Download A 10.0 cm length of wire carries a current of 4.0 A in the positive z

HW11 Solutions (due Tues, Apr 14)
1. T&M 26.P.024
A 10.0 cm length of wire carries a current of 4.0 A in the positive z
direction. The force on this wire due to a magnetic field B is = (-0.2
+ 0.2 ) j N. If this wire is rotated so that the current flows in the
positive x direction, the force on the wire is F = 0.2 k N. Find the
magnetic field vector.
Solution:
2. T&M 26.P.028
A 4.5-keV electron (an electron that has a kinetic energy equal
to 4.5 keV) moves in a circular orbit that is perpendicular to a
magnetic field of 0.325 T. (a) Find the radius of the orbit. Find
the (b) frequency and (c) period of the orbital motion.
Solution:
Picture the Problem (a) We can apply Newton’s 2nd law to the orbiting electron
to obtain an expression for the radius of its orbit as a function of its mass m,
charge q, speed v, and the magnitude of the magnetic field B. Using the definition
of kinetic energy will allow us to express r in terms of m, q, B, and the electron’s
kinetic energy K. (b) The period of the orbital motion is given by T = 2πr/v.
Substituting for r (or r/v) from Part (a) will eliminate the orbital speed of the
electron and leave us with an expression for T that depends only on m, q, and B.
The frequency of the orbital motion is the reciprocal of the period of the orbital
motion.
(a) Apply Newton’s 2nd law to the
orbiting electron to obtain:
mv
v2
qvB = m ⇒ r =
qB
r
Express the kinetic energy of the
electron:
K = 12 mv 2 ⇒ v =
Substituting for v in the expression
for r and simplifying yields:
r=
m 2K
=
qB m
2K
m
2 Km
qB
Substitute numerical values and evaluate r:
' 1.602 !10 (19 J $
""
2(4.5 keV ) 9.109 !10 (31 kg %%
eV
&
#
= 0.696 mm = 0.70 mm
(19
1.602 !10 C (0.325 T )
(
r=
(
)
)
(b, c) Relate the period of the
electron’s motion to the radius of its
orbit and its orbital speed:
Because r =
mv
:
qB
Substitute numerical values and
evaluate T:
T=
2!r
v
2!
T=
T=
mv
qB 2! m
=
v
qB
(
)
2# 9.109 ! 10"31 kg
1.602 ! 10"19 C (0.325 T )
(
)
= 1.099 ! 10"10 s = 0.11 ns
The frequency of the motion is known
as the cyclotron frequency and is the
reciprocal of the period of the
electron’s motion:
f =
1
1
=
= 9.1 GHz
T 0.110 ns
3. T&M 26.P.035
A velocity selector has a magnetic field that has a magnitude
equal to 0.28 T and is perpendicular to an electric field that has
a magnitude equal to 0.46 MV/m. (a) What must the speed of a
particle be for that particle to pass through the velocity selector
undeflected? What kinetic energy must (b) protons and (c)
electrons have in order to pass through the velocity selector
undeflected?
Solution:
Picture the Problem Suppose that, for positively charged particles, their
motion is from left to right through the velocity selector and the electric field
is upward. Then the magnetic force must be downward and the magnetic
field out of the page. We can apply the condition for translational
equilibrium to relate v to E and B. In (b) and (c) we can use the definition of
kinetic energy to find the energies of protons and electrons that pass through
the velocity selector undeflected.
(a) Apply
!F
y
= 0 to the
particle to obtain:
Felec ! Fmag = 0
or
qE ! qvB = 0 ⇒ v =
Substitute numerical values
and evaluate v:
v=
E
B
0.46 MV/m
= 1.64 ! 106 m/s
0.28 T
= 1.6 ! 106 m/s
(b) The kinetic energy of
protons passing through the
velocity selector undeflected is:
K p = 12 mpv 2
(
)(
= 12 1.673 ! 10" 27 kg 1.64 ! 106 m/s
= 2.26 ! 10"15 J !
= 14 keV
1 eV
1.602 ! 10"19 J
)
2
(c) The kinetic energy of electrons
passing through the velocity
selector undeflected is:
K e = 12 me v 2
=
1
2
(9.109 !10
" 31
= 1.23 !10 "18 J !
kg
)(1.64 !10
6
m/s
)
2
1eV
1.602 !10 "19 J
= 7.7 eV
4. T&M 26.P.042
Before entering a mass spectrometer, ions pass through a
velocity selector consisting of parallel plates that are separated
by 2.0 mm and have a potential difference of 160 V. The
magnetic field strength is 0.42 T in the region between the
plates. The magnetic field strength in the mass spectrometer is
1.2 T. Find (a) the speed of the ions entering the mass
spectrometer and (b) the difference in the diameters of the
238
235
235
orbits of singly ionized U and U. The mass of a U ion is
–25
3.903 × 10 kg.
Solution:
Picture the Problem We can apply a condition for equilibrium to ions passing
through the velocity selector to obtain an expression relating E, B, and v that we
can solve for v. We can, in turn, express E in terms of the potential difference V
between the plates of the selector and their separation d. In (b) we can apply
Newton’s 2nd law to an ion in the bending field of the spectrometer to relate its
diameter to its mass, charge, velocity, and the magnetic field.
(a) Apply
!F
y
= 0 to the ions in
Felec ! Fmag = 0
the crossed fields of the velocity
selector to obtain:
or
Express the electric field between the
plates of the velocity selector in
terms of their separation and the
potential difference across them:
E=
qE ! qvB = 0 ⇒ v =
V
d
E
B
Substituting for E yields:
v=
V
dB
Substitute numerical values and
evaluate v:
v=
160 V
= 1.905 ! 105 m/s
(2.0 mm)(0.42 T )
= 1.9 ! 105 m/s
(b) Express the difference in the
diameters of the orbits of singly
ionized 238U and 235U:
Apply
!F
radial
= mac to an ion in
the spectrometer’s magnetic
field:
Express the diameter of the orbit:
The diameters of the orbits for 238U
and 235U are:
Substitute in equation (1) to obtain:
"d = d 238 ! d 235
qvB = m
d=
mv
v2
⇒r =
qB
r
2mv
qB
d 238 =
2m238v
2m235v
and d 235 =
qB
qB
2m238v 2m235v
!
qB
qB
2v
(m238 ! m235 )
=
qB
"d =
Substitute numerical values and evaluate Δd:
' 1.6606 ! 10 (27 kg $
""
2 1.905 ! 105 m/s (238 u ( 235 u )%%
u
&
# = 1cm
Äd =
1.602 ! 10 (19 C (1.2 T )
(
)
(
)
(1)
5. T&M 26.P.071
An alpha particle (charge +2e) travels in a circular path of radius
0.50 m in a magnetic field of 0.10 T. Take m = 6.65 10-27 kg for
the mass of the alpha particle.
(a) Find the period of the alpha particle.
(b) Find its speed.
(c) Find its kinetic energy.
Solution:
6. T&M 26.P.080
A circular loop of wire that has a mass m and a constant
current I is in a region with a uniform magnetic field. It is initially
in equilibrium and its magnetic moment is aligned with the
magnetic field. The loop is given a small angular displacement
about an axis through its center and perpendicular to the
magnetic field and then released. What is the period of the
subsequent motion? (Assume that the only torque exerted on
the loop is due to the magnetic field and that there are no other
forces acting on the loop.
Solution:
7. T&M 26.P.078
A circular loop of wire that has a mass m and a constant current
I is in a region with a uniform magnetic field. It is initially in
equilibrium and its magnetic moment is aligned with the
magnetic field. The loop is given a small angular displacement
about an axis through its center and perpendicular to the
magnetic field and then released. What is the period of the
subsequent motion? (Assume that the only torque exerted on the
loop is due to the magnetic field and that there are no other
forces acting on the loop. Use pi for π, m, I, and B as necessary.)
Solution:
8. T&M 27.P.019
A small current element at the origin has a length of 2.0 mm and
carries a current of 2.0 A in the +z direction. Find the magnitude and
direction of the magnetic field due to the current element at the point
(0, 3.0 m, 4.0 m).
Solution:
"
"
Picture the Problem We can substitute for I and d ! " ! ! in the Biot-Savart law
"
" µ0 Id ! ! rˆ
( dB =
), evaluate r and r̂ for (0, 3.0 m, 4.0 m), and substitute to
4" r 2
!
find dB.
The Biot-Savart law for the given
current element is given by:
"
" µ 0 Id ! ! rˆ
dB =
4" r 2
Substituting numerical values yields:
!
(2.0 A )(2.0 mm)kˆ ! rˆ = 0.400 nT " m 2 kˆ ! rˆ
dB = 1.0 ! 10 #7 N/A 2
r2
r2
(
)
Find r and r̂ for the point whose
coordinates are (0, 3.0 m, 4.0 m):
!
Evaluate dB at (0, 3.0 m, 4.0 m):
(
)
!
r = (3.0 m )ˆj + (4.0 m )kˆ ,
3
4
r = 5.0 m , and rˆ = ˆj + kˆ
5
5
!
dB (0, 3.0 m, 4.0 m ) = 0.400 nT ) m 2
(
4 $
'3
kˆ ( % ˆj + kˆ "
5 #
&5
= ! (9.6 pT )iˆ
(5.0 m )2
)
9. T&M 27.P.076
Find the magnetic field at point P in the figure below, where I = 15
A and R = 20 cm.
Solution:
Picture the Problem Because point P is on the line connecting the straight
segments of the conductor, these segments do not contribute to the magnetic field
at P. Hence, we can use the expression for the magnetic field at the center of a
current loop to find BP.
Express the magnetic field at the
center of a current loop:
µ0 I
2R
where R is the radius of the loop.
Express the magnetic field at the
center of half a current loop:
B=
Substitute numerical values and
evaluate B:
( 4! " 10
B=
B=
1 µ0 I µ0 I
=
2 2R
4R
#7
)
N/A 2 (15 A )
4 ( 0.20 m )
= 24 µ T into the page
10. T&M 27.P.081
A long, straight wire carries a current of I1 = 20 A as shown in
the figure. A rectangular coil with two sides parallel to the
straight wire has sides 5 cm and 10 cm with the near side a
distance 2 cm from the wire. The coil carries a current of I2 = 5
A.
Solution:
!
Picture the Problem Let I1 and I2 represent the currents of 20 A and 5.0 A, Ftop ,
!
!
!
!
Fleft side , Fbottom , and Fright side the forces that act on the horizontal wire, and B1 ,
! !
!
B2 , B3 , and B4 the magnetic fields at these wire segments due to I1. We’ll need
!
!
to take into account the fact that B1 and B3 are not constant over the segments 1
and 3 of the rectangular coil. Let the +x direction be to the right and the +y
direction be upward. Then the +z direction is toward you (i.e., out of the page).
!
! !
!
Note that only the components of B1 , B2 , B3 , and B4 into or out of the page
contribute to the forces acting on the rectangular coil. The +x and +y directions
are up the page and to the right.
!
(a) Express the force dF1 acting on a
"
current element I 2 d ! in the top
segment of wire:
( )
"
Because I 2 d ! = I 2 d! ! iˆ in this
segment of the coil and the
magnetic field due to I1 is given by
"
µ I
B1 = 0 1 ! kˆ :
2" !
( )
!
! !
dFtop = I 2 d " ! B1
( )
( )
"
µ I
Ftop = I 2 d! " iˆ # 0 1 " kˆ
2! !
µ I I d! ˆ
=" 0 1 2
j
2! !
!
Integrate dFtop to obtain:
"
µ II
Ftop = ' 0 1 2
2)
='
7.0 cm
d! ˆ
j
!
2.0 cm
(
µ 0 I 1 I 2 & 7.0 cm # ˆ
ln$
!j
2)
% 2.0 cm "
!
Substitute numerical values and evaluate Ftop :
!
Ftop
(
!7 N %
& 4) " 10
# (20 A )(5.0 A )
A2 $
( 7.0 cm % ˆ
'
!5
=!
ln&
# j = ! 2.5 " 10 N ˆj
2)
2.0
cm
'
$
(
!
Express the force dFbottom acting on
"
a current element I 2 d ! in the
horizontal segment of wire at the
bottom of the coil:
()
"
Because I 2 d ! = I 2 d! iˆ in this
segment of the coil and the
magnetic field due to I1 is given by
" µ I
B1 = 0 1 ! kˆ :
2" !
( )
!
Integrate dFbottom to obtain:
)
!
! !
dFbottom = I 2 d " ! B3
( )
"
µ I
dFbottom = I 2 d!iˆ # 0 1 " kˆ
2!!
µ I I d! ˆ
= 0 1 2
j
2! !
"
µ II
dFbottom = 0 1 2
2(
=
7.0 cm
d! ˆ
j
!
2.0 cm
'
µ 0 I 1 I 2 & 7.0 cm # ˆ
ln$
!j
2(
% 2.0 cm "
!
Substitute numerical values and evaluate Fbottom :
(
!7 N %
& 4) " 10
# (20 A )(5.0 A )
!
A2 $
( 7.0 cm % ˆ
'
Fbottom =
ln&
#j =
2)
' 2.0 cm $
!
Express the forces Fleft side and
!
!
Fright side in terms of I2 and B2 and
!
B4 :
(2.5 " 10
!
!
!
Fleft side = I 2 " 2 ! B2
and
!
!
!
Fright side = I 2 " 4 ! B4
!5
)
N ˆj
!
!
Express B2 and B4 :
!
!
µ 2I
µ 2I
B2 = " 0 1 kˆ and B4 = " 0 1 kˆ
4! R1
4! R4
!
!
Substitute for B2 and B4 to obtain:
"
' µ 2I
Fleft side = ( I 2 ! 2 ˆj ) %% ( 0 1
& 4! R1
$ µ ! I I
kˆ "" = 0 2 1 2 iˆ
2!R2
#
and
"
( µ 2I
Fright side = I 2 ! 4 ˆj ) && " 0 1
' 4! R4
%
µ ! II
kˆ ## = " 0 4 1 2 iˆ
2!R4
$
!
!
Substitute numerical values and evaluate Fleft side and Fright side :
(
)
!
4# " 10 !7 N/A 2 (0.100 m )(20.0 A )(5.00 A ) ˆ
Fleft side =
i = 1.0 " 10 ! 4 N iˆ
2# (0.0200 m )
(
and
!
4# " 10 !7 N/A 2 (0.100 m )(20.0 A )(5.00 A ) ˆ
Fright side = !
i=
2# (0.0700 m )
(
)
(b) Express the net force acting on
the coil:
(! 0.29 " 10
)
!4
)
N iˆ
!
!
!
!
!
Fnet = Ftop + Fleft side + Fbottom + Fright side
!
!
!
!
Substitute for Ftop , Fleft side , Fbottom , and Fright side and simplify to obtain:
!
Fnet = ! 2.5 " 10 !5 N ˆj + 1.0 " 10 !4 N iˆ + 2.5 " 10 !5 N j + ! 0.29 " 10 !4 N iˆ
(
) (
= (0.71" 10 N )iˆ
!4
) (
) (
)