Physics 114 Exam 3 Spring 2015
Name: ___________________________________________________________
For grading purposes (do not write here):
Question
Problem
1.
1.
2.
2.
3.
3.
4.
Answer each of the following questions. Points for each question are indicated in red. Unless
otherwise indicated, the amount being spread among parts (a,b,c etc) are equal. Be sure to show all
your work.
Use the back of the pages if necessary.
Question 1. (10 points) Find the direction of the minimum magnetic field acting on the positively charged
particle moving in the various situations shown in the figure below if the direction of the magnetic force acting
on it is as indicated. By minimum magnetic field we mean the smallest one capable of producing the force
shown.
Solution.
. Since the particle is positively charged, use the right hand rule. In this case, start with the
fingers of the right hand in the direction of and the thumb pointing in the direction of . As you start closing
the hand, the fingers point in the direction of after they have moved
(the minimum field to provide a
given force would be perpendicular to the velocity). The results are:
(a) into the page; (b) toward the right; (c) toward the bottom of the page.
Question 2. (10 points) (a) In the figure below find the direction of the current in the wire that would produce
a magnetic field directed as shown. State its direction (to the right, left, top of the page, bottom of the page, out
of the page, into the page)
Solution: Imagine grasping the conductor with the right hand so the fingers
curl around the conductor in the direction of the magnetic field. The thumb
then points along the conductor in the direction of the current. The results
are:
(a) toward the left;
(b) In the figure below find the direction of the current in the wire that would produce a magnetic field directed
as shown. State its direction (to the right, left, top of the page, bottom of the page, out of the page, into the
page)
(b) out of the page;
(c) In the figure below find the direction of the current in the wire that would produce a magnetic field directed
as shown. State its direction (towards the upper right, lower left, top of the page, bottom of the page, out of the
page, into the page, towards the upper left, lower right).
(c) lower left to upper right.
Question 3. (10 points) (a) A rectangular conducting loop is placed near a long wire carrying a current I
as shown in the figure below. If I decreases in time, what can be said of the
current induced in the loop (is there one, if so is clockwise or counter
clockwise as viewed from above)?
Solution. The field through the loop is into the page and it decreases in time
due to the decreasing current. Thus, the flux decreases into the page and the
counter flux needs to be into the page and so the current is clockwise.
(b) A circuit consists of a conducting movable bar and a lightbulb connected to two conducting rails as shown
in the figure below. An external magnetic field is directed perpendicular to the plane of the circuit. Answer the
following (True or False)
i) When the bar is moved to the left, the bulb lights up
ii) When the bar is moved to the right, the bulb lights
up
iii) When the magnitude of the magnetic field increases
the light bulb lights up
iv) When the magnitude of the magnetic field decreases
the light bulb lights up
v) An applied force is required to maintain the motion
of the bar to the right one it is given some initial
velocity.
Solution. These are all true.
(c) Two rectangular loops of wire lie in the same plane as shown in the figure below. If the current I in the outer
loop is counterclockwise and increases with time, is there an induced
current in the inner loop and (if so) what is its direction (clockwise or
counter clockwise). Does the magnitude of the induced field (if there is
one) depend on the dimensions of the loops?
Solution: There is an induced current. The field due to the outer loop
within the inner loop is increasing out of the page so the flux is
increasing out of the page through the inner loop. The induced current
will be into the page, which means the current must be clockwise. The
dimensions do matter. The flux depends on the area of the inner loop and on how close the outer loop is to the
inner one.
Problem 1. A particle with a charge q = 3 C moves through a region where there is a magnetic field

ˆ .
B = (3iˆ + k)T
(a) (6 points) Determine the force on the charge when it has a velocity of

ˆ m/s
v = (3j)

 
F

qv
xB
Solution:
ˆ X (3iˆ + k)
ˆ ]= 3(-9kˆ  3i)
ˆ
=3[v(3j)
ˆ
ˆ
 (9i-27
k)
(b) (3 points) What is the angle between this force (calculated in part a) and the magnetic field?
Solution: The angle is always 90 degrees.
(c) (6 points) Now say that the velocity of the particle is (somehow) changed so that it is moving
with a speed (magnitude of the velocity) of 5 m/s. It now experiences a force given by

F = (27 ˆj) N .
Determine the y-component of the velocity of the charge when it experiences
this force (that is, find vy, - if this component is undetermined state so).
Solution:

ˆ  (3iˆ + k)
ˆ
27 ˆj = 3 ( v x ˆi + v y ˆj  vz k)
ˆ v ˆi  3v ˆj  0)
 3(0  v x ˆj-3v y k+
y
z

For both sides to be equal, vy must be zero.
(d) (3 points extra credit) Find the other two components (that is, find vx, and vz - if one component
is undetermined state which).
27 ˆj = 3( vx ˆj+3vz ˆj)
 9  vx  3vz
. vx = 3 vz -9
Since the speed is 5 m/s, we have 25 = vx2 + vz2.
25 = 9 vz2 + 81-54vz+ vz2
25 = 10 vz2 + 81-54vz
0 = 10 vz2 + 56- 54vz
=(vz - 4)(10vz -14). vz = 4 or vz = 1.4; vx = 3 or vx = -4.8 m/s
Problem 2. (15 points) A conductor consists of a circular loop of radius R= 0.2m and two straight, long
sections as shown below. The wire lies in the plane of the paper and carries a current I = 5A.
R
a. What is the magnitude and direction of the magnetic field due to the long wire at the center of
the circular loop
b. What is the magnitude and direction of magnetic field due to current in the loop at the center of
the loop?
c. What is the magnitude and direction of the total magnetic field at the center of the loop?
Solution
(a) The magnetic field due to the long straight wire has magnitude
= 5 A and R = 0.2 m we have B = 5T.
(b) For the circular loop is obtained from the Biot Savart law. B 
and is directed into the page. With I
0 I ds  rˆ
. Here, ds is perpendicular to r̂
4  r 2
0 I
2
so the cross product is just ds and since r is constant we get that integral is just 2r/r = 2 /r. Thus B is
15.7 T and is into the page.
(c)
= 20.7 T (directed into the page)
2R
=
Problem 3. (15 points) Consider the long cylindrical shell for which a cross section is shown below.
X
X
X
X
0.03 m
X
X
X
X
There is a uniform current in the shell (blue) that is going into the paper as indicated by the yellow X.
The magnitude of the current is 2 A. The inner radius of the conducting shell is 0.03 m and the
outer radius of the outer conductor is 0.035 m. Use Ampere’s law to find the magnetic field a
distance r = (a) 0.05 m, and (b) 0.02 m.
Solution
(a) Generally, if we apply Ampere’s law ( 
 B  ds  0 I ) at a point at radius ra, we have the integral will
give B*2ra and
, where
is the net current through the area of the circle of radius
. In
this case, Ia = 2A into the page (the current in the shell), r = 0.05 m so
(4  107 T  m/A)(2A)
B
= 8 T
2 (0.05m)
(b) From Ampere's law, the magnetic field is
 B  ds   I , where I refers to the current enclosed.
0
If we
draw an Amperian loop with radius 0.02 m, then the total current enclosed is zero and so is the magnetic
field.
Problem 4. (15 points) Consider A conducting bar of length moves to the right on two frictionless rails as
shown in the figure below. A uniform magnetic field directed into the page has a magnitude of 0.30 T. Assume
R = 9.00 Ω and = 0.350 m.
(a) When the bar is 0.1 m from the resistor, what is the flux through the
square loop enclosed by the circuit comprised of the resistor, rails, and
moving bar?
(b) At what constant speed should the bar move to produce an 8.50-mA
current in the resistor?
(c) What is the direction of the induced current?
(d) What is ∮ ∙
around square loop enclosed by the circuit
comprised of the resistor, rails, and moving bar?
Solution
 
 B   B. dA
(a)
, and here we just have  = B*A = (0.3)(0.35)(0.1) = 0.011 Wb.
(b) The motional emf induced in the bar must be
where I is the current in this series circuit. The induced
emf is given by the change in flux which will be BdA/dt = Bldw/dt, so
, the speed of the moving bar
must be
(c) The flux through the closed loop formed by the rails, the bar, and the resistor is directed into the page and is
increasing in magnitude. To oppose this change in flux, the current must flow in a manner so as to produce flux
out of the page through the area enclosed by the loop. This means the current will flow counterclockwise.
(d) ∮ ∙
, so it is just Blv = (0.3)(0.35)(0.729) = 0.0765 V.
Possibly Useful Information
1 | q 1 || q 2 |
4  0 r 2
e = 1.6 X 10-19 C
 0  885
. X 10-12 (C2 / N  m2 )
| q|
E
, E = /
4 0 r 2
x = x2 - x1, t = t2 - t1
 
 0    0  E. dA  q enc
s = (total distance) / t
v = dx/dt
a = v / t
a = dv/dt = d2x/dt2
v = vo + at
g = 9.8 m/s2

r = xi + yj + zk
  
r = r2 - r1
F
x-xo = vot + (½)at2
v2 = vo2 + 2a(x-xo)
 
E  Fq
0
v = x / t
x-xo = vt -1/2at2


a = dv / dt

r = (x 2 - x 1 ) i + (y 2 - y 1 ) j + (z 2 - z 1 ) k




v = r / t , v = dr / dt


a = v / t
U = Uf - Ui = -W
V = Vf - Vi = -W/q0 = U/q0
f
 
Vf  Vi    E. ds
U=-W∞
V = -W∞/q0
f
 
V    E. ds
x-xo = ½( vo+ v)t
i
V
1 q
4  0 r
Uf + Kf = Ui +Ki
K = ½ mv2
V
Es 
s
V
E
s
Q = CV
l
ln(b / a)
C  4  0 R
C  2 0
1
1

(series)
C eq
Cj
u  12  0 E 2
I= dQ/dt
i
n
1
4 0
i 1
1
dq
V

4  0 r
V   Vi 
n
q
 ri
i 1
i
V
V
V
; Ey 
;Ez 
x
y
z
1 q1q2
U  W 
4 0 r12
 A
C 0
d
ab
C  4  0
ba
Ex 
C eq   C j (parallel)
Q2 1
 2 CV 2
2C
C =  C0
U
  1
R
L
P = IV
A
Pemf = I
1
1

(parallel)
R eq
Rj
I = (R)e-t/RC
I = (QRC)e-t/RC, I0 = (QRC)
E = o 

F  qv x B
r = mv/qB, = qB/m

 
dF  Ids x B
V = IR
P = I2R=V2/R
I 
(R  r)
R eq   R j (series)
q(t)= Q(1-e-t/RC)
q(t) = Qet/RC
 = Q/L, = Q/A,  = Q/V

 
F  IL xB

  xB



  NIA
B = I/(2 r)
F/l = (I1I2)/2 a
 
d B
   E. ds  N
dt
=

U   B
 
 
Ids x r
dB  0
4
r3
 = 4 x 10-7 T.m/A
B = nI (solenoid)
 
B
 .ds   0 I enc
= Blv
 
 B   B. dA
 
B  
B
 .dA  0
,