RASHIDATU KARGBO, 41290
ELECTROMAGNETIC FIELDS AND WAVES (EENG322)
ASSIGNMENT
2018 & 2020
Past Papers Solutions
2018
1. (a)
Explain Coulomb’s law of electrostatic forces.
7 marks
This law state that the force between two charged bodies, Q1 and Q2, that are very
small compared with the distance of separation, R12, is proportional to the product of
the charges and inversely proportional to the square of the distance, the direction of the
force being along the line connecting the charges.
Mathematically written as,
F12= aR12𝐾𝑄1𝑄2/𝑅122
Where F12 is the Vector Force exerted by Q1 and Q2, aR12, is a unit vector in direction
from Q1 and Q2 and K is a proportionality constant depending on the medium of the
system of units.
(b)
Three identical point charges each 𝑄 coulombs, are placed at the vertices of an
equilateral triangle 10 𝑐𝑚 apart. Calculate the force on each charge. 6 marks
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(c)
Two charges 𝑄 coulombs each are placed at two opposite corners of a square.
What additional charge 𝑞 placed at each of the other two corners will reduce
the resultant electric force on each of the charges 𝑄 to zero?
7 marks
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2. (a)
Explain the term electric field strength.
6 marks
Is defined as the Force per unit charge that a very small Stationary test charge
experiences when it is placed in a region where an electric Field exist, ie
(b)
Derive the electric field equation due to a point charge
(c)
Briefly explain the Gauss’s electric flux theorem.
.
7 marks
7 marks
The total of the electric flux out of a closed surface is equal to the charge enclosed divided by
the permittivity.
The electric flux through an area is defined as the electric field multiplied by the area of the
surface projected in a plane perpendicular to the field. Gauss's Law is a general law applying
to any closed surface. It is an important tool since it permits the assessment of the amount of
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enclosed charge by mapping the field on a surface outside the charge distribution. For
geometries of sufficient symmetry, it simplifies the calculation of the electric field.
Another way of visualizing this is to consider a probe of area A which can measure the electric
field perpendicular to that area. If it picks any closed surface and steps over that surface,
measuring the perpendicular field times its area, it will obtain a measure of the net electric
charge within the surface, no matter how that internal charge is configured.
3. (a)
Define electric potential of an electric field.
3 marks
Electric potential at a point in an electric field, is defined as the amount of work done
in order to move a unit charge from infinity to that point along any path, against the
electrostatic forces, with acceleration zero.
(b)
Show that the work done in bringing a charge from infinity to a point under
consideration is given as
𝑄
𝑉 = 4𝜋𝜀
0
𝑄
= 9 × 109 𝑑 𝑣𝑜𝑙𝑡
𝑑
9 marks
Consider a positive point charge of Q coulombs placed in air. At a point x metres
from it, the force on one coulomb positive charge is Q/4 πε0 x2 (Fig. above).
Suppose, this one coulomb charge is moved towards Q through a small distance dx.
Then, work done is
The negative sign is taken because dx is considered along the negative direction of x.
The total work done in bringing this coulomb of positive charge from infinity to any
point D, which is d metres from Q is given by
By definition, this work in joules in numerically equal to the potential of that point in
volts.
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(c)
A point charge of 10−9 𝐶 is placed at a point A in free space.
Calculate
i.
ii.
The intensity of electrostatic field on the surface of sphere of radius
5 𝑐𝑚 and centre A.
The difference in potential between two points 20 𝑐𝑚 and 10 𝑐𝑚 away
from the charge at A
8 marks
(i). E = Q/4πε0 r2 = 10−9/4π × 8.854 × 10−12 × (5 × 10−2)2 = 3,595 V/m
(ii) Potential of first point =Q/4πε0d = 10−9/4π × 8.854 × 10−12 × 0.2 = 45 V
Potential of second point = 10−9/4π × 8.854 × 10−12 × 0.1 = 90 V
∴ P.d. between two points = 90 − 45 = 45 V
4. (a)
Explain the capacitance of a capacitor
5 marks
The capacitance of a capacitor is the ability of a capacitor to store an electric charge per
unit of voltage across its plates of a capacitor. Capacitance is found by dividing electric
charge with voltage by the formula C=Q/V. Its unit is Farad.
(b)
Show that the total potential of a spherical capacitor consisting of two
concentric spheres is given by
𝑎𝑏
𝐶 = 4𝜋𝜀0 𝜀𝑟 𝑏−𝑎 𝐹
7 marks
Consider a spherical capacitor consisting of two concentric spheres of radii ‘a’ and ‘b’
metres as shown in Fig. above Suppose, the inner sphere is given a charge of + Q
coulombs. It will induce a charge of −Q coulombs on the inner surfaces which will go
to earth. If the dielectric medium between the two spheres has a relative permittivity
of εr, then the free surface potential of the inner sphere due to its own charge
Q/4πε0εra volts. The potential of the inner sphere due to −Q charge on the outer
sphere is −Q/4πε0εrb (remembering that potential anywhere inside a sphere is the
same as that its surface).
Total potential difference between two surfaces is
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(c)
A capacitor consists of two similar square aluminium plates
each 10 × 10 𝑐𝑚 mounted parallel and opposite to each other. What is their
capacitance when distance between them is 1 cm and the dielectric is air? If
the capacitor is given a charge of 500 𝑝𝑖𝑐𝑜 𝑐𝑜𝑢𝑙𝑜𝑚𝑏, what will be the
difference of potential between plates? How will this be affected if the space
between the plates is filled with wax which has a relative permittivity of 4?
8 marks
The capacity of a flat capacitor is C = ϵ⋅ϵ0⋅S / d
Where
ϵ =1
ϵ0=8.85x10−12 F/m
S=0.1⋅0.1m2=1x10−2m2
d=1x10−2m
Then
C = 1x8.85x10−2x1x10−2/1x10-2 = 8.85x10−12F = 0.00885μF
Q = CV; V = Q/C ; Q = 500x10-12C
V = 500/8.85 = 56.5V
ϵ= 4
V = 500/(4x8.85) = 14.1V
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5. (a)
Explain Bio-Savart law.
6 marks
The expression for the magnetic field strength dH produced at point P by a vanishingly small
length dl of a conductor carrying a current of I amperes (Fig. below) is given by
(b)
Show that the magnetic field strength along the axis of a square coil is given by
𝐻=
2𝑎2 .𝐼
𝜋𝑎2.√2.𝑎
=
√2.𝐼
𝐴𝑇⁄𝑚
𝜋𝑎
14 marks
Given four conductors each of length say, 2a metres and carrying a current of I
amperes as shown in Fig. above. The Magnetic Field Strengths at the axial point P due
to the opposite sides ab and cd are Hab and Hcd directed at right angles to the planes
containing P and ab and P and cd respectively. Now, Hab and Hcd are numerically
equal, hence their components at right angles to the axis of the coil will cancel out,
but the axial components will add together. Similarly, the other two sides da and bc
will also give a resultant axial component only
The above diagram gives,
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But
Sub r in (1) gives,
Its axial components is
Now,
In case, value of H is required at the centre O of the coil, then putting x = 0 in the
above expression, we get
6. (a)
Derive the equation of the force between two infinitely long parallel current carrying
conductors.
5 marks
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Consider two infinitely long straight conductors 1 and 2 carrying currents I1 and I2
respectively are suspended by a distance d. The conductor 1 produces the same magnetic field
at all points along the conductor 2
The second conductor experiences a force F21
The force between two infinitely long parallel currently-carrying conductors is given by the
expression
The force per metre run of the conductors is
If I1 = I2 = 1 ampere (say) and d = 1 metre, then F = 2× 10−7 N
(b)
Define the Ampere
4 marks
one ampere current can be defined as that current which when flowing in each of the two
infinitely long parallel conductors situated in vacuum and separated 1 metre between centres,
produces on each conductor a force of 2× 10−7 N per metre length.
(c)
Two long straight parallel wires standing in air 2 𝑚 apart, carrying currents 𝐼1
and 𝐼2 in the same direction. The magnetic intensity at a point midway between the
wires is 7.95 𝐴𝑇⁄𝑚. if the force on each wire per unit length is . 2.4 × 10−4 𝑁,
evaluate 𝐼1 and 𝐼2 .
11 marks
The magnetic intensity of a long straight current-carrying conductor is
when the two currents flow in the same direction, net field strength midway between the two
conductors is the difference of the two field strengths.
Now, H1 = I1/2π and H2 = I2/2π because r = 2/1 = 2 metre
Force per unit length of the conductors is F = 2×10−7 I1I2/d newton
2.4 ×10-4 =2×10−7 I1I2/2 ∴ I1I2 = 2400 … (3)
Substituting the value of I1, we get
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(50 + I2) I2 =2400 or I22 + 50I2 − 2400 = 0
Or (I2 + 80) (I2 − 30) = 0 ∴ I2 = 30 A and I1 = 50 + 30 = 80 A
7. (a)
Show that the force on a carrying-current conductor lying in a magnetic field
is given by
𝐹 = 𝐵𝐼𝐿 𝑠𝑖𝑛 𝜃
8 marks
It is found that whenever a current-carrying conductor is placed in magnetic field, it
experiences a force which acts in a direction perpendicular both to the direction of the current
and the field. In the Fig. above is shown a conductor XY lying at right angles to the uniform
horizontal field of flux density B Wb/m2 produced by two solenoids A and B. If l is the length
of the conductor lying within this field and I ampere the current carried by it, then the
magnitude of the force experienced by it is
F = BIl= μ0μrHIl newton
Using vector notation F=Il×B and F = IlBsinθ where θ is the angle between l and B which is
90º in the present case
Or
(b)
State the law of Ampere’s work done or Ampere’s circuital law.
5 marks
The law states that m.m.f.* (magnetomotive force corresponding to e.m.f. i.e. electromotive
force of electric field) around a closed path is equal to the current enclosed by the path.
Mathematically,
is the vector representing magnetic field strength in dot product with vector ds of the enclosing
path S around current I ampere and that is why line integral of dot product H. ds is taken.
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(c)
Obtain the equation of the magnetomotive force around a long straight
conductor.
7 marks
In the Fig. above is shown a straight conductor which is assumed to extend to infinity in
either direction. Let it carry a current of I amperes upwards. The magnetic field consists of
circular lines of force having their plane perpendicular to the conductor and their centres at
the centre of the conductor.
Suppose that the field strength at point C distant r metres from the centre of the conductor is
H. Then, it means that if a unit N-pole is placed at C, it will experience a force of H newtons.
The direction of this force would be tangential to the circular line of force passing through C.
If this unit N-pole is moved once round the conductor against this force, then work done i.e.
Obviously, if there are N conductors (as shown in the Fig. below), then
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8. (a)
(b)
Derive equation for the capacitance between two parallel wires.
12 marks
The conductors of a two-wire transmission line (4 𝑘𝑚 𝑙𝑜𝑛𝑔) are spaced 45 𝑐𝑚
between centre. If each conductor has a diameter of 1.5 𝑐𝑚, calculate the capacitance
of the line.
8 marks
12
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