PH317 MAGNETISM
WINTER Quarter 2009-2010
Solution to Homework Assignment 1
Assigned: Monday 11/30/09
Due: Monday 12/07/09
A point charge q1=-4.00 mC travels along a prescribed path in
three-dimensional space. When it passes through the point P1(3.00 m,4.00 m,5.00 m) its instantaneous velocity is
v1=
æ è
3.00
^
x
+
^
y
+6.00
^
z
ö ø
105 m/s
(a) Calculate the magnitude and direction of the magnetic field
at the point P2(5.00 m,12.0 m,13.0 m).
SOLUTION
The magnetic field equation is
Bat 2
=
moq1
4p
v1× r12
r312
where,
r12
=
r2-r1
=
2
^
x
+8
^
y
+8
^
z
r12
=
2
Ö
33
Substitution gives.
Bat 2 =
æ è
1.055
^
x
+0.3165
^
y
-0.5803
^
z
ö ø
10-9 T
(b) At the exact instant charge q1 passes through point P1,
another point charge q2=+6.60 mC passes through point
P2 with a velocity
v2 =
æ è
-6.00
^
x
+3.00
^
y
+5.00
^
z
ö ø
105 m/s
Calculate the instantaneous magnetic force charge q1 exerts on charge
q2, Fq1,q2. Ignore the electric force.
SOLUTION
Fq1,q2
=
q2v2×Bat 2
=
æ è
-2.193
^
x
+1.184
^
y
-3.342
^
z
ö ø
10-8 N
(c) Reverse the situation and calculate the instantaneous magnetic
force which charge q2 exerts on charge q1,
Fq2,q1. Again, ignore the electric force.
SOLUTION
The results are:
Bat 1
=
moq2
4p
v2× (-r12)
r312
=
æ è
1.741
^
x
+0.5222
^
y
-0.9574
^
z
ö ø
10-9 T
Fq2,q1
=
q1v1×Bat 1
=
æ è
1.636
^
x
-5.327
^
y
-0.06963
^
z
ö ø
10-9 N
(d) Does Newton's third law hold?
ANS: NO. The magnitudes of the two forces are different and the direc"nowrap" align="left">
5.573×10-9 N
Fel
=
kq1q2
r212
=
1.798×10-3 N
Fel/Fq1,q2
=
4.31×105
Fel/Fq2,q1
=
3.23×105
Helix Problem: The purpose is to determine the position vector of
a point charge which is traveling at an angle to the direction of a uniform
magnetic field. The values for the field components and the initial conditions
(assume at t=0) are:
Ex=0
Ey=0
Ez=0
Bx=0
By=0
Bz=0.800 T
x(0)=0
y(0)=0
z(0)=0
vox=0
voy=4.33×105 m/s
voz=2.50×105 m/s
The values of the charge and it's mass are
q=+4.00 mC
m=6.50×10-12 kg
(a) Determine the instantaneous position coordinates
x(t), y(t), z(t) of the charge by solving the differential equations given
in class and applying the initial conditions. You should not have to do much
more than I did in class.
SOLUTION
Use the results I obtained in class. The only change occurs in the
z equation.
x(t)
=
R(1-coswt)
y(t)
=
Rsinwt
z(t)
=
vozt
(b) Calculate the radius of the imaginary cylinder containing the
helical path. This was set up for you in class.
SOLUTION
R
=
mvoy
qB
=
(6.50×10-12 kg)((4.33×105 m/s)
(4.00×10-6 C)(0.800 T)
=
0.880 m
(c) Calculate the period, T, (one revolution around the cylinder).
SOLUTION
T
=
2pR
voy
=
2p(0.880 m)
4.33×105 m/s
=
1.28×10-5 s
(d) Make a Maple plot over a time interval
0 £ t £ 3T
DON'T use the unit method which I used to get my plots. USE the ACTUAL values
given. Since only three periods are being graphed, the scale numbers will be
reasonable. Take a good look at the graph and make sure all initial conditions
are obeyed.
SOLUTION
See the Maple file helicalpath.mws in the homework folder.
(e) Extra: NOT ASKED FOR. The cyclotron frequency and pitch are
wc
=
2p
T
=
2p
1.28×10-5 s
=
4.92×105 rad/s
pitch
=
vozT
=
(2.50×105 m/s)(1.28 ×10-5 s)
=
3.19 m
5.4 Text
SOLUTION
Do the top and bottom wires first.
B=(+ka/2)
^
x
I=(-I)
^
y
Since the magnetic field is constant along the length of the straight wire
F=(I×B)L
Ftop
=
é ë
(-I)
^
y
ù û
×
é ë
(+ka/2)
^
x
ù û
a
=
(+kIa2/2)
^
z
B=(-ka/2)
^
x
I=(+I)
^
y
Since the magnetic field is constant along the length of the straight wire
F=(I×B)L
Fbottom
=
é ë
(+I)
^
y
ù û
×
é ë
(-ka/2)
^
x
ù û
a
=
(+kIa2/2)
^
z
B=(kz)
^
x
I=(-I)
^
y
0 £ z £ a/2
The magnetic field varies over the length of the wire.
dFLT
=
é ë
(-I)
^
z
ù û
×
é ë
(kz)
^
x
ù û
dz
FLT
=
-kI
^
y
ó õ
a/2
0
zdz
=
-(kIa2/8)
^
y
B=(kz)
^
x
I=(-I)
^
y
0 £ z £ a/2
dFLT
=
é ë
(-I)
^
z
ù û
×
é ë
(kz)
^
x
ù û
dz
FLT
=
-kI
^
y
ó õ
0
-a/2
zdz
=
+(kIa2/8)
^
y
B=(kz)
^
x
I=(+I)
^
y
0 £ z £ a/2
dFLT
=
é ë
(+I)
^
z
ù û
×
é ë
(kz)
^
x
ù û
dz
FLT
=
+kI
^
y
ó õ
a/2
0
zdz
=
+(kIa2/8)
^
y
B=(kz)
^
x
I=(+I)
^
y
0 £ z £ a/2
dFLT
=
é ë
(+I)
^
z
ù û
×
é ë
(kz)
^
x
ù û
dz
FLT
=
+kI
^
y
ó õ
0
-a/2
zdz
=
-(kIa2/8)
^
y
The totaL force on the loop is
Ftot
=
Ftop+Fbottom+FLT +FLB+FRT+FRB
=
(kIa2/2+kIa2/2)
^
z
+( -kIa2/8+kIa2/8 +kIa2/8-kIa2/8)
^
y
=
(kIa2)
^
z
5.6 Text
SOLUTION
(a) Using the equation for K,
K
=
sv
=
sw×r
=
swr
^
q
(b) Using the equation for J,
J
=
rv
=
rw×s
=
rws
^
j
The length s is the perpendicular distance from the axis of rotation to
any point in the sphere. Its relation to the distance of the point from the
center of the disk (r) is
s=rsinq
Also, r = 3Q/4pR3. Therefore,
J
=
rw(rsinq)
^
j
=
3Qwrsinq
4pR3
^
j
File translated from
TEX
by
TTH,
version 3.85. On 11 Dec 2009, 16:38.