Homework 4 — Solutions
Kelly Douglass
Phys 115 — Contemporary Physics iii
Drexel University
May 1, 2015
21.X.35
An alpha particle (consisting of two protons and two neutrons) is moving at constant speed in a circle,
perpendicular to a uniform magnetic field applied by some current-carrying coils, making one clockwise
revolution every 80 nanoseconds. If the speed is small compared to the speed of light, what is the numerical
magnitude of B of the magnetic field made by the coils? What is the direction of this magnetic field?
The force the alpha particle experiences due to the magnetic field is a result of the magnetic field; this
has the same effect as the centripetal force on the particle.
mv 2
r
mv
B=
qr
F = qvB =
Since the particle is travelling in a circle, we can rewrite
is the period, we see that
B=
v
r
= ω. And since we know that ω =
2π
T ,
where T
2π4(1.7 × 10−27 kg)
2πm
=
qT
2(1.6 × 10−19 C)(80 × 10−9 s)
B = 1.67 T
If the particle is orbiting clockwise as shown, then the magnetic field is pointing out of the page.
21.P.38
An electron and a proton are both in motion near each other as shown in Figure 21.84; at this instant they
are a distance r apart. The proton is moving with speed vp at an angle θ to the horizontal, and the electron
is moving straight up with speed ve .
(a) Calculate the x and y components of the force that the proton exerts on the electron. First calculate the
electric field and magnetic field that the proton produces at the location of the electron, then calculate
the forces that these fields exert on the electron.
~ p = 1 e2 to the right. Therefore, the electric force on the
The electric field due to the proton is E
4πε0 r
electron due to the proton is
~
F~e = q E
F~e =
1 e2
to the left
4πε0 r2
1
The magnetic field due to the proton is
~ p = µ0 e~v × ~r = µ0 evp sin θ into the page
B
4π r3
4π
r2
Therefore, the magnetic force on the electron is
~ = eve Bp to the right
F~B = q~v × B
µ0 e2 ve vp sin θ
to the right
F~B =
4π
r2
So, on the electron,
Fx = −
µ0 e2 ve vp sin θ
1 e2
+
4πε0 r2
4π
r2
Fy = 0
(b) Calculate the x and y components of the force that the electron exerts on the proton. First calculate the
electric field and magnetic field that the electron produces at the location of the proton, then calculate the
forces that these fields exert on the proton. (In a completely correct quantitative calculation we would
have to use the relativistically correct fields.)
~ e = 1 e2 to the right. Therefore, the electric force on the
The electric field due to the electron is E
4πε0 r
proton due to the electron is
~
F~p = q E
F~p =
1 e2
to the right
4πε0 r2
The magnetic field due to the electron is
~ e = µ0 e~v × ~r = µ0 eve into the page
B
4π r3
4π r2
Therefore, the magnetic force on the proton is
~ = evp Be
F~B = q~v × B
µ0 e2 ve vp
F~B =
h− sin θ, cos θ, 0i
4π r2
So, on the proton,
Fx =
1 e2
µ0 e2 ve vp sin θ
−
4πε0 r2
4π
r2
Fy =
µ0 e2 ve vp
cos θ
4π r2
(c) Consider carefully your results. Are the magnetic forces on electron and proton equal and opposite? Does
reciprocity hold for magnetic forces?
The electric force is equal and opposite, but the magnetic force is not. Reciprocity does not hold for
magnetic forces.
(d) Will the total momentum of the two particles remain constant? Is this a violation of conservation of
momentum for an isolated system?
The total momentum of the two particles will not remain constant. However, the total momentum of
the system does not change, since the fields have momentum.
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21.P.49
A Nichrome wire has a shape of two quarter-circles of radius a and b, connected by straight sections as shown
in Figure 21.93. The wire carries conventional current I in the direction shown.
(a) Calculate the direction and magnitude of the magnetic field at point C, the center of the quarter-circles.
Briefly, explain your work, including a diagram.
Since the two straight sides are parallel with r̂, the only components of the circuit that contribute to
the magnetic field at C are due to the two quarter-circles.
~ =B
~a + B
~b
B
B = Ba − Bb
The magnetic field for a half-circle of wire of radius R is
Z
µ0
BR =
4π
π/2
0
µ0 I
BR =
4π R
µ0 I
BR =
8 R
Z
I d~l × ~r
r3
π/2
dθ
0
Therefore,
µ0
I
B=
8
1 1
−
a b
~ = µ0 I b − a out of page
B
8
ab
(b) At a particular instant, an electron is at point C, traveling to the right with speed v. Calculate the
direction and magnitude of the magnetic force on the electron. Briefly explain your work, including a
diagram.
The force on the electron due to the magnetic field is
~
F~B = q~v × B
µ0 b − a
F~B = ev I
up
8
ab
21.P.51
A metal rod of length 12 cm and mass 70 grams has metal loops at both ends, which go around two metal
poles. The rod is in good electrical contact with the poles but can slide freely up and down. The metal poles
are connected by wires to a battery, and a 5-ampere current flows through the rod. A magnet supplies a large
uniform magnetic field B in the region of the rod, large enough that you can neglect the magnetic fields due
to the 5-ampere current. The magnetic field is oriented so as to have a maximum effect on the rod. The rod
sits at rest 4 cm above the table. What are the magnitude and direction of the magnetic field in the region
of the rod? Briefly explain your work, including a diagram.
3
In order for the rod to sit at rest above the table, the gravitational force on it must be equal and opposite
to the magnetic force it experiences. Therefore,
Fg = FB
mg = ILB
2
B=
(70 × 10−3 kg)(9.8 m/s )
mg
=
IL
(5 A)(0.12 m)
B = 1.14 T
Since the gravitational force on the rod is downwards, the magnetic force must be up. Using the righthand rule, the magnetic field must be into or out of the page (depending on the direction of the current in
the rod).
21.X.52
An object of charge q is moving in the x direction with speed v. Throughout the region there is a uniform
electric field in the y direction of magnitude E. Determine a direction and magnitude B for a uniform
~ on a
magnetic field such that the net electric and magnetic force on the moving charge is zero. Draw B
diagram of the situation.
The magnetic force on the charged object must be down, to cancel the force from the electric field. This
requires the magnetic field to point out of the page.
FE = FB
qE = qvB
~ = E out of the page
B
v
21.P.72
In Figure 21.112 a bar 11 cm long with a rectangular cross section 3 cm high and 2 cm deep is connected
to a 1.2-volt battery and an ammeter. The resistance of the copper connecting wires and ammeter, and the
internal resistance of the battery, are all negligible compared to the resistance of the bar.
Using large coils not shown on the diagram, a uniform magnetic field of 1.8 teslas was applied perpendicular to the bar (out of the page, as shown). A voltmeter was connected across the bar, with the connections
across the bar carefully placed directly across from each other.
The mobile charges in the bar have charge +e, their density is 7×1023 per cubic meter, and their mobility
is 3 × 10−5 (m/s)/(V/m).
Predict the readings of the voltmeter and ammeter, including signs. Explain carefully, using diagrams
to support your explanation. Remember that a voltmeter reads positive if the “+” terminal is connected to
higher potential, and that an ammeter reads positive if conventional current enters the “+” terminal.
The direction of the conventional current through the circuit here is counter-clockwise. As a result, the
ammeter will read positive.
Since the mobile charges in the bar have positive charges, they will be moving in the same direction as
the conventional current, or to the right in the figure. Using the right-hand rule, with a current to the right
and a magnetic field pointing out of the page, the direction of the magnetic force on the mobile charges is
down. Therefore, the bottom of the bar will accumulate a net positive charge, while the top will become
negatively charged. Therefore, the voltmeter will read negative.
4
Since the only two components in the circuit are the battery and the bar, from Kirchoff’s loop rule, we
know that the potential difference across the bar is equal to the emf of the battery.
∆V = IR
∆V
I=
R
From the definition of resistance, we can determine what the current is.
R=
L
σA
where L is the length of the bar, A is the cross-sectional area of the bar, and σ = nuq is the conductivity of
the bar. Therefore, the current in circuit is
∆V qnuA
(1.2 V)(1.6 × 10−19 C)(7 × 1023 m−3 )(3 × 10−5 m/s/V/m)(0.02 m)(0.03 m)
=
L
0.11 m
I = 0.0219 A
I=
Since the charges separate in the bar, there will be a net electric field present, pointing up. This results
in an electric force on the charged particles in the opposite direction of the magnetic force separating them.
In steady state, these two forces will cancel out.
FE = FB
qE = qvB
E = vB
The velocity of the charges can be found from the current.
∆V qnuA
L
∆V u
v=
L
I = qnAv =
Therefore, the electric field in the bar is
E=
B∆V u
L
The Hall potential across the bar is then
∆VHall = Eh =
B∆V uh
(1.8 T)(3 × 10−5 m/s/V/m)(0.03 m)
=
L
0.11 m
∆VHall = 1.77 × 10−5 V
21.X.78
A metal rod of length L slides horizontally at constant speed v on frictionless insulating rails through a region
of uniform upward magnetic field of magnitude B. On a diagram, show the polarization of the rod and the
direction of the Coulomb electric field inside the rod. Explain briefly. What is the magnitude EC of the
Coulomb electric field inside the rod? What is the potential difference across the rod? What is the emf across
the rod? What are the magnitude and direction of the force you have to apply to keep the rod moving at a
constant speed v?
5
With the velocity of the bar to the right and the magnetic field pointing out of the page, the magnetic
force on the charged particles in the bar is down. Therefore, the bottom of the bar will have a net postive
charge, and the top of the bar will have a net negative charge. This separation of charge results in an electric
field pointing up through the bar.
In steady state, the magnitude of the magnetic force is equal to the magnitude of the electric force.
FB = FE
qvB = qE
EC = vB
From this electric field, we can solve for the magnitude of the potential difference across the bar.
∆V = EC L
∆V = vBL
The emf across the rod is equal to the potential difference across the rod.
E = ∆V = vBL
Since the rails are insulating, this emf is not going to result in a current (since there is no complete
circuit). Therefore, there is no macroscopic magnetic force on the bar, so we do not need to apply any force
to the bar to keep it moving at a constant speed v.
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