Download Electromagnetism and Optics Problems and more Exams Calculus in PDF only on Docsity!
Final Exam Solutions
- Three charges form an equilateral triangle of side length d = 2 cm. The top charge
is q3 = 3 μC, while the bottom two are q1 = q2 = -‐ 6 μC. What is the magnitude of the
net force acting on q3?
(1) 700 N (2) 350 N (3) 810 N (4) 405 N (5) 0 N
By symmetry the net force will be aligned vertically downward (attractive force). So
we need to project the force from each of the two lower forces onto the vertical axis:
F
net
= k
q 1
q 3
d
2
cos 30
q 2
q 3
d
2
cos 30
= 2 9 × 10
9
3 × 10
− 6
( ) 6 ×^10
− 6
2
- A glass rod forms a semi-‐circle of radius r = 4 cm with a charge of -‐q distributed
uniformly along the right quadrant and + q distributed along the left quadrant,
where q = 5 pC. What is the magnitude and the direction (as the polar angle relative
to the direction of the x -‐axis) of the electric field at the center P of the semi-‐circle.
(1) 29 N/C, θ = 0 ° (2) 45 N/C, θ = 27 0 ° (3) 0 N/C (4) 29 N/C, θ = 90 ° (5) 45 N/C, θ =
The diagram shows the direction of the electric field contribution from each
quadrant. By symmetry, the net field points down in the
i direction, which is θ = 0
The magnitude (not really needed to choose the correct answer!) is given by:
λ =
q
π
R
dq = λ ds = λ Rd θ
E
y
= k
dq
R
2
sin θ =
π / 2
π
k
λ Rd θ
R
2
sin θ
π / 2
π
= − k
λ
R
cos θ | π / 2
π
= k
λ
R
⇒ E
y
tot
= 2 k
λ
R
4 k
π
q
R
2
- A parallel plate capacitor with capacitance 0.5μF is connected to a 6V battery. If
the plates are squeezed to half of their original separation, what will be the charge
stored on the capacitor?
(1) 6μC (2) 3μC (3) 1.5μC (4) 0.5μC (5) 12μC
The capacitance of a parallel plate capacitor is C = ε 0
A
d
, so the capacitance
increases two times when the plates are squeezed together. So the new capacitance
is 1 μF. The charge is thus
q = CV = 6 μ C
- Part of a long insulated wire carrying current i = 4A is bent into a circular section
of radius R = 1.5 cm as shown in the figure. What is the magnitude of the magnetic
field (in T) at the center of curvature C if the circular section lies in the plane of the
page as shown?
(1) 2.2 x 10
-‐ 4 (2) 5.3 x 10
-‐ 5 (3) 1.7 x 10
-‐ 4 (4) 1.1 x 10
-‐ 4 (5) 5.0 x 10
-‐ 6
The field from the long straight wire is given by
B
w
μ 0
i
2 π R
out of the page at the center of the circle.
The field from the circular loop itself is given by the Biot-‐Savart law:
B
loop
μ 0
i
4 π R
φ =
μ 0
i
2 R
also out of the page
Both contributions add, given a field B=2.2 x 10
-‐ 4
- A series RLC circuit is driven by a sinusoidally-‐varying EMF source. The current
lags the EMF by 30°. What can be concluded about the driving frequency ω?
(1) ω >
LC
(2) ω <
LC
(3) ω =
LC
(4) ω = 0 (5) ω = ∞
ε ( t ) = ε
m
sin( ω t )
i ( t ) = i
m
sin( ω t − φ)
So the current lags the EMF when Φ>
tan φ =
X
L
− X
C
R
X
L
> X
C
ω L >
ω C
ω
2
LC
⇒ ω >
LC
- A constant current of i = 3A is used to charge a parallel plate capacitor with
circular plates of radius R = 1cm. What is the magnitude of the magnetic field at a
radius of r = 0.3 cm, which is less than R , in the region between the plates?
(1) 1.8 x 10
-‐ 5 T (2) 6 x 10
-‐ 5 T (3) 1.2 x 10
-‐ 4 T (4) 3.8 x 10
-‐ 6 T (5) 0 T
The total displacement current between the plates is id=i=3A. The effective
displacement current density is j d
i d
π R
2
So using Maxwell’s Law of Induction:
B ⋅ ds
= μ 0
i ′ d
2 π rB = μ 0
j d
π r
2
= μ 0
i d
π R
2
π r
2
B =
μ 0
2 π
i d
R
2
r = 1.8 × 10
− 5
T
- Light traveling horizontally enters a right prism as shown in the figure. The index
of refraction of the prism is n=1.6, and it is surrounded by air. What is the measure
of the angle θ 2 that the light deflected from horizontal when it exits the prism?
There are 2 refractions from the 2 surfaces. And since the second surface is not parallel
to the first, the light does not return to the incident direction.
θ 1
θ 2
0
2 =0 for minimum thickness
490nm
87.5 nm
n
n
t m m
t
n
- A point particle with charge q is at the center of a Gaussian surface in the form of a
cube. The electric flux through any one face of the cube is:
(1) q/ 6 ε 0 (2) q/ 4 πε 0 (3) q/ 4 ε 0 (4) q/ε 0 (5) q/ 16 ε 0
A cube has six faces. The electric flux through one face is one sixth of total flux through
the surface of the cube.
A: q /6ε 0
- A total charge of 7 × 10
C is uniformly distributed throughout a non-conducting
sphere with a radius of 5 cm. The electric potential at the surface, relative to the potential
far away, is about:
(1) 1.3x 10
4
V (2) - 1 .3x 10
4
V (3) 7 x 10
5
V (4) - 6 .3x 10
4
V (5) 0
Outside the sphere, the electric potential is identical to that of a point charge.
= 1. 3 × 10
! 𝑉
A: 1.3 × 10
4
V
- Five cylindrical wires are made of the same material. Their lengths and radii are
wire 1: length radius r
wire 2: length radius r /
wire 3: length radius r /
wire 4: length radius r /
wire 5: length radius r /
Rank the wires according to their resistances, least to greatest.
(1) 1, 3, 4, 2, 5 (2) 5, 4, 3, 2, 1 (3) 1 and 2 tie, then 5, 3, 4 (4) 1, 3, 5, 2, 4 (5) 1, 2, 4, 3, 5
The resistances of the five wires are
!
!
!
!!
!
!!
!!
!
!!
!!
!
!!
!!
!
!"!
!!
!
respectively.
A: 1, 3, 4, 2, 5
l ,
3 / 2,l
l / 2,
l ,
5 ,l
- The current in the 5.0-Ω resistor in the circuit shown is:
(1) 1.5 A (2) 0.67 A (3) 0.42 A (4) 2.4 A (5) 3.0 A
Apply loop rule to the loop consisting of battery, 3.0-Ω and 5.0-Ω resistors. We have
Solve it to get i=1.5 A.
A: 1.5 A
- A rod with resistance R lies across frictionless conducting rails in a constant uniform
magnetic field B , as shown. Assume the rails have negligible resistance. The magnitude
of the force that must be applied by a person to pull the rod to the right at constant speed
v is:
(1) B
2
L
2
v/R
(2) BLv
(3) BLv/R
(5) B
2
Lxv/R
Induced emf in the loop is ℇ = 𝐵
!"
!"
!"
!"
= 𝐵𝐿𝑣. Induced current is 𝑖 =
ℇ
!
!"#
!
𝑐𝑐𝑤. Magnetic force on the rod is 𝐹 !
!
! !
! !
!
to the left. Pulling force has the
same magnitude as the magnetic force and points to the right because the rod moves with
constant velocity.
A: B
2
L
2
v / R
- The diagram shows a single slit with the direction to a point P on a distant screen
shown. At P, the pattern has its second minimum (from its central maximum). If X and Y
are the edges of the slit, what is the path length difference (PX) – (PY)?
(1) 2λ (2) λ (3) 3 λ / 2 (4) λ /2 (5) 5 λ/ 2
Second minimum of single slit occurs when 2 𝑎𝑠𝑖𝑛𝜃 = 2 𝜆. (PX) – (PY)= 2 𝑎𝑠𝑖𝑛𝜃 = 2 𝜆.
A: 2 λ
- Monochromatic light is normally incident on a diffraction grating that is 1 cm wide
and has 10,000 slits. The first order line is deviated at a 30° angle. What is the
wavelength, in nm, of the incident light?
The grating spacing is 𝑑 =
!×!"
!!
!"
!
!!
𝑚. We have 𝑑𝑠𝑖𝑛𝜃 = 𝜆 for 1
st
order maximum.
Solve it to get wavelength.
A: 500
