" = " f # " 0 = #
+
+
#
.
2m
4m 2 mr 3
mr 3