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Air University
Mid Term Examination
(Spring-2011)
Course: PH-106 Physics-II
Max. Marks: 45
Date: 04-06-2011
Time Allowed: 3 hrs
Instructors: A. Sadiq, M. Hussain, A. Minhas, K. Ali, T. Zehra and T. Firdous
Note: Attempt all 9 questions. Each question carries equal marks. Please encircle your final answers. You
may use basic scientific non-programmable calculator. DO NOT write in pencil or red ink on answer sheet.
DO NOT write anything on the question paper.
P
x
1. Fig 1 shows an electric dipole of moment
p
 q d , where q is the charge of the dipole and d is the
separation between its two charges.
Fig 1
p
Find the expression for the electric field intensity at a point P in terms of
q
x
q
d
that is at a distance x from
its center for x  d when the point P a) lies on the axis of the dipole, b) it lies on the plane
perpendicular to the dipole and passing through its center.

c) Derive the expression for the torque
Solution: Electric field intensity
E
acting on a dipole
[3]
p
in a uniform electric field
E
.
[2]
at a point P is situated at a distances r1 and r2 from two point
charges q1 and q 2 is given by,
E

q1

r1 

q2
r2
40 r1
40 r2
Here, q1  q , q2  q , and
a) When P lies on the line joining the two charges and at a distance x from the center of this line segment,
2
2
them, we have,


r1  x  d / 2 , r2  x  d / 2 , and r 1  r 2 
E

q

j
4 x  d / 2
2
0
j
, giving,
q
j
4 x  d / 2
2
0

q

2 xd
40 x 2  d 2 / 4

2
j

x

p
20 x 2  d 2 / 4

2
For x  d
E

p
20 x 3
b) When P lies on the line perpendicularly bisecting the line joining the two charges and at a
distance x from the center of this line segment, them, we have,
q
x
q
d
r1  x 2  d 2 / 4  r2
In this case the components of the electric field along the perpendicular bisector of the line joining P to
the origin cancel out and its components perpendicular to this line add up, giving,
E

E


2qd / 2
40 x 2  d 2 / 4
For x  d , we have,

3/ 2
j


p
40 x 2  d 2 / 4

3/ 2
p
40 x 3
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P
P
c) Two forces each of magnitudes qE act on a dipole
electric field
E
p
 q d making an angle  with a uniform
. One of these that act on the positive end of the dipole points in the direction of the field.
The other one that acts on the negative end of the dipole points in the opposite direction. Their resultant
vanishes. They form a couple providing a torque that orients the dipole along the field. Its magnitude is
given by,
  qEd sin   pE sin  , or



p E
2. a) A 1g charged water drop is freely suspended between the plates of capacitor with an electric field
of 10kV / m between its plates. Calculate the value of the charge on the drop.
[2.5]
mg
E
b) Find the charge and the energy stored in the capacitor if its capacitance in the absence of water drop
is 0.1F and the separation between its plates d  0.1mm .
[2.5]
Solution: a) If the charge of on the water drop is q , then the magnitude of electric force acting on it is,
Fq  qE
Magnitude of gravitational force acting the drop is given by,
Fg  mg
Since the gravitational force acts downward, the electric force must act in the upward direction to balance
it. This means that the charge on the drop must be positive. Equating these two forces we get,
qE  mg , or,
q
mg 10 9  10

 10 12 C  10 6 e
4
E
10
b) Charge on the capacitor Q  CV ,
C  0.1F and V  Ed  10 4  10 4  1volt , giving,
Q  CV  10 7  1  0.1C
Energy stored in the capacitor U 
1
CV 2  0.5  10 7  50nJ
2
3. Consider introducing an uncharged spherical conductor in a uniform horizontal electric field of
p
intensity E as shown in Fig 3.
a) What will be the values of the electric field intensity inside the conductor and outside it very far from
its center?
P'
E
[1]
Fig 3
b) What is the direction of the electric field very close to the surface of the conductor? Give its sketch. 2]
c) Find the value of the induced charge at P and P ' , with the diameter PP' parallel to the field.
[2]
Solution: a) Electric field intensity inside the conductor is zero everywhere.
Electric field everywhere intensity outside the conductor far from its center is E the same as that of the
uniform electric field.
b) The electric field close to the surface of a conductor is always perpendicular to it as shown in the
E
Fig 3
sketch.
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c) The value of the electric field at the point P is E , pointing into the conductor. Therefore, the charge
induced at the point P ,  P   0 E
The value of the electric field at the point P is E , pointing out of the conductor. Therefore, the charge
induced at the point P ,  P   0 E


4. A 1MeV proton m  1.67  10 27 kg, q  1.60  10 19 C enters a region of uniform magnetic field of
strength B  0.01T at an angle of   45 0 with the direction of the field.
a) Calculate its speed.
[1]
b) Calculate the radius and the pitch of its spiral trajectory in the field.
[3]
c) What will be its energy when it exits the region of magnetic field?
[1]
2  1.6  10 13
 1.4  10 7 m / s
 27
1.67  10
Solution: a) Speed of proton v 
2E / m 
b) Radius of proton trajectory r 
mv
1.67  10 27  1.4  10 7

 2.1km
eB sin  1.6  10 19  10 2  0.71
Time to complete one revolution T 
2r
2  3.14  2058

 11.4ms
v sin  1.4  10 7  0.71
Pitch of the proton trajectory p  v cos   T  1.4  10 7  0.71 11.4  10 3  113km
c) Since the magnetic field only changed the direction of motion of a charge particle with changing the
magnitude of its velocity, the proton will exit the region of the magnetic field with its original energy
of 1MeV .
5. Assume free electrons in a conductor loose all their energy that they have gained while accelerating in
an applied electric field E when they collide with ion cores of the material.
a) Find the drift velocity of free electrons as a function of E in a material with inter-atomic spacing a [1].
b) Find the expression for the density J in the a material with free electron density n as a function of E [2]
c) Calculate the drift velocity and the current density J for E  1volt / cm in copper for
which a  0.23nm and n  8.5  10 28 m 3 .
Solution: a) Acceleration of electron in an electric field E ,
[2]
eE
m
Maximum drift velocity of an electron after starting from rest and covering distance a vd 
2eEa
m
0  vd vd
eEa


2
2
2m

eEa
b) Current density J  ne v d  ne
2m

Average drift velocity v d 
c)
2  1.6  10 19  10 2  2.3  10 9
vd 
 20ms 1
32
1.67  10
J  nev d  8.5  10 28  1.6  10 19  21  2.9  1011 Am 2
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r
6. a) Find the magnitude of the magnetic field intensity B at a point P at a distance r  0 from a axis of a
long straight wire of radius R  4mm carrying a current i  10A. shown in Fig 5a. Give a sketch of B as a
function of r .
[3]
b) What is the magnitude and direction of the magnetic field intensity at a point P due to the current
i  10 A in the wire? The curved part of the wire is an arc of a circle of radius R  10cm shown in. [2]
Solution: a) Ampere’s law gives the magnetic field intensity
,
i
B
 B . dl
i
  0 i pen
R
clo loop
Here
dl
P
i
is a line element of the loop, the integral is over any closed loop and i pen is the electric current
that penetrates an open surface bounded by that loop. To find the magnetic field intensity at a point P that
is at a distance r from the axis of a long current conductor carrying current i we choose a closed circular
loop passing through the point P with the axis of the wire as its axis. In this case r is the radius of the loop
and magnetic field is constant along the circumference of this loop and points along it. Thus the integral
on the left hand side of the above equation reduces to,
 B . dl
 2rB
clo loop
We get different values for the right hand sides of the above equation depending whether point P lies
outside or inside the wire. When point P lies outside the wire, r  R , the entire current i we passing
through the wire penetrates the open surface bounded by the closes loop of radius r , and we have,
i pen  i , giving,
B
0i
2r
B
When point P lies inside the wire, r  R , only a part of the current penetrates the open surface bounded
by the closed loop, giving,
i
ir 2
2
i pen 
r  2 , and we get,
R 2
R
i
B  0 2 r , or,
2R
0i
rR
2

r
B
0i
rR
R2
R
The maximum value of the magnetic field intensity is at he surface of the wire when r  R . Its value is,
B  B0 
 0 i 2  10 7  10

 0.5  10 3 T , which is about ten times the value of earth magnetic field
3
2R
4  10
near its surface.
The direction of the magnetic field intensity in both cases, given by the right hand rule, is into the paper
on the right side of the wire and out of the paper on the left of the wire. Fig shows a sketch of the of B as a
function of r .
b) In this case it is convenient to use Biot-Savart law;

dB

Here
dl
 0 i dl  r
4 0 r 2

is n element of the wire along the current in it and r is unit vector pointing from this element
towards the point where the magnetic field
dB
is to be evaluated.
In Fig 3b the curved portion of the current carrying wire is a quarter of an arc of a circle of radius R , the
magnetic field is to be evaluated at the center of the circle of which this arc is a part and the current in the
straight portions of the wire are either pointing into or away from the center. The contribution to the
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magnetic field due to the current in the straight portion of the wire is zero since in this case the cross
product vanishes. Therefore, the magnetic field is only due to the curved portion of the wire for which the
current is along the circumference of the circle and the unit vector, being along its radius, is
perpendicularly to it. Thus,
B
0i  / 2
 0 i 10 7  10
6
R
d



 5  10  T  5T
2 
1
8R
4R 0
2 1 0
7. The plane of a rectangular loop of length l and breadth b is making an angle  with a uniform magnetic
field of strength B with its long side l always perpendicular to. B as shown in Fig 6.
Fig 6
a) Find the magnitude and directions of the forces on sides of the loop and calculate the expression for the
magnitude of the torque acting on the coil when current i passes through it.
[3]
b) What is the angular acceleration of the coil if it has a moment of inertia I ?
[1]
c) What can we say about the rotational frequency of the loop in the magnetic field B?
[1]
Solution: a) Since the long side of the loop is always perpendicular to the uniform magnetic field equal
and opposite force act on these two sides, each of magnitude given by,
Fl  ilB
Fb  ilB sin 
These forces always lie in the plane of the loop and, depending on the direction of the current in the loop,
are either point towards the center of the loop or point away from the center of the loop.
Therefore only the force acting on the long sides of the loop act as a couple that tends to align the plane of
the loop perpendicular to the magnetic field. The magnitude of the torque of this couple is given by,
  ilBb sin   ilbB sin   iAB sin 
b) If I is the moment of inertia of the loop and  is its angular acceleration due to the torques acting on it,
then,
  iAB sin   I , or,
iAB
sin 
I
The angular acceleration  is maximum for    / 2 , it becomes zero for   0 and    , and it is
negative for 2     .
c) The loop will oscillate in the magnetic field with a frequency that increases with the current i , its
area A and the intensity of the magnetic field B and decreases with its moment of inertia I .
8) Consider a cylindrical solenoid of length l  30cm , radius r  4cm and number of turns N  2000 .
a) Find the magnetic field intensity B inside the solenoid when current i  1A is passed through it.
[1]
b) Calculate the corresponding total magnetic flux through the solenoid.
[1]
c) What is the induced emf in the solenoid when i  cos 6t A ?
d) Find the self-inductance of the solenoid.
[2]
[1]
N
10 7  2000  1
i  4 
 8.4  10 3 T
l
0.3
b) Magnetic flux through the solenoid   NAB  2000    16  10 4  8.4  10 3  8.4  10 2 W
c) Flux through the solenoid for i  cos 6t   8.4  10 2 cos 6t
d
 6  8.4  10 3 sin 6t  0.51sin 6t volt
emf through solenoid=  
dt
di
di
 8.4  10  2 ,
d) Self inductance of solenoid is given by emf  L
dt
dt
2
Therefore L  8.4  10 H
Solution: a) Magnetic field intensity B   0
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l

b
For   0 the other two sides of the loop are parallel to the uniform magnetic field and for   90 0 these
sides are perpendicular to the uniform magnetic field. Therefore for any value of the angle  the
magnitude of each of these equal and opposite forces acting on these two sides is given by,

P
5
B
9. A parallel plate capacitor with C  0.1F and plate separation d  1mm is being charged with
current i  i0 e  t /   e 0.1t A .
a) Find the charge q on the capacitor and the electric field E between its plates as a function of time.
b) Calculate the electric flux through an open surface bounded by a closed circular loop of
radius r between and parallel to the plates of the capacitor.
c) Find the magnetic field intensity at a distance r from the axis of the capacitor for r  R / 2 ,
and r  2 R . Here R is the radius of the circular plates of the capacitor.
t

1


0
0
[2]
[1]
[2]

Solution: a) q  idt  10 6  e 0.1dt  10 5 1  e 0.1t C
q 10 5
Potential difference across the plates of the capacitor V   7 1  e 0.1t  10 2 1  e 0.1t volt
C 10
2
V 10
Electric field across the plates E   3 1  e 0.1t  10 5 1  e 0.1t volt / m
d 10
b)
Electric flux through an open surface bounded by a closed circular loop of radius r in between the
plates  E  r 2 E  10 5 r 2 1  e 0.1t
c)
Magnetic field intensity B at a distance r from the axis of capacitor plates,
d
B   0 0 E   0 0 10 5  0.1e 0.1t r 2   0 0 10 4 e 0.1t r 2T
dt
For r  R / 2 ,
2
4  0.1t R
B  10 e
T
4
For r  2R,
B   0  0 10 4 e 0.1t R 2T



© Air University, 2010







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