Physics 114 Exam 3 Fall 2014
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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). Determine the initial direction of the deflection of charged
particles as they enter the magnetic fields shown in the figure below.
(b)
(a)
v × B is down, but
since this is an electron
the force is upward, so it is
deflected upward
Perpendicular,
Into the
page
o
(c)
(d)
No deflection
Perpendicular,
In to the
page
Positive charge moving up at 45 degrees with
respect to the horizontal.
Question 2 (10 points).
(a) Two long and parallel wires each carry the same current, I, in the same direction as
shown below. Is the total magnetic field at the point P midway between the wires (i) zero,
(ii) directed into the page, (iii) directed out of the page, (iiii) directed to the left, or (iiiii)
directed to the right?
(i) The field is zero as that
due to the wire is into the
page and that due to the
bottom wire is out of the page
Now (for parts b-f), consider these true or false questions regarding the figure below (more than
one can be true)
(b) In region I, the magnetic field is into the page and is never zero (no matter what the
distances between the wires are)
(c) In region II, the magnetic field is into the page and can be zero
(d) In region III, it is possible for the field to be zero
(e) In region I, the magnetic field is out of the page and is never zero
(f) There are no points where the field is zero.
(g) Still considering the picture above, what is the direction of the force on the 10.0 A wire?
(b) False. The field is out of the page since that due to the 20 A current is stronger
(c) False. It cannot be zero since both wires produce fields into the page
(d) True. The top wire produces a field into the page. The bottom wire produces a field out
of the page. The 10 A current is closer but smaller than the 20 A one
(e) True. Even though the field due to the 10 A wire is into the page, it can never overcome
the field due to the 20 A wire which is out of the page
(f) False

 
(g) The field due to the 20 A wire is into the page. F  IL x B gives that the force on the
bottom wire is downward.
Question 3 (10 points). (a) A long, straight wire carries a steady current I. A
rectangular conducting loop lies in the same plane as the wire, with two
sides parallel to the wire and two sides perpendicular. Suppose the loop is
pushed toward the wire as shown. What is the direction of the current
induced in the loop (clockwise, counter clockwise, or is there actually no
induced current)?
The field from the wire is
increasing into the page
through the loop. Thus, the
induced field will be out of
the page so the current is
Counterclockwise
. (b) Will dropping a magnet through a long copper tube produce currents in the tube?
Explain. Will the magnet fall faster or slower as it falls than if it were released
outside the tube (so that it free-falls)? Explain.
Currents will be produced. The flux through the tube changes as the magnet falls so
counterfields are set up by induced currents that oppose the motion.
It will fall slower as the counterfields oppose the motion (see animated slide from
class in Ch 31).
Now consider the bar shown in the figure below that moves on rails to the right with a velocity
, in a uniform, constant magnetic field that is directed out of the page.
(c) Is there an induced current? If so is it clockwise or counterclockwise?
There is an induced current. Flux is increasing outward so the induced field is
inward and the induced current will be clockwise.
(d) Is there a force required to keep the bar moving?
Yes, there is a required
applied force to keep the bar
moving.
As there is a current, the field
will exert a force on the bar to
the right. The applied force
must balance this to the left.
Also, considering the energy,
energy must go onto the
resistor so this must come
from an applied force.
Problem 1 A particle with a charge = 3.2 × 10-19 C moves with a velocity

ˆ m/s
v = (2iˆ + 3j)

B = (2iˆ + 4jˆ )T .
through a region where there is a magnetic field
(a) Determine the force on the particle due to the field (9 points).
(b) What is the angle between this force and the magnetic field? (3 points).
(c) What is the angle between the velocity and the magnetic field? (3 points).

 
ˆ X (2iˆ + 4jˆ )
(a) F  qv x B = 3.2 × 10-19 [ (2iˆ + 3j)
 (2iˆ  2iˆ)  (2iˆ  4 ˆj )  
= 3.2 × 10-19 
 =3.2 × 10-19 0  8kˆ  6kˆ  0
ˆ
ˆ
ˆ
ˆ
(3
j

2
i
)

(3
j

4
j
)


=3.2 × 10-19  2kˆ   6.4  1019 kˆ
(b) It is 90o, this angle is always 90o.
 
(c) We know that the magnitude of v × B is given by v B sin( ) . We have also
already found that v × B = 2 k̂ , which has a magnitude of 2. Thus,
 
2  v B sin( )  4  9 4  16 sin( )  16.12sin( )
sin( )  2 / 16.12
  7.13o
Problem 2 . A loop of wire consists of two semi circles of radii R = 2 m and 2R = 4 m,
both centered at a point P, and connected with wires going radially from one to the other.
A current I = 10 A flows in the loop.
(a) Can Ampere’s law be used to solve for the field at point P? Why or why not? (3
points)
(b) Find the magnetic field at the point P (by whatever means necessary)? (12 points)
I
P
R
2R
(a) We cannot use Ampere’s law as the geometry does not have enough symmetry.
(b) Do one side at a time. First do the straight segments. ds and r-hat are parallel. No
contribution to the integral, so we get zero contribution from the straight segments. Now
do inner half loop (assume x-y).
I
ds
 I 2 R 0 I ˆ

k
ds  rˆ  kˆ ds . Binner  0  kˆ 2  0 2 kˆ
R
4
4R
4 R
2
I
Outer loop opposite direction, similar B outer   0 kˆ
8R
0 I ˆ 010 ˆ
B
k
k  7.85  107 Tkˆ
8R
16
Problem 3 (15 points). A circular loop of wire of resistance R = 0.5  and radius r =
8.00 cm is in a uniform magnetic field directed out of the page as shown below. If the
field is changing in time so that a clockwise current is induced in the loop, I = 2.5 mA,
(a) is the magnetic field increasing or decreasing in time?
(b) Find the rate at which the magnetic field is changing with time?
(c) Find the direction and magnitude of electric field at the point P which is 5 cm from
the center of the loop.
(a) If the magnetic field were increasing, the flux would be
increasing out of the page, so the induced current would
tend to oppose the
increase by generating a field into the page. The direction
of such a current would be clockwise. This is the case
here, so the field is
increasing .
P
(b)
(
 
d B
(c)    E. ds =  N
E 2 r   r 2
dB
dt
dt
E = (2.5 × 10-2)(62.2 × 10-3) = 1.555 × 10-3N/C
r dB
E
2 dt
The direction is tangent to the electric field lines which are circular and in the direction of the
induced current (clockwise). Thus the electric field points to the right at Point P.
1 | q 1 || q 2 |
4  0 r 2
e = 1.6 X 10-19 C
F
Possibly Useful Information
.
 0  8.85 X 10 -12 (C 2 / N  m 2 )
 
E  Fq
0
| q|
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
x-xo = vot + (½)at2
v2 = vo2 + 2a(x-xo)
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
1 q
V
4  0 r
V
s
V
E
s
Q = CV
Es 
C  2 0
l
ln(b / a)
i
n
1 n qi

4 0 i1 ri
i 1
1
dq
V

4 0 r
V
V
V
; Ey  
; Ez  
Ex  
x
y
z
1 q1q 2
U  W 
4 0 r12
 0A
C
d
ab
C  4  0
ba
V   Vi 
C  4  0 R
C eq   C j (parallel)
1
1
(series)

C eq
Cj
Q2 1
 2 CV 2
2C
C =  C0
U
u  12  0 E 2
I= dQ/dt
  1
L
R
A
P = IV
V = IR
P = I2R=V2/R
I 
(R  r)
Pemf = I
R eq   R j (series)
1
1
(parallel)

R eq
Rj
q(t)= Q(1-e-t/RC)
I = (R)e-t/RC
-t/RC
I= (QRC)e
, I0 = (QRC)
 
F  qv x B
r = mv/qB,  = qB/m

 
dF  Ids x B
q(t) = Qet/RC

 
F  IL x B

=  x B


  NIA
U   B
 
 
Ids x r , 0= 4 x 10-7 T.m/A
dB  0
4
r3
B = 0I/2r
F/l = (0I1I2)/2a
B = 0nI (solenoid)
 
 B.d s   0 I enc


 
d B
   E. ds =  N
= Blv
dt
 
 B   B. dA
 
B  
B
 .dA  0