Example 1
Given that D̅ = r2a̅ + 2sinθar ̅̅̅ in Spherical Coordinate System, where D is the electric flux θ density, find the charge density ρ?
Solution:
Given:
ρv= 4 (r +1
rcos θ)
Electric Flux Density, D⃗⃗ = Dra⃗ r+ Dθa⃗ θ+ Dϕa⃗ ϕ In Spherical Co-ordinate System,
Where Dr= r2 , Dθ= 2 sin(θ) , Dϕ = 0
The relation between D⃗⃗ and volume charge density ρ, is given by
∇ . D⃗⃗ = ρv… … … (1) Which is point form of Gauss law
∇ . D⃗⃗ in Spherical Co-ordinates is given by
∇ . D⃗⃗ = 1 r2
∂
∂r(r2Dr) + 1 r sinθ
∂
∂θ(Dθsin θ) + 1 r sinθ
∂Dϕ
∂ϕ
∴ ρv = 1 r2
∂
∂r(r2r2) + 1 r sinθ
∂
∂θ(2 sin2θ) + 0 = 1
r24r3+ 1
r sinθ 2 × 2 sinθ cos θ = 4r + 4
rcos θ
∴ ρv = 4 (r +1
rcos θ)
Bm = −a Hm+ b Bm
B
Hm H
Example 2
Evaluate the integral, ∫ 𝑟 C⃗⃗ . d⃗ r, where C is the helical path described by, x = cos t, y = sin t, z = t, joining the points given by t = 0 and t = π/2
Solution:
Let I = ∫ r
c
. dr
Where C is the helical path described by x = cos (t), y = sin (t), z = t joining the point P at t = 0 and the point, Q at t = π/2.
Position vector, r in Cartesian co-ordinates is given by r = x a⃗ x+ y a⃗ y+ z a⃗ z and dr = dx a⃗ x+ dy a⃗ y+ dz a⃗ z
∴ I = ∫(x dx + y dy + z dz) =1
2[x2+ y2+ z2]PQ
c
With t = 0, P(x, y, z) = (1, 0, 0)
With t = π/2, Q(x, y, z)=(0,1,π/2) I =1
2[x2|10+ y2|01+ z2|0π/2] =1
2[−1 + 1 +π2 4] =π2
8 Example 3
Three electrostatic point charges are located in the xy-plane as given below +Q at (−a/2, 0), +Q at (a/2, 0) and −2Q at (0, a√3/2)
Calculate the co-ordinates of the point, P, on the y-axis, where the potential due to these charges is zero. Also, calculate the magnitude of the electric field strength at p. At the point, P, what is the angle between the equi-potential passing through p and the y-axis?
Solution:
Three point charges + Q at A (−a
2, 0) , +Q at B (a
2, 0) and − 2Q at C (0,√3 2 a) are located in the xy-plane as shown in figure below.
(i) Let P(0, y) be the point on y-axis, where the potential, V = V1+ V2+ V3 due to these charges is zero.
V = V1+ V2+ V3= 0 … … … . . (1) (a
2, 0) OC = √3
2 a P(0, y)
+Q +Q
–2Q y
O x
A
C
(−a 2, 0)
B
z
V1= Q
Find equation (5), V at(0,y) is given by
V = 2Q
(iii) The direction of E⃗⃗ at P is the direction of the normal to the equipotential surface (V = 0) at that point, in the direction of the decreasing values of V.
From equation (6) it follows that the direction of E⃗⃗ is in the –ve y-direction.
∴ The angle between the equipotential surface and the y-axis is zero.
Example 4
Given an irrotational vector field
F⃗ = (k1 xy + k2z3)a⃗ x + (3x2− k3z)a⃗ y+ (3xz2− y)a⃗ z Find ∇. F⃗ at(1,1, −2)
Solution:
Given: Irrotational vector field, F⃗ = Fxa⃗ x+ Fya⃗ y+ Fza⃗ z,
Where Fx = (k1x y + k2z3), Fy= (3 x2− k3z), Fz= (3 x z2− y) In Cartesian coordinates, Div (F⃗ ) = ∇. F⃗ is given by
∇. F⃗ = ∂
∂x(Fx) + ∂
∂y(Fy) + ∂
∂z(Fz)
∂
∂xFx = k1y , ∂
∂yFy= 0, ∂
∂zFz = 6 x z At the point (x, y, z). ∇ . F⃗ = k1y + 6 x z At (1, 1, −2), ∇ . F⃗ = (k1− 12)
Example 5
Given E = 10e−j(4x−kt)ŷ V/m in free space.
(A) Write all the four maxwell’s equations in free space (B) Find ∇ × E
(C) Find H
Solution: Maxwell’s equations are Electro magnetic equations relating the field variables, which are vectors: E⃗⃗ , H⃗⃗ , B⃗⃗ and D⃗⃗
Point or Differential Form Integral Form
∇ x H⃗⃗ = ∂
∂t D⃗⃗ + J ∮ H⃗⃗ . dl⃗⃗⃗ = ∫ (∂
∂t D + J )
s
. (ds) (i)
∇ x E⃗⃗ = − ∂
∂t B⃗⃗ ∮ E⃗⃗ . dl⃗⃗⃗ = ∫ − ∂
s ∂t
B⃗⃗ . ds⃗⃗⃗⃗ (ii)
∇ . D⃗⃗ = ρ ∮D.
s d s = ∫ρdV
v (iii)
∇ . B⃗⃗ = 0 ∮B. d s
s
= 0 (iv) Contained in the above is the equation osf continuity,
∇ . J = − ∂
∂t ρ ∮ J . ds⃗⃗⃗⃗ = ∫ −∂
∂tρ dv
Characteristics of the medium in which the fields exist given rise to the following Constitutive relation:
D⃗⃗ = ϵ E⃗⃗ , B⃗⃗ = μ H⃗⃗ , J = σ E⃗⃗
For a homogenous, isotropic and source free (no impressed voltages or currents) medium
In free space medium i.e., perfect dielectric containing no charges, and no conduction currents
Put σ = 0, J = 0, ρ = 0, ε = ε0 = 1
36π×10−9F
m , μ = μ0 = 4π × 10−7H/m In the above equations.
(a) The given field, E⃗⃗ = 10 e−j(4x−kt) a 𝑦̂ V/m ………..(1) represents a uniform plane wave travelling in the x-direction with velocity v = k/4 and having components, Ex = 0, Ez= 0, Ey= 10e−j(4x−kt)
∇ × E⃗⃗ = ||
a⃗ x a⃗ y a⃗ z
∂
∂x
∂
∂y
∂
∂z Ex Ey Ez
|| = ||
a x̂ a ŷ a ẑ
∂
∂x
∂
∂y
∂
∂z
0 Ey 0
||
= a⃗ x(0) − a⃗ y(0) + a⃗ z ∂
∂xEy
∴ ∇ × E⃗⃗ = −j 40 e−j(4x−kt)= 40 e−j(4x−kt+90o)aẑ
(b) For uniform plane wave E
H= η
As E⃗⃗ = Ey a⃗ y , H⃗⃗ can have only z-component so that wave travels in the x-direction.
∴ Ey
Hz = η = η0= (120π)Ω for free space Hz = Ey
120π= 10
120πe−j(4x−kt), from equation (2) H⃗⃗ = Hz a⃗ z= 1
12πe−j(4x−kt) a ẑ A m
∴ H⃗⃗ = 26.5 ej(kt−4x) aẑ mA/m Example 6
A system of three electric charges lying in a straight line is in equilibrium. Two of the charges are positive with magnitudes Q and 2Q, and are 50 cm apart. Determine the sign, magnitude and position of the third charge.
Solution:
Let Q3 be the third charge located at a distance, x from the first charge, Q
V1= 2 Q
4πε50+ Q3 4πεx V2= Q
4πε50+ Q3
4πε(X − 50) V3= Q
4πεx+ 2Q
4πε(x − 50)
P. E of the field = Q V1+ 2Q V2+ 3Q V3= 0, for equilibrium 50 cm
Q 2Q Q3
1 2 3
x
Q [ 2Q
4πε50+ Q3
4πεx] + 2Q [ Q
4πε50+ Q3
4πε(x − 50)] +Q3[ Q
4πεx+ 2Q
4πε(x − 50)] = 0 2Q
50+Q3 x +2Q
50+ 2Q3
x − 50+ Q3[1
x+ 2
(x − 50)] = 0 4Q
50+ Q3[1
x+ 2
x − 50+1
x+ 2
x − 50] = 0 4Q
50 = −Q3[2
x+ 4
x − 50] = −Q3[6x − 100 x(x − 50)] 4Q
50 = −Q3, Q3= − 2 25Q x2− 50 x = 6x − 100 x2− 56 x + 100 = 0 x =56 ± √562− 400
2 =56 ± 52.3
2 = 54.15, 1.85
∴ x = 54.15 cm
∴ Q3 should be a negative charge with magnitude 0.08 Q and located at x = 54.15 cm from Q and 4.15 cm from 2Q.
Example 7
Consider a circular cylinder of radius 1 meter and length 0.75 meters in free space (intrinsic impedance η = 120 π ohms) with its oriented along the z – direction. Let ȧx, ȧy and ȧz denote the unit vectors in the x, y and z directions respectively. An EM wave is propagating in the positive z-direction with its electric field E = cos [2π f(t − z/c )] ȧx volts/m, where f = 100 MHz and c is the velocity of light 3 × 108m/s). Determine (i) the pointing vector P, and (ii) the net power flux intering the cylinder.
Solution:
Given E⃗⃗ = Ex a⃗ x
With Ex = 1 cos[2π f(t − z/c)]
|Ex| = 1 V/m
(i) As E⃗⃗ is having only a⃗ x component with |Ex| = 1, the corresponding magnetic field should have only y component so that the EM Wave travels in +ve z-direction.
H⃗⃗ = |Hy| cos [2πf (t −z c)] aŷ
|Ey| and |Hy| are related by
0.75m
1m z
x
y
|Ex|
|Hy|= η0= (120π)Ω
∴ H⃗⃗ = 1
120πcos [2πf (t −z c)] aŷA
m The pointing vector, P⃗⃗ is given by P⃗⃗ = E⃗⃗ × H⃗⃗ = 1
120πcos2[2πf (t −z
c)] aẑ W/m2
(ii) Power enters from the bottom plane area = (π) m2 of the cylinder and leaves from the top plane area of the cylinder.
Power flow through the curved surface is zero.
∴ Net average power flux entering the cylinder P = 1
T 1
120π π ∫ cos2
T
t=0
(2πf (t −z
c)) dt, where T = 1/f sec = 1
120 T T 2= 1
240 W
Assignment 1
1. Maxwell’s divergence equation for the electric field is
(A) ∇.E = ρ
electromagnetic wave passing from quartz (μ= μ0, ε = 4ε0) into air is
turns has an inductance ‘L’. If a new coil is formed by having 2n turns while retaining the coil length and diameter at the same value as before, the inductance of the new coil will be (A) L
4. A parallel plate capacitor of plate area A and plate separation ‘t’ has a capacity C.
If a metallic plate P of area A and of negligible thickness is introduced in the capacitor at a distance t
2 from either of the two plates, as shown in the given figure, then the capacity of the capacitor will come uniformly distributed charge of density 2 nC/m2. The electric field intensity at
(C) Single magnetic pole cannot exist.
(D) B is solenoidal
8. The displacement flux density at a point on the surface of a perfect conductor is D⃗⃗ = 2(âx− √3âz) C/m2 and is pointing away form the surface. The surface charge density at that point (C/m2) well be sense of increasing ϕ in a magnetic field B̂=10−5 (2âx − 2ây + â ) Wb/mz 2 is:
10. The given figure shown the surface charge distribution of q Coulombs/m2. What is the force on a unit charge placed at the centre of circle?
(A) Zero
(Applications) and select the correct answer using the codes given below the lists:
2. Force due to a current carrying conductor
3. Electric flux density at a point.
4. Magnetic flux density at a point Codes
13. Two coils have self-inductances of 0.09 H and 0.01 H and a mutual inductance
The electric field intensity at point C(4, 0, 0) is in the
(A) Negative x –direction (B) Negative z-direction (C) Positive x-direction (D) Positive z-direction
16. The force between two point charges of 1 nC each with a 1mm separation in air is
constant current of 2 μA for 6 seconds.
The voltage across the capacitor at the inductance 0.4 H is connected to a 50 V d.c supply. The energy stored in the
19. The circulation of H⃗⃗ around the closed material measurement yield a SWR of 5 and the appearance of an electric field minimum at 4λ in front of the interface.
The impedance of material is (A) 100−j300 Ω
(A) The divergence of the field is equal to zero
9. Two long parallel wires in free space are separated by a distance R and carry currents of equal magnitude but opposite in direction. At any general point, the Z – component of
(A) The magnetic vector potential is
μ0I
4πln(d22/d12)
(B) The magnetic induction is
μ0I
2π(d2/d1)
(C) The magnetic induction is zero (D) The magnetic vector potential is
μ0I
4π(d22/d12)
10. On either side of a charge – free interface between two media
(A) The normal components of the electric field are equal
(B) The tangential components of the electric field are equal
(C) The normal components of the electric flux density are equal (D) The tangential components of the
electric flux density are equal 11. Vector potential is a vector
(A) Whose curl is equal to the magnetic flux density 100 mV/meter. The point charge is now
enclosed by a perfectly conducting hollow metal sphere with its centre at the origin, 0. The electric field strength at the point, P,
(A) Remains unchanged in its magnitude and direction (B) Remains unchanged in its
magnitude but reverse in direction (C) Would be that due to a dipole
14. The incoming solar radiation at a place
on the surface of the earth is 1.2 kW/m2. The amplitude of the
electric field corresponding to this incident power is nearly equal to
(A) 80 mV/m
(B) 2.5 V/m (C) 30 V/m (D) 950 V/m
15. Given V⃗⃗ = x cos2 y î + x2 ez ĵ + z sin2 y k̂
and S the surface of a unit cube with one corner at the origin and edges parallel to the coordinate axes, the value of the
E (r=2R) is __________
d1 d2