Download Name: Roll No: Final Exam: Part A List of Physical Constants

Phy 204: Electricity and Magnetism
Name:
Fall Semester 2014
Roll No:
Final Exam: Part A
Maximum Marks: 30
Total Time: 1 Hour 15 Minutes
Instructions. Write your name and roll number in the space specified above. Attempt all questions. This exam comprises of two parts, A and B and will last for 3 hours. The recommended
time for part A is 1 hour. Part A has 10 Multiple Choice Questions. The most appropriate answer
is to be circled on the question paper. Part A is to be filled on the question paper and returned
within 1 hour 15 minutes. You can use the blue answer books for rough work relevant to Part A.
Please turn off mobile phones.
List of Physical Constants
Bohr Magneton, µB = 9.27 x 10−24 J K−1
Boltzmann Constant, kB = 1.38 x 10−23 J K−1
Vacuum Magnetic Permeability, µ0 = 4π x 10−7 T m A−1
Elementary Charge, e = 1.6 x 10−19 C
Vacuum Permittivity, εo = 8.85 x 10−12 C2 N m−2
Useful Formulae
⃗ = µ0 q⃗v x ⃗r
B
4π r2
⃗
⃗j = σ E
(relation between current density and electric field);
j = nqvd
σ=
(Biot-Savart Law);
(current density depends on the charge density and drift velocity);
nq 2 τ
m
(conductivity depends on average scattering time);
⃗
⃗vd = µE
(definition of mobility);
⃗ = µ0 nI
|B|
(magnetic field inside an infinite solenoid);
⃗ ⃗
⃗= ExB
S
µ0
(Poynting Vector as a cross-product of electric and magnetic fields);
Maxwell’s Equations:
∫
⃗ dA
⃗ = Qenc
E.
ε0
(Gauss’s Law for Electric Charges);
Closed Area
∫
⃗ dA
⃗= 0
B.
(Gauss’s Law for Magnetism);
Closed Area
∫
⃗ dA
⃗
B.
(Faraday’s Law);
Open Area
Closed P ath
{
∫
⃗ d⃗l = µ0
B.
Closed P ath
∫
⃗ d⃗l = − d
E.
dt
d
Ienc + ε0
dt
∫
}
⃗ dA
⃗
E.
Open Area
(Maxwell-Ampere’s Law);
MCQs
Part A:
1. A wire is fashioned into a right angled triangle with two sides of equal length l. Initially, the
triangle, which is at rest, is fully immersed in a uniform magnetic field B, normal to the plane
of the wire. Its vertical side coincides with the edge of the magnetic field. At t = 0, the loop
starts moving out of the field with constant speed v. This is shown in the diagram. What can
we say about the induced emf ε(t) at time t?
Region of
non-zero B
B = 0 here
t=0
l
l
v
(a) |ε(t)| = Bv(l − vt) for all t > 0
l
l
, ε(t) = 0 for t >
v
v
l
l
(c) |ε(t)| = Bv(l − vt) for 0 < t < , ε(t) = 0 for t >
v
v
(b) |ε(t)| = Bvl for 0 < t <
(d) |ε(t)| = Bvl for t > 0
(e) |ε(t)| =
Bvl2
for t > 0
t
t>0
2. A coil of N turns is placed inside an electromagnet. The coil has radius r and is connected to
a resistor. The total resistance inside the circuit is R. The magnetic field is confined to the
area of the coil (as shown). The electromagnet is switched off and the field decays to zero in
time τ . How much total charge Q flows through the resistor?
N turns
r
Field B is only
inside the loop
R
(a) An emf and a current will be induced but Q = 0.
N Bπr2 τ
R
N Bτ
(c) Q =
R
N Bπr2
(d) Q =
Rτ
N Bπr2
(e) Q =
.
R
(b) Q =
3. An infinite solenoid of diameter D carries a linearly increasing current I(t) = kt, where k > 0.
Point P is inside the solenoid at a distance r ≪ D from the axis of the solenoid. Point Q is
outside the solenoid. Which of the following graphs show the magnitude of the electric field at
points P and Q? Assume the field is uniform inside the solenoid.
Q
Cross-section of
the infinite solenoid
P
r
D
EP
EQ
EQ is constant
and non-zero
EP is constant
and non-zero
0
t
0
t
(a)
EP
EQ
EP is constant
and non-zero
0
(b)
t
EP
EQ= 0 always
0
EQ
EQ= 0 always
EP = 0 always
0
(c)
t
t
0
t
EP
EQ
EP increases
quadratically
0
0
t
(d)
EP
t
EQ
EQ= 0 always
0
0
t
(e)
t
4. An infinite solenoid carries a time-varying current I(t) = I0 cos(ωt). A flexible conducting ring
with smaller radius (as shown) is placed inside and coaxially with the solenoid. What will
happen to the ring?
Ring
I(t)
x
x
x
x
x
shrink
stretch
x
x
x
x
x
Current
coming out
Current
going in
(a) The ring will continuously stretch and increase its radius.
(b) The ring will alternately stretch and shrink.
(c) The ring will continuously shrink and decrease its radius.
(d) The ring will twist because it will experience a torque.
(e) The ring will start rotating about the axis of the solenoid.
5. A constant current I flowing through a conducting wire terminates on a point charge Q,
depositing more and more charge as time passes by. What is the rate of change of electric flux
dΦe
through a closed surface (represented by the dashed line in the figure) enclosing the point
dt
charge?
Q
I
µ0 I
4πε0 r2
I
(b)
4πε0 r2
µ0 ε0 I
(c)
4πr2
I
(d)
ε0
(a)
(e) I(4πr2 )
r
6. A positive charge of +Q is uniformly distributed over the surface of a hollow, insulating sphere
of radius L. A point charge +Q is located at a distance 4L from the center. Point A is located
at a distance L from the sphere’s surface, on the axis joining the point charge and the center
L
of the sphere. Point B is located at a distance
from the center of the sphere. What can be
2
said about the electric fields at A and B?
+
+
charge
+Q
2L
A
+
L +
+
L
+
+
+Q
distributed
+ + uniformly
+
+
+
L/2
+
B
+
+
+
(a) EA = 0, EB = 0
(b) EA = 0, EB ̸= 0
(c) EA ̸= 0, EB ̸= 0
(d) EA ̸= 0, EB = 0
(e) |EA | < |EB |
7. In experiment A, three point charges, each of magnitude +Q are placed symmetrically on the
perimeter of a circle. In experiment B, a charge 3Q is uniformly distributed over the same
circle. Which of the following statements is true about the magnitudes of the electric fields at
different points on the dashed circle, concentric with the bold circle?
Charge 3Q
+Q
+
+ + +
+
+
+Q
+Q
+
+
+ + +
+
B
A
(a) Magnitude of the electric field varies from point to point on the dashed circles both in
experiment A and B.
(b) Magnitude of the electric field is the same at all points on the dashed circle in A and at all
points on the dashed circle in B, but it differs between A and B.
(c) Magnitude of the electric field is the same at all points on the dashed circle in A and at all
points on the dashed circle in B, and it is the same in A and B.
(d) Magnitude of the electric field is the same everywhere on the dashed circle only in experiment A.
(e) Magnitude of the electric field is the same everywhere on the dashed circle only in experiment B.
8. A high frequency alternating current flows through a uniform wire. What is an approximate
pattern of the electric field inside the wire at the particular time t0 , shown in the accompanying
diagram? The arrows represent electric field vectors. Dots represent zero electric field. [HINT:
Skin Effect]
I (t)
t0
time
(a)
(b)
(c)
(d)
(e)
9. A switch inside an RC circuit is closed at time t = 0. The ideal capacitor builds up charge with
⃗ in the region between the capacitor
time. Which of the following shows the Poynting Vector S
plates while the capacitor charges?
ε
a
b
a
S
b
(a)
a
S
b
(b)
a
X
X
X
S
b
(c)
a
S
b
(d)
⃗ = 0)
(e) There is no energy flow into or out of the capacitor (S
10. A metallic conductor has a hot and a cold end. The resistivity depends on temperature. How
⃗ A and E
⃗ B at points A and B compare?
do the electron mobilities µ
⃗ A and µ
⃗ B and electric fields E
Hot
Cold
A
Heating by laser
⃗ A | > |E
⃗B|
(a) |⃗
µA | < |⃗
µB |, |E
⃗ A | = |E
⃗B|
(b) |⃗
µA | = |⃗
µB |, |E
⃗ A | > |E
⃗B|
(c) |⃗
µA | > |⃗
µB |, |E
⃗ A | < |E
⃗B|
(d) |⃗
µA | > |⃗
µB |, |E
⃗ A | = |E
⃗B|
(e) |⃗
µA | > |⃗
µB |, |E
B
Phy 204: Electricity and Magnetism
Name:
Fall Semester 2014
Roll No:
Final Exam: Part B
Solution
Maximum Marks: 40
Total Time: 1 hour 45 minutes
Instructions. Write your name and roll number in the space specified above. Attempt all questions. This exam comprises of two parts, A and B. There are 6 comprehensive questions in Part
B. Use the blue answer books provided to answer Part B questions as completely as possible.
Please write your name and roll number on ALL answer books.
List of Physical Constants
Bohr Magneton, µB = 9.27 x 10−24 J K−1
Boltzmann Constant, kB = 1.38 x 10−23 J K−1
Vacuum Magnetic Permeability, µ0 = 4π x 10−7 T m A−1
Elementary Charge, e = 1.6 x 10−19 C
Vacuum Permittivity, εo = 8.85 x 10−12 C2 N m−2
Useful Formulae
⃗ = µ0 q⃗v x ⃗r
B
4π r2
⃗
⃗j = σ E
(relation between current density and electric field);
j = nqvd
σ=
(Biot-Savart Law);
(current density depends on the charge density and drift velocity);
nq 2 τ
m
(conductivity depends on average scattering time);
⃗
⃗vd = µE
(definition of mobility);
⃗ = µ0 nI
|B|
(magnetic field inside an infinite solenoid);
⃗ ⃗
⃗= ExB
S
µ0
(Poynting Vector as a cross-product of electric and magnetic fields);
Maxwell’s Equations:
∫
⃗ dA
⃗ = Qenc
E.
ε0
(Gauss’s Law for Electric Charges);
Closed Area
∫
⃗ dA
⃗= 0
B.
(Gauss’s Law for Magnetism);
Closed Area
∫
⃗ dA
⃗
B.
(Faraday’s Law);
Open Area
Closed P ath
{
∫
⃗ d⃗l = µ0
B.
Closed P ath
∫
⃗ d⃗l = − d
E.
dt
d
Ienc + ε0
dt
∫
}
⃗ dA
⃗
E.
Open Area
(Maxwell-Ampere’s Law);
Part B:
Comprehensive Questions
⃗ pointing downwards exists only in a thin columnar region of radius
1. A uniform magnetic field B
b. A positively charged ring of uniform charge density λ is coaxial with this column. The field
dB
dB
is switched off slowly at a rate
. Note that
< 0.
dt
dt
B
b
+
+ +
+ +
+
+
+ λ = Q/2πr
r
+
+ +
+ + + +
(a) What happens to the ring of charge?
(i) It will start rotating in the clockwise direction (as seen from above).
(ii) It will start rotating in the anticlockwise direction (as seen from above).
[Correct Answer= 2 marks, Incorrect = -1 mark, Blank = 0 marks.]
⃗ has dropped to 0. First
(b) Find the total angular momentum L imparted to the ring when B
find the torque τ exerted on the charged ring due to the changing magnetic field. [7 marks]
∫
(HINT: L = τ dt where L is the angular momentum and τ is the torque.)
Answer
Step 1: A time-varying magnetic field creates a non-conservative electric field (Faraday’s
Law). We first find the electric field generated at the distance r. The right-hand curl rule
suggests that this electric field will be circular and in the clockwise direction (as seen from
above).
∫
∫
⃗ d⃗l = − d
E.
dt
Closed P ath
⃗ dA
⃗
B.
Open Area
dB
E.2πr = −
π b2
dt
dB πb2
E = −
.
dt 2πr
Step 2: We can now find the electric force acting on the ring of charge. Torque exerted on
the ring τ (t) is then calculated from the electric force:
dF
= Edq
dτ
τ
= |d⃗τ | = |⃗r x dF⃗ | = rdF = Erdq
∫
= Er dq = ErQ
τ
= −
dB πb2
rλ(2πr)
dt 2πr
dB
= −
(πb2 ) rλ.
dt
Step 3: Calculate the total angular momentum L imparted to the ring using the formula
provided:
∫
L =
τ dt
∫0
−(πb2 ) rλdB
=
B
= (πb2 ) rλB.
2. We have two infinite, thin sheets, each carrying a uniform current of sheet density κ (where κ
is in Amperes per meter). The currents are in opposite directions, as shown below. Copy the
diagram and mark the direction of the magnetic field due to this current distribution on your
diagram. Plot the magnitude of this magnetic field as a function of x. Calculate and identify
the maximum and minimum field on your plot. Explain your reasoning. [7 marks]
Sheet A
Current
flowing out
x is
negative
x= 0
Sheet B
x
x
x
x
x Current
x flowing in
x
x
x
x
x
x= d
x
Answer
We can calculate the magnetic field in each region using Ampere’s Law. For example, consider
the Amperian loop shown below for Sheet A. The loop is of width w and length l and is
traversed in the direction shown, starting from point P . Assuming that the magnetic field
produced by Sheet A is B0 , we can calculate B0 as follows:
Sheet A
w
B0
Current
flowing out
l B
0
P
∫
⃗ ⃗l = µ0 Ienc
B.d
B0 (l) + 0 + B0 (l) + 0 = µ0 (κl)
2B0 (l) = µ0 (κl)
B0 =
µ0 κ
.
2
Using the right hand curl rule, we can verify that the direction of the magnetic fields produced
by both sheets (labeled as sheet A and B) are as shown below. Note that the magnetic fields
are independent of the distance from the sheet. The magnetic fields cancel out in regions x < 0
and x > d and add in the region 0 < x < d due to superposition.
Sheet A
BA
A
B x BA
x
x
x
x Current
x flowing in
x
x
x
x
x
Current
flowing out
x is
negative
Sheet B
x
x= d
x= 0
From the Principle of Superposition:
BT otal = B0 − B0 = 0 (x < 0)
BT otal = B0 + B0 = µ0 κ (0 < x < d)
BT otal = B0 − B0 = 0 (x > d).
This is shown in the graph below. The maximum field is µ0 κ and the minimum field is zero.
BTotal
µ0κ
0
0
d
x
3. A cylindrical conductor carries a uniform current of density ⃗j. Find the ratio of the magnitudes
of the Poynting Vectors at the surface and half-way between the surface and the central axis
of the cylinder. [5 marks]
j
R/2
R
Answer
⃗ =
S
S =
⃗ xB
⃗
E
µ0
EB
.
µ0
Using Ampere’s Law, calculate the magnetic field at the surface (B1 ):
B1 (2πR) = µ0 j(πR2 )
B1 =
µ0 j(πR2 )
.
2πR
Similarly, calculate the magnetic field half-way from the surface (B2 ):
B2 (2π
R
R2
) = µ0 j(π )
2
4
µ0 j(πR2 )
B2 =
.
4πR
⃗ is the same throughout the conductor (since ⃗j is constant and ⃗j = σ E).
⃗ Therefore, S1 =
E
S2
4
B1
= = 2.
B2
2
4. You will not receive any credit for using the Biot-Savart Law in this question.
A U-shaped wire is made up of two parallel straight wire segments of semi-infinite length placed
a distance 2R apart and a semi-circular segment of radius R connecting the straight wires, as
shown in the diagram below. A constant current I passes through this wire. Using Ampere’s
Law and the Principle of Superposition, find the magnitude and direction of the magnetic field
⃗ at point C, the center of the semi-circular segment of the wire. [6 marks]
B
You are given that the the magnetic field at the center of a circular loop of wire of radius R,
µ0 I
carrying current I, is
.
2R
I
C R
R
Answer Consider the figures drawn below, showing A and B which are identical with respect
to the field they produce at C and C ′ .
I
C
C'
=
B
A
I
I
C
R
A+B = P
I
Let’s first find the field at C in the configuration P as shown above. Later we will halve this
result. The configuration P is a sum of two infinitely long wires and a circular loop of wire,
all three carrying a current I. Using Ampere’s Law it can be found that the magnetic field
µ0 I
at distance R due to an infinite wire carrying current I is
. Using the Right Hand Grip
2πR
rule, it is found that all three fields are in the same direction, hence are added (Principle of
Superposition).
P
=
=
=
µ0 I
µ0 I
µ0 I
+
+
2πR 2πR
2R
µ0 I
µ0 I
+
πR ( 2R )
µ0 I 1 1
+
.
R π 2
Field in configuration A is half of the field in configuration P (Principle of Superposition),
(
)
µ0 I 1 1
+
.
hence the required field is
2R π 2
5. A conducting sphere of radius R initially has a uniform surface-charge density σ0 . Beginning
(
)
t
at t = 0, this charge is drained off over a period t0 such that σ = σ0 1 −
.
t0
R
σ
Ic
(a) Find the magnitude of the displacement current Id just outside the surface of the sphere.
[6 marks]
(b) Calculate the current Ic carried off by the wire. Compare the displacement current Id to
Ic . [4 marks]
Answer
Id
R
σ
Ic
We use Maxwell-Ampere’s Law in this question. We choose a very small circular path encircling
the wire carrying the conduction current. This path is shown in red in the diagram above. For
part (a) we consider the spherical surface just outside the conducting sphere as the open surface
corresponding to this path. Note that Ic through this surface (shown as a dashed circle in the
diagram above) is zero. In part (b) we should be able to verify that Id = Ic , since the only
current piercing the flat circular surface enclosed by the red path is Ic .
(a)
Id
d
= ε0
dt
∫
⃗ dA
⃗
E.
Open Area
= ε0
dE
(4πR2 ).
dt
dE
. Using Gauss’s Law for Electric Charges we find the Electric Field outside
dt
the sphere in terms of the electric charge density:
We need to find
∫
⃗ dA
⃗ =
E.
Qenc
ε0
Closed Area
E(4πR2 ) =
Q
ε0
= (4πR2 )
⇒E =
σ
ε0
σ
.
ε0
This gives us
dE
dt
Id = −
1 dσ
ε0 dt
(
)
1
1
(σ0 ) −
=
ε0
t0
σ0
.
= −
ε 0 t0
=
(4πR2 )σ0
(4πR2 )σ0
and |Id | =
.
t0
t0
(b)
dQ
dt
(
)
d
t
2
=
(4πR )σ0 1 −
dt
t0
2
(4πR )σ0
= −
.
t0
Ic =
Therefore, |Ic | = |Id |. Note that this was expected from the Maxwell-Ampere’s Law. Both
currents equal the path integral of the magnetic field around the small circular red path,
therefore must be equal.
⃗ = B0 xî, that is, a uni-directionally
6. Using a Maxwell’s Equation show that a magnetic field B
varying magnetic field (as shown below) cannot exist. [3 marks]
B
y
x
z
Answer
B1
B2
B
y
x
z
Using Gauss’s Law for Magnetism it can be shown that a uni-directionally varying magnetic
field cannot exist. We draw a Gaussian Surface as shown above. Note that since there are no y
or z components of the magnetic field, the flux through the curved surface area of the cylinder
is zero. Taking A to be the area of the end caps we find:
∫
⃗ dA
⃗ =
B.
0
Closed Area
B2 A − B1 A = 0
⇒ B1 = B2 .
Hence B1 cannot be different from B2 . Therefore a uni-directionally varying magnetic field,
which changes with x, cannot exist because it violates Gauss’s Law for magnetism.