MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Spring 2013
Problem Set 7 Solutions
Problem 1
a) Two infinitely long wires, each carrying current I in the same direction (directed out
of the plane of the figure), are separated by a distance s (see figure below). What are the
magnitude and direction of the magnetic field at point P , which is a distance s from each
wire?
b) At what position in the xy - plane would you have to place a third parallel infinitely
long wire carrying the same current I in the same direction to make the sum of the
magnetic fields from the three wires at point P equal to zero?
Solution:
(a) We can use Ampere’s Law for each wire individually. When we add the two
contributions only the horizontal component is non-zero.
For the wire on the left, the magnetic field points tangent to the circle as shown in the
figure below.
Ampere’s Law becomes
B2p s = m0 I Þ B =
m0 I
2p s
In the figure above, cos(q ) = 3 / 2. Therefore the x -component of the magnetic field is
given by
Bx,L = -
m0 I cos(q )
3m0 I
=2p s
4p s
In similar fashion, for the wire on the right, the magnetic field points tangent to the circle
as shown in the figure below.
Ampere’s Law becomes
B2p s = m0 I Þ B =
m0 I
2p s
In the figure above, cos(q ) = 3 / 2 . Therefore the x -component of the magnetic field is
given by
Bx, R = -
m0 I cos(q )
3m0 I
=2p s
4p s
We now use the superposition principle as shown in the figure below to find the x component of the magnetic field
Bx = Bx, L + Bx, R = -
3m0 I
3m0 I
3m0 I
=4p s
4p s
2p s
(b) We need to place the third current a distance d above the point
of the x -components of the three magnetic fields is zero.
P such that the sum
The x-component of the magnetic field due to the top current is Bx,T = m0 I / 2p d . We
require that the sum is zero:
3m0 I m0 I
+
= 0.
2p s
2p d
Bx = Bx, L + Bx, R + Bx,T = Thus we can solve for d and find that
d=
s
3
.
Problem 2: Force and Magnetic Field A very long straight wire carrying current I
directed into the plane of the figure below is suspended a distance R above a nearly
infinite plane surface of thickness d with uniform current density J directed out of the
plane of the paper (below figure on left).
a) What force per unit length acts on the wire? Assume the current in the wire does
not change the uniform current density J .
b) The wire is replaced by a particle of charge q that is initially traveling parallel to
the direction of the uniform current density J and is a distance R from the
surface (above figure on right shown from a side view). What is the direction and
magnitude of the force acting on the charged particle?
c) Sketch the trajectory of the particle. What magnitude of current density J is
required in order that the charged particle just grazes the infinite surface?
Solutions:
a) What force per unit length acts on the wire? Assume the current in the wire does
not change the uniform current density J .
Answer: We began by choosing coordinates and Amperian loop shown in the figure
below for the region outside the slab. For an infinite plane the magnetic field is zero
along the plane z = 0 . Above the slab the direction of B is in the negative x-direction
and below the slab the direction of B is in the positive x-direction.
The enclosed current is
Ienc = ò J × dA = ò (-Jĵ) × (- ĵda) = Jld .
The line integral of the magnetic field around the closed square loop is therefore
ò B × d s = 2Bl .
Therefore Ampere’s Law
ò B×d s = m I
0 enc
becomes
2Bl = m0 Jld
Therefore the magnetic field outside the slab is given by
ì m0 Jd
î; z > d / 2
ïï2
B=í
ï+ m0 Jd î; z < -d / 2
ïî
2
The force on a length dL of current carrying wire above the slab is
æ m Jd ö m I(dL)Jd
dF = IdL ´ B = IdLĵ ´ ç - 0 î ÷ = 0
k̂ .
2 ø
2
è
Therefore the force per unit length on the wire is
dF m0 IJd
=
k̂ .
dL
2
b) The wire is replaced by a particle of charge q that is initially traveling parallel to
the direction of the uniform current density J and is a distance R from the
surface (above figure on right shown from a side view). What is the direction and
magnitude of the force acting on the charged particle?
Answer:
The force on the charge particle is given by
æ m Jd ö
m qvJd
F = qv ´ B = qv(- ĵ) ´ ç - 0 î ÷ = - 0
k̂ .
2 ø
2
è
c) Sketch the trajectory of the particle. What magnitude of current density J is
required in order that the charged particle just grazes the infinite surface?
Answer: Because the magnetic field outside the slab is uniform the particle will undergo
circular motion.
If the particle just grazes the surface then by Newton’s Second Law
-
m0qvJd
2
k̂ = -
mv 2
k̂ .
R/2
We can now solve for the magnitude of the current density
J=
4mv
.
m0 dqR
Problem 3 Nested Solenoids: Two long solenoids are nested on the same axis, as in the
figure below. The inner solenoid has radius R1 and n1 turns per unit length. The outer
solenoid has radius R2 and n2 turns per unit length. Each solenoid carries the same
current I flowing in each turn, but in opposite directions, as indicated on the sketch.
Use Ampere’s Law to find the direction and magnitude of the magnetic field in the
following regions. Be sure to show your Amperian loops and all your calculations.
i)
ii)
iii)
0 < r < R1
R1 < r < R2
R2 < r
Solution: Everywhere outside the two solenoids, B = 0 .
B = 0;
R2 < r
Use Ampere’s Law to find the direction and magnitude of the magnetic field in the
following regions:
(a) 0  r  R1 ;
To solve for the magnetic field in this case, we take the top rectangular loop shown in the
figure. We circulate counterclockwise so current directed out of the plane of the figure is
positive and into the plane of the figure is negative. The current through the loop is
Ienc = n1l1 I - n2l1 I = (n1 - n2 )l1 I
The loop has four segments. Everywhere outside the two solenoids, B = 0 . Along two of
those segments (top and bottom horizontal segments) inside the solenoids, B is
perpendicular to d s , and B × d s = 0 . Thus,
ò B×d s = Bl + 0+ 0+ 0 = Bl .
1
We now can apply Ampere’s law
1
ò B×d s = m I
0 enc
, yielding
Bl1 =m0 (n1 - n2 )l1I Þ B = m0 (n1 - n2 )I .
Because the circulation direction for the segment inside the inner solenoid is directed in
the - k̂ -direction this implies that the vector description of the magnetic field is given by
B = m0 I(n1 - n2 )(-k̂) ;
0 < r < R1.
If n1 > n2 , then the magnetic field points in the + k̂ -direction; if n1 < n2 , then the magnetic
field points in the - k̂ -direction.
(b) R1  r  R2
To solve for the magnetic field in this case, we take the bottom rectangular loop shown in
the figure. Notice that we now are circulating clockwise, so current directed into the page
is positive. The current through the loop is
Ienc = n2l2 I
By the same reasoning as above, the loop has four segments. Everywhere outside the two
solenoids, B = 0. Along two of those segments (top and bottom horizontal segments)
inside the solenoids, B is perpendicular to d s , and B × d s = 0 . Thus,
ò B×d s = Bl
2
We again apply Ampere’s law
+ 0+ 0+ 0 = Bl2 .
ò B×d s = m I
0 enc
, yielding
Bl2 =m0 n2l2 I Þ B =m0 n2 I
Because the circulation direction for the segment between the solenoids is directed in the
k̂ -direction this implies that the vector description of the magnetic field is given by
B = m0 n2 I(k̂) ;
R1 < r < R2 .
In summary:
æ m0 I(n1 - n2 )(-k̂) ;
ç
B=ç
m0 n2 I (k̂) ;
ç
0;
è
0 < r < R1
R1 < r < R2
R2 < r
Problem 4 Magnetic Field of a Toroid
A toroid has N turns, and an inner radius a , outer radius b , and height h . The toroid
has a rectangular cross section shown in the figures below.
When a current I is flowing through the toroid, what are the magnitude and direction of
the magnetic field inside the toroid as a function of distance r from the axis of the
toroid?
Solution: One can think of a toroid as a solenoid wrapped around such that its ends are
connected. Thus, the magnetic field is completely confined inside the toroid and the field
points in the azimuthal direction (clockwise as seen from above) due to the way the
current flows. We identify three different regions: (i) r < a , (ii) a < r < b , and (iii) r > b.
We begin with (ii) a < r < b : We draw a circular Amperian loop of radius r as shown in
the figure below.
If we consider the circular disk of radius r then each turn of the toroid carrying a current
I cuts through the disk at r = a so I enc = NI . ò B×d s = B2p r . Applying Ampere’s law,
ò B×d s = m I
0 enc
, we obtain
B(2p r) = m0 NI Þ B =
m0 NI
.
2p r
Choose the unit vector to point tangential to the circle of radius r in the direction of
circulation (clockwise in the figure above). Therefore
(1)
.
(2)
Unlike the magnetic field of a solenoid, the magnetic field inside the toroid is nonuniform and decreases as 1 / r .
For (i) r < a , we again choose a circle of radius r , but now no current cuts through the
disk of radius r so I enc = 0 .
Therefore Ampere’s Law becomes
B(2p r) = 0 ,
(3)
.
(4)
therefore
For (iii) r > b, we again choose a circle of radius r , but now each turn cuts the plane
twice, once up and then down and the pair sum to zero. Thus extending this to all turns,
we have that I enc = 0 .
Therefore Ampere’s Law becomes
B(2p r) = 0 ,
(5)
.
(6)
therefore
Combining our results we have that
(7)
Problem 5: Magnet Moving Through a Coil of Wire Suppose a bar magnet is pulled
through a stationary conducting loop of wire at constant speed, as shown in the figure
below. Assume that the positive direction for flux is to the right. Assume that the north
pole of the magnet enters the loop of wire first, and that the center of the magnet is at the
center of the loop at time t = 0.
(a) Sketch qualitatively a graph of the magnetic flux F B through the loop as a
function of time.
(b) Sketch qualitatively a graph of the current I in the loop as a function of time.
Take the direction of positive current to be clockwise in the loop as viewed from
the left.
(c) What is the direction of the force on the permanent magnet due to the current in
the coil of wire just before the magnet enters the loop?
(d) What is the direction of the force on the magnet just after it has exited the loop?
(e) Do your answers in (c) and (d) agree with Lenz's law?
(f) Where does the energy come from that is dissipated in ohmic heating in the wire?
Solution:
a) Sketch qualitatively a graph of the magnetic flux F B through the loop as a function of
time.
When flux is positive to the left, the plot of flux vs. time looks like:
t
(b) Sketch qualitatively a graph of the current I in the loop as a function of time. Take
the direction of positive current to be clockwise in the loop as viewed from the left.
Solution: For the choice of sign in part (a), the plot of current vs. time looks like:
t
Remark: If you choose the opposite sign convention for flux, then your plots in (a) and
(b) are flipped.
(c) What is the direction of the force on the permanent magnet due to the current in the
coil of wire just before the magnet enters the loop?
Solution: As the magnet approaches the loop, the force on the magnet must be resistive
so the force on the magnet points to the left. Note that the induced current is directed
counterclockwise (as seen from the left hence I ind < 0 ) as the magnetic approaches the
loop, so the current acts as a magnetic dipole with the dipole moment pointing to the left.
We can model the current loop as a magnet, with the north pole facing to the left, hence
the permanent magnet and the dipole have north poles facing each other indicating that
the magnetic force is repulsive.
(d) What is the direction of the force on the magnet just after it has exited the loop?
Solution: The force is still resistive so it points to the left on the permanent magnet. The
induced current has changed direction and now flows clockwise (as seen from the left
hence I ind > 0). Thus dipole moment of the loop points to the right. Again, we model the
current loop as a magnet, with the north pole facing to the right, hence the permanent
magnet and the dipole have opposite poles facing each other indicating that the magnetic
force is now attractive.
(e) Do your answers in (c) and (d) agree with Lenz's law?
Solution: See solutions to part (c) and (d).
(f) Where does the energy come from that is dissipated in ohmic heating in the wire?
Solution: The magnet is pulled a constant speed so the energy is coming from the source
that is pulling the magnet.
Problem 6 A conducting rod with zero resistance and length w slides without friction on
two parallel perfectly conducting wires. Resistor 1 (with resistance R1 ) and resistor 2
(with resistance R1 ) are connected across the ends of the wires to form a circuit, as
shown. A constant and uniform magnetic field B is directed out of the page. In
computing magnetic flux through any surface, take the surface normal to be out of the
page, parallel to B .
a) What is the current flowing through the resistor R1 in the left hand loop of the
circuit shown? Give its magnitude and indicate its direction on the figure.
b) What is the magnitude and direction of the magnetic force exerted on this rod?
Solution:
a) What is the current flowing through the resistor R1 in the left hand loop of the
circuit shown? Give its magnitude and indicate its direction on the figure.
Choose coordinates such that the position of the rod as a function of time is given by
x(t) then
vx =
dx(t)
= -V .
dt
If we circulate counterclockwise around the left loop, then the flux through that loop is
positive F(t) = BA = Bx(t)w . The time derivative of the flux through the left loop is
then
dF(t)
dx(t)
= Bw
= Bwvx = -BwV .
dt
dt
Therefore the induced electromotive force (emf) is given by
e1 = -
dF(t)
= BwV .
dt
Because the emf is positive, that means the direction of the induced current is in the
same counterclockwise direction that we circulated around the loop. Alternatively, the
flux out of the plane of the figure is decreasing so by Lenz’s Law, the current is in the
counterclockwise direction to make an induced flux out the plane of the figure to oppose
the decrease in flux out of the plane of the figure. The magnitude of the current I1 in
the left loop is given by
I1 =
e1
R1
Therefore
I=
BwV
R1
b) What is the magnitude and direction of the magnetic force exerted on this rod?
In order to determine the magnetic force on the rod, F = IL ´ B , we must determine the
current through the rod. The current through the rod is the sum of the two currents
through the left loop and the right loop.
We repeat our argument above for determining the current in the right loop. If we
circulate counterclockwise around the right loop, then the flux through that loop is
positive F(t) = BA = B(d - x(t))w . The time derivative of the flux through the left loop
is then
dF(t)
dx(t)
= -Bw
= - Bwvx = BwV .
dt
dt
Therefore the induced electromotive force (emf) is given by
e2 = -
dF(t)
= - BwV .
dt
Because the emf is negative, that means the direction of the induced current is in the
opposite the counterclockwise direction that we circulated around the loop, so it is
clockwise. Alternatively, the flux out of the plane of the figure is increasing so by
Lenz’s Law, the current is in the clockwise direction to make an induced flux into the
plane of the figure to oppose the increase in flux out of the plane of the figure. The
magnitude of the current I 2 in the right loop is given by
I2 =
e2
R2
Therefore
I2 =
BwV
R2
Both currents are directed in the + ĵ -direction through the rod, so the current through the
rod is
æ 1
1ö
I = I1 + I 2 = BwV ç + ÷
è R1 R2 ø
The magnetic force on the rod is then
æ 1 1ö
æ 1 1ö
F = IL ´ B = BwV ç + ÷ (wĵ ´ Bk̂) = B2 w2V ç + ÷ î ,
è R1 R2 ø
è R1 R2 ø
the force is to the right. You could also get this directly from Lenz’s Law because the
induced currents always flow in such a direction as to oppose the motion. The
magnitude of the force is:
æ 1
1ö
F = B 2 w2V ç + ÷
è R1 R2 ø
Problem 7: An infinitely long wire of radius a carries a current density J 0 , which is
uniform and constant. The current is directed out of the plane of the figure below.
a)
Calculate the magnitude and direction of the magnetic field B(r) for (i) r < a , and
(ii) r > a . For both cases show your Amperian loop and indicate (with arrows) the
direction of the magnetic field.
b)
What happens to your answers above if the direction of the current is reversed so
that it flows "into" the page?
c)
Consider now the same wire but with a cylindrical hole bored throughout. The hole
has radius b (with b < a ) and is shown in the figure below. We have also indicated
four special points: O, M, N, and P. The point O is at the center of the original wire
and the point M is at the center of the hole. The point N is a distance a from the
center of the wire, lying on the line connecting the center of the wire and hole. The
point P is an arbitrary point inside the hole. In this new wire, the current density
points out of the plane of the figure and has magnitude equal to J 0 over the
remainder of the cross section of the wire (and of course is zero in the hole).
Calculate the magnitude and direction of the magnetic field at (i) the point M, and
(ii) the point N. Hint: Try to represent the current carrying wire as the superposition
of two types of current carrying wires.
d)
Now calculate the magnitude and direction of the magnetic field at the point P .
Does your answer depend on where the point P is located in the hole?
Solutions:
a) Calculate the magnitude and direction of the magnetic field B(r) for (i)
r £ a, and (ii) r ³ a . For both cases show your Amperian loop and indicate
(with arrows) the direction of the magnetic field.
Answer: (This is a repeat of the first in-class problem from W09D1.)
(i) We first consider the region r ³ a .
As always, the first step is to think about the problem a little. Using the right hand rule
the magnetic field will circle the wire counter-clockwise. We use the circular symmetry
to choose the Amperian loops, pictured above, and write down Ampere’s law:
ò B×d s = m I
0 enc
Note that along the loop the magnetic field is constant and parallel to ds, which makes the
integral easy to do, and so we have that
ò B × d s = 2p r
All of the current is enclosed so
I enc = I = J 0p a 2
Therefore Ampere’s Law becomes B2p r = m0 J 0p a 2 . Let’s choose to be a unit vector
tangent to the Amperian loop pointing in the direction of circulation (“counterclockwise”). The magnetic field outside the wire is then
(ii) We now consider the region r £ a . We choose our Amperian loop inside the wire as
shown in the figure below.
The current enclosed in our Amperian loop is now
I enc = òò J × d A = òò J dA =
r ¢ =r
ò
J 0 (2p r ¢ dr ¢ ) = J 0p r 2 .
r ¢=0
The line integral is the same as in part (i),
Therefore Ampere’s Law becomes
ò B × d s = 2p r
B2p r = m0 J0p r 2 .
The magnetic field inside the wire is then
.
In summary:
.
b)
What happens to your answers above if the direction of the current is reversed so
that it flows "into" the page?
Answer: If the direction of current flips then so does the direction of the magnetic field, so
it is clockwise rather than counterclockwise. The magnitude of the field remains the same.
c) Consider now the same wire but with a cylindrical hole bored throughout. The hole
has radius b (with b < a ) and is shown in the figure below. We have also indicated
four special points: O , M , N , and P . The point O is at the center of the original
wire and the point M is at the center of the hole. The point N is a distance a from
the center of the wire, lying on the line connecting the center of the wire and hole.
The point P is an arbitrary point inside the hole. In this new wire, the current density
points out of the plane of the figure and has magnitude equal to J 0 over the
remainder of the cross section of the wire (and of course is zero in the hole).
Calculate the magnitude and direction of the magnetic field at (i) the point M , and
(ii) the point N . Hint: Try to represent the current carrying wire as the superposition
of two types of current carrying wires.
Answer: We use the superposition principle. The wire with a cylindrical hole is equivalent
to two solid wires offset by the distance d with equal current densities J 0 but directed in
opposite directions as shown in the figure below.
The magnetic field B at any point is the superposition of the magnetic fields from these
two wires, B = B1 + B2 . We can now use the results of part a) to find the magnetic fields for
these individual wires.
For the point M :
B1 ( M ) =
m0 J 0 d
2
ĵ and B2 ( M ) = 0 . Therefore
B( M ) =
For the point N :
m0 J 0 d
2
ĵ .
B1 (N ) =
m0 J 0 a
2
ĵ and B 2 (N ) =
m0 J 0 b
2
(- ĵ). Therefore
B(N ) =
m0 J 0
2
(a - b) ĵ.
d) Now calculate the magnitude and direction of the magnetic field at the point P . Does your
answer depend on where the point P is located in the hole?
This is slightly more complicated due to the vector addition. Let r1 be the distance from O to
P , let
be a unit vector at P pointing in the counterclockwise direction. Then
.
Let r2 be the distance from M to P , let
direction. Then
be a unit vector at P pointing in the clockwise
.
Therefore
(1)
Now we need to use vector decomposition to express the unit vectors
and
in terms of î
and ĵ in order to add these two vectors. Let (x, y) denote the coordinates of the point P .
From the geometry shown in the figure below sinq1 = y / r1 and cosq1 = x / r1 .
Therefore
.
(2)
Similarly, from the geometry shown in the figure below sinq 2 = (d - x) / r2 and cosq 2 = y / r2 .
Therefore
(3)
Substituting Eqs. (2) and (3) into Eq. (1) yields
B(P) =
B(P) =
m0 J 0 r1 æ y
2
m0 J 0
2
x ö m0 J 0r2 æ y
d-x ö
ç - r î + r ĵ÷ + 2 ç r î + r ĵ÷
è 1
è 2
ø
1 ø
2
( - yî + xĵ) +
m0 J 0
2
( yî + (d - x) ĵ) =
m0 J 0 d
2
The magnetic field is uniform inside the hole with magnitude
B=
m0 J 0 d
2
,
and points in the + ĵ -direction.
Note that the current in the wire with the cylindrical hole is given by
I = J 0p (a 2 - b2 ) .
Therefore the magnitude of the magnetic field can be expressed as
.
ĵ
B=
m0 Id
2p (a 2 - b2 )
.
Problem 8: Rail Gun In his book “From the Earth to the Moon”, Jules Verne has a space
capsule shot out of a giant cannon as such a velocity that it flies all the way to the moon.
This is impossible. A cloud of gas from an explosion cannot expand much faster than the
speed of sound in the gas, and the speed of sound goes as the T 1/2 . In a chemical reaction
the temperature cannot get hotter than about 3000 K , and so the speed of sound is about 3
times higher than the speed of sound at room temperature, or about 1 km × s-1 . The escape
velocity from the earth is 11 km × s-1 . An alternative scheme is to use a rocket, which uses
that 1 km × s-1 relative to the rocket itself, and therefore is capable (in a Newtonian World)
of unlimited velocity. This works, of course, but it very expensive because the rocket must
accelerate not only itself, but also the fuel that it will use later on.
A third possibility is to use what is called a “rail gun” or “mass driver”. Imagine that you
have two conducting rails separated by a distance w . At one end you have a big
generator, which can maintain an emf e . A car of mass m rides the rails. The car
conducts current between the two rails and has resistance R . In order to make things
happen, we establish a uniform magnetic field B everywhere between the tracks and
pointing upward (out of the plane of the figure below).
a) What is the emf induced in the loop formed by the generator, rails, and the car
when the car is moving at speed v(t) ? What is the magnitude of the induced current
and in what direction does it point?
b) How much current I(t) will flow in the circuit when the car is moving at speed
v(t) ?
c) When the car is moving at speed v(t) , what is the force on the car?
d) Using Newton’s Second Law, find a differential equation involving v(t) and
dv(t) / dt . Set v(0) = 0 , and look for a solution of the form a + be-a t , where you
must find the constants a , b and a . What is the terminal speed vy,term that the car
will approach?
e) Suppose that e = 104 V , R = 10-3 W , w = 0.5 m , B = 1 T , and m = 102 kg . What
is the terminal speed?
f) When the car is traveling at terminal speed, how much power is the voltage source
providing and how much power is dissipated as joule heating? Does your answer
agree with your expectations? Briefly explain.
Solutions:
a) What is the emf induced in the loop formed by the generator, rails, and the car
when the car is moving at speed v(t) ? What is the magnitude of the induced
current and in what direction does it point?
Answer: Choose coordinates as shown in the figure below. We shall apply Faraday’s
Law to calculate the emf. Choose counterclockwise circulation direction for the closed
path formed by the voltage source, rails and car. Thus the unit normal for the planar
surface points out of the plane of the figure and is given by n̂ = k̂ .
At time t the bar has moved to a position y(t) . So the area of the loop is A = y(t)w. The
magnetic flux becomes
Fmag =
òò B × n̂da = òò Bk̂ × k̂da = òò Bda = BA = By(t)w
loop
loop
loop
The time derivative of the magnetic flux through the loop at time t is then
dFmag
dt
=B
dy(t)
w = Bv y w .
dt
So there is a non-zero electromotive force in the wire loop.
e ind = -
d
òò B × dA = -Bvyw
dt open
surface
The induced current is given by
Iind = eind / R = -Bvyw / R .
Because I ind < 0, it points in the clockwise direction opposite our choice of circulation
direction. We can check this by Lenz’s Law. As the bar moves to the left. The magnetic
flux is positive and is increasing. Hence an induced current must flow clockwise to
generate induced flux to oppose this change.
b) How much current I(t) will flow in the circuit when the car is moving at speed v(t) ?
Answer: The voltage source provides a current I1 = e / R in the counterclockwise
direction. From part a), there is an induced current I ind = -Bvy w / R . So the current in the
circuit is
1
I = I1 + I ind = e / R - Bvy w / R = (e - Bvy w)
R
c)
When the car is moving at speed v(t) , what is the force on the car?
Answer: We can use the magnetic force law to find the force on the car
F = Iw ´ B =
1
1
(e - Bvy w)w(- î) ´ Bk̂ = (e - Bvy w)wBĵ
R
R
d)
From the differential equation you found for v(t) from part c). Set v(0) = 0 , and
you should look for a solution of the form a + be-a t , where you must find the constants
a , b and a . What is the terminal speed vy,term that the car will approach?
Answer: We can now use Newton’s Second Law (in the y-direction) to find the
differential equation satisfied by vy
dvy
1
(e - Bvy w)wB = m
.
R
dt
Assume a solution of the form vy (t) = a + be-a t . Then
dvy
dt
= -a be-a t .
The differential equation becomes
1
(e - B(a + be-a t )w)wB = -a mbe-a t .
R
Expanding this we have that
1
1
(e - awB)wB - w 2 B 2be-a t = -a mbe-a t
R
R
In order for vy (t) = a + be-a t to be a solution, we require that
1
(e - awB) = 0
R
-
1 2 2 -a t
w B be = -a mbe-a t
R
Thus
a=
e
wB
,
and
a=
1 2 2
w B.
mR
From our initial condition that the car starts from rest we have that
0 = vy (0) = a + b .
Thus
b = -a
and so our solution for the speed of the car is given by
vy (t) = a(1 - e
-a t
)=
e æ
ç1 - e
wB è
In the limit that t ® ¥ ,
vy,term = lim vy (t) =
t ®¥
e)
e
wB
-
w2 B2
t
mR
ö
÷.
ø
.
Suppose that e = 104 V , R = 10-3 W , w = 0.5 m , B = 1 T , and m = 102 kg .
What is the terminal speed?
Answer:
vy,term
e
(10 4 V)
=
=
= 20 km × s-1
wB (0.5 m)(1 T)
f)
When the car is traveling at terminal speed, how much the power is the voltage
source providing and how much power is dissipated as joule heating? Does your answer
agree with your expectations?
Answer: At terminal speed, the current is zero, because the induced emf exactly cancels
the voltage source. Therefore the voltage source is no longer providing power. Because
there is no current there is also no joule heating.