Air University
Final Exam Summer - 2012
Course: PH-106 Physics-II
Date: 09-08-2012
Max. Marks: 45
Time Allowed: 3 hrs
Q1. (a) A long, rigidly supported wire carries a current ia of 96 A. Directly above it and parallel to it is a fine
wire that carries a current ib of 23 A. How far above the lower wire should this second wire be hang if the force
per unit length is 0.073 N/m? (μo = 1.26 x 10-6 T.m/A)
(3)
(b) An electrical pulse of i = 10 µA is sent through a short conducting path of length 1 mm of the brain for
recording its activity. Estimate the magnetic field at a distance r = 2 cm from this path.
(3)
Solution: a) Solving the magnetic force per unit length on second wire for d yields
d = μoiaib / 2π(F/L) = [(4π x 10-7 Tm/A)(96 A)(23 A)] / [2π(0.073 N/m)] = 6 x 10-3 m
b) The magnetic field intensity B at a point P at a distance r from a current element dl is given by Biot-Savart law;

B

 0i dl  r
4r 2

Here r is a unit vector pointing from the current element towards the point P . We

have i  10 5 A , dl  10 3 m and r  2  10 2 m . If we assume that the current element is perpendicular to r , then,
B
4 10 7 10 5 10 3
 2.5 10 12 T
4  4 10 4
i
Q2. (a) Magnetic field B = 0.01 T is applied to a strip of material of width d = 5 mm carrying electric
current i = 1 A as shown in Fig. 1. i) Find the expression for direction and magnitude of the force on
the current in terms of i and B. ii) Find the drift velocity of the charge carriers when a voltage V =1
μVolt, is measured across the strip. iii) What is the sign of the charge carriers if the left side of the
strip is positive?
(1+1.25+0.75)
B


b
Figure 1
(b) A straight, horizontal length of copper wire has a current i = 28 A through it. What are the magnitude and
direction of the minimum magnetic field B needed to suspend the wire - that is, to balance the gravitational
force on it? The linear density (mass per unit length) of the wire is 46.6 g/m.
(3)
Solution: a)
i) We chose x-axis point from left to tight and y-axis pointing upwards in the direction of the current. Since
Figthe
1
magnetic field is pointing into the plane of the paper, the x-y plane, the Lorentz force is of the strip is given by,
F
 q vd  B
Here vd is the drift velocity of the charge carriers. With  as the density of charge carriers in the strip and A is
the are of cross-section of the strip perpendicular to the current flow, i  Avd .
Since the magnetic field is perpendicular to the drift velocity of the charge carriers, the magnitude of the of the
magnetic force acting on them is given by,
F  qvd B
The direction of this force will be in the –x direction for positive charge carriers and it will be in the x direction
for negative charge carriers.
ii) From qE  qvd B , we get,
vd 
E V
10 6


 2mm / s
B bB 5 10 3 10 2
iii) As mentioned is part (a) above, positive charge carrier will be pushed to the right by the Lorentz force.
Therefore, when the left side of the strip is positive then the charge carrier must have positive charge.
b) The magnitude of FB is FB = iLB sinθ. Because we want FB to balance Fg, we want
iLB sinθ = mg
where mg is the magnitude of Fg and m is the mass of the wire. We also want the minimal field magnitude B for
FB to balance Fg. Thus, we need to maximize sinφ in above equation. To do so, we set φ = 90o, thereby
arranging for B to be perpendicular to the wire. We then have sinφ = 1, so above Eq. yields
B = mg / iL sinφ = (m/L)g / i
We write the result this way because we know ml L, the linear density of the wire.
Substituting known data then gives us
B = (46.6 x 10-3 kg/m)(9.8 m/s2) / 28 A
B = I.6 X 10-2 T.
This is about 160 times the strength of Earth's magnetic field.
Q3. (a) Fig. 2 shows the cross section of a long conducting cylinder with inner radius a = 2 cm and outer radius
b = 4 cm. The cylinder carries a current out of the page, and the magnitude of the current density in the cross
section is given by J = cr2, with c = 3 x 106 A/m4 and r in meters. What is the magnetic field B at a point that is
3 cm from the central axis of the cylinder?
(3.5)
(b) Fig. 3 gives, as a function of radial distance r, the magnitude B of the magnetic field inside and outside four
wires (a, b, c, and d), each of which carries a current that is uniformly distributed across the wire's cross section.
Overlapping portions of the plots are indicated by double labels. Rank the wires according to i) radius, ii) the
magnitude of the magnetic field on the surface, and iii) the value of the current, greatest first. iv) Is the
magnitude of the current density in wire a greater than, less than, or equal to that in wire c?
(2.5)
Solution: a) We write the integral as
The direction of integration indicated in Fig.2 is (arbitrarily) clockwise. Applying the right-hand rule for
Ampere's law to that loop, we find that we should take ienc as negative because the current is directed out of the
page but our thumb is directed into the page. We next evaluate the left side of Ampere's law and we obtain
Solving for B and substituting known data yield
Thus, the magnetic field E at a point 3.0 cm from the central axis has magnitude
Figure 2
and forms magnetic field lines that are directed opposite our direction of integration, hence counterclockwise
in Fig. 2.
b)
i) c, a, d, b.
ii) a, c, b, d.
iii) a, c, b, d.
iv) J is greater in wire a as compare to wire c.
Figure 3
Q4. (a) Prove that in Fig. 4, the rate at which the mechanical work is done on the
loop by an external agent is dissipated as Joule heating of the loop?
(4)
(b) A coil has an inductance of 53 mH and resistance of 0.35Ω. i) If a 12 V emf is
applied, how much energy is stored in the magnetic field after the current has built up
to its maximum value? ii) In terms of τL, how long does it take for the stored energy
to reach half of its maximum value?
(2+2)
Figure 4
Solution: a)
To pull the loop at a constant velocity /, we must apply a constant force F to the loop because a magnetic force
of equal magnitude but opposite direction acts on the loop to oppose you. The rate at which we do work - that
is, the power - is then
P = Fv
where F is the magnitude of force. We wish to find an expression for P in terms of the magnitude B of the
magnetic field and the characteristics of the loop-namely, its resistance R to current and its dimension L.
As we move the loop to the right in Fig.4, the portion of its area within the magnetic fleld decreases. Thus, the
flux through the loop also decreases and, according to Faraday's law, a current is produced in the loop. It is the
presence of this current that causes the force that opposes the pull.
To find the current, we first apply Faraday's law. When x is the length of the loop still in the magnetic field, the
area of the loop still in the field is Lx. Then the magnitude of the flux through the loop is
As x decreases, the flux decreases. Faraday's law tells us that with this flux decrease, an emf is induced in the
loop. Dropping the minus sign in Eq. 30-4 and using Eq. 30-7, we can write the magnitude of this emf as
To find the magnitude of the induced current, we cannot apply the loop rule for potential differences in a circuit
because, as you will see in Section 30-6, we cannot define a potential difference for an induced emf. However,
we can apply the equation i :81R, as we did in Sample Problem 30-2.With Eq.30-8, this becomes
Because three segments of the loop in Fig. 30-10 carry this current through the magnetic field, sideways
deflecting forces act on those segments. From Eq. 28-26 we know that such a deflecting force is, in general
notation,
In Fig. 4, the deflecting forces acting on the three segments of the loop are marked F1, F2, and F3. Note,
however, that from the symmetry forces F2 and F3 are equal in magnitude and cancel. This leaves only force F1,
which is directed opposite your force F on the loop and thus is the force opposing you. So, F = - F1,
Substituting for i in above Eq., then gives us
Because B, L, and R are constants, the speed v at which you move the loop is constant if the magnitude F of the
force you apply to the loop is also constant.
By substituting, we find the rate at which you do work on the loop as you pull it from the magnetic field:
To complete our analysis, let us find the rate at which thermal energy appears in the loop as you pull it along g
at constant speed. We calculate it from
Substituting for i, we find
which is exactly equal to the rate at which you are doing work on the loop. Thus, the work that you do in
pulling the loop through the magnetic field appears as thermal energy in the loop.
b) i) Thus, to find the energy UB∞ stored at equilibrium, we must first find the equilibrium current.
Then substitution yields
ii) Given condition is
By canceling ξ/R and rearranging, we can write this as
which yields
Q5. (a) An electric current of 1 A passes through a conducting wire of length l = 10 m
and diameter d = 0.2 mm when it is connected to a 2 V power source. i) What is the electric
resistance R of the wire? ii) Calculate the electric resistivity ρ of the material of the wire.
(1+1)
(b) In Fig. 5, take R = 8.5 cm and dB/dt = 0.13 T/s. Find an expressions for the magnitude
E of the induced electric field at points within and outside the magnetic field, at radius r
from the center of the magnetic field. Evaluate the expression for r =5.2 cm and r = 12.5 cm.
Also plot induced electric field E as a function of r.
Figure 5
(2+2)
Solution: a)
V 2
  2
i) R 
i 1
AR 3.14  10 6  2

 6.2  10 5 m
ii)  
l
10
b)
Assume from the symmetry that E has the same value at all points along the circular path. Then the left side of
above Eq. becomes
Right side of Eq. becomes
Substituting this in the first Eq., we find that
This Eq. gives the magnitude of the electric field at any point for which r < R (that is, within the magnetic
field). Substituting given values yields, for the magnitude of E at r = 5.2cm,
We can now write
Substituting this and above 2nd Eq. and solving for E yield
Because E is not zero here, we know that an electric field is induced even at points that are outside the
changing magnetic field, an important result that makes transformers possible.
With the given data, above equ. yields the magnitude of B at r = 12.5 cm:
Q6. (a) A parallel plate capacitor with A = 1 m2, with an air gap d = 1 mm between its plates has a capacitance
of C = 10 pF. i) Find the value of the dielectric constant of the material between the plates. ii) Find the values
of the free charge densities σ on the plates and the induced charge densities on the dielectric material when it is
connected to a 12 volt battery. iii) How much energy is stored in the capacitor and what is the value of the
energy density?
(1+2+2)
(b) Describe why charging of a thin conducting layer to a given surface density requires twice as much charge
as that given to an insulating sheet to have the same surface charge density. Also evaluate the electric field near
a charged conducting sheet.
Solution: a)
i)
C  10 8 F 
8
3
 0 A
d

8.8  10 12  1
, giving,
10 3
10  10
 1.1
8.8  10 12
7
ii) q  CV  1.2  10 C  0.12C

(2)
f 
q
 0.12 Cm  2
A
If E n is the net electric field between the plates, E f is the electric field due to the free charge density and E i is the electric
field due to the induced charge density, then,
Ei 
f V
i
 E f  En 

0
0 d
i   f  o
, giving,
V
12
 1.2  10 7  8.8  10 12  3  1.2  10 7  1.05  10 7  15  10 9 Cm  2
d
10
1
10 8  12  12
2
iii) Energy stored in the capacitor U  CV 
 7.2  10 7 J
2
2
b) This is bcz we have to supply enough charge to cover both surfaces of the conducting sheet in contrast to the
single surface of an insulating sheet.
If we regard each face of conductor as a sheet of charge, it will give an electric field E = σ/2εo
Then at a point outside the conducting sheets the EL from left face and ER from right face add up to give a total
electric field near conductor.
E = EL + ER = σ/2εo + σ/2εo = σ/εo
Q7. (a) Fig. 6 shows two particles fixed in place: a particle of charge q1 = +8q at the origin and a particle of
charge q2 = -2q at x = L. At what point (other than infinitely far away) can a proton be placed so that it is in
equilibrium (the net force on it is zero)? Is that equilibrium stable or unstable?
(3)
(b) Fig. 7 shows three point charges held in fixed positions by forces that are not shown. What is the electric
potential energy U of this system of charges? Assume that d = 12 cm and that q1 = +q, q2 = -4q, and q3 = +2q,
in which q = 150 x 10-9 C.
(3)
Figure 6
Figure 7
Solution: a)
If F1 is the force on the proton due to charge ql and F2 is the force on the proton due to charge q2, then the point
we seek is where F1 + F2 = 0. Thus,
This tells us that at the point we seek, the forces acting on the proton due to the other two particles must be of
equal magnitudes,
and that the forces must have opposite directions.
If the proton is at any point on the x axis to the right of q2, such as point R in below Fig., then F1 and F2, are
again in opposite directions. However, because now the charge of greater magnitude (q1) is farther away from
the proton than the charge of lesser magnitude, there is a point at which F1 is equal to F2. Let x be the coordinate
of this point, and let qp be the charge of the proton.
By using Coulomb law, we can now rewrite
Rearranging above equ.
After taking the square roots of both sides, we have
which gives us
The equilibrium at x = 2L is unstable; that is, if the proton is displaced leftward from point R, then F1 and F2
both increase but F2 increases more (because q2 is closer than q1), and a net force will drive the proton farther
leftward. If the proton is displaced rightward, both F1 and F2 decrease but F2 decreases more, and a net force
will then drive the proton farther rightward. In a stable equilibrium, if the proton is displaced slightly, it returns
to the equilibrium position.
b) Let's mentally build the system of Fig. 7, starting with one of the point charges, say q1, in place and the
others at infinity. Then we bring another one, say q2, in from infinity and put it in place. The potential energy
U12 associated with the pair of point charges q1 and q2 is
We then bring the last point charge q3 in from infinity and put it in place. The work that we must do in this last
step is equal to the sum of the work we must do to bring q3 near q1 and the work we must do to bring it near q2.
The total potential energy U of the three-charge system is the sum of the potential energies associated with the
three pairs of charges. This sum (which is actually independent of the order in which the charges are brought
together) is
The negative potential energy means that negative work would have to be done to assemble this structure,
starting with the three charges infinitely separated and at rest. Put another way, an external agent would have to
do 17 mJ of work to disassemble the structure completely, ending with the three charges infinitely far apart.