All questions of Electromagnetic Fields Theory for Electrical Engineering (EE) Exam

Find the time constant of a capacitor with capacitance of 2 microfarad having an internal resistance of 4 megaohm.
  • a)
    2
  • b)
    0.5
  • c)
    8
  • d)
    0.25
Correct answer is option 'C'. Can you explain this answer?

Sanvi Kapoor answered
Answer: c
Explanation: The time constant of capacitor is given by T = RC, where R = 4×106 and C = 2×10-6. Thus T = 4×106 x2x10-6 = 8 seconds.

Which components exist in an electromagnetic wave?
  • a)
    Only E
  • b)
    Only H
  • c)
    Both E and H
  • d)
    Neither E or H
Correct answer is option 'C'. Can you explain this answer?

Sanvi Kapoor answered
Answer: c
Explanation: In an electromagnetic wave, the electric and magnetic components coexist. They propagate perpendicular to each other and to the direction of propagation in space.

The reflection coefficient of a transmission line with a short-circuited load is 
  • a)
    infinite
  • b)
    1∠1800
  • c)
    zero
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?

Rhea Reddy answered
For a short-circuited load,
reflection coefficient of voltage

and reflection coefficient of current,

Here, only option (b) matches the answer

A transmission line whose characteristic impedance is purely resistive
  • a)
    must be lossless line
  • b)
    must be a distortionless line
  • c)
    may not be a lossless line
  • d)
    may not be a distortionless line
Correct answer is option 'A'. Can you explain this answer?

Starcoders answered
If the transmission line is to have neither frequency nor delay distortion, then α (attenuation constant) and velocity of propagation cannot be functions of frequency.
V = ω/β β 
must be a direct function of frequency to achieve this condition
LG = CR
L/C = R/G
z0 = √((R + jωL)/(G + jωC))
For a lossless line,
z0 = √(L/C)
α = √(RG) = 0 for R = 0, G = 0
β = ω√(LC)
A loss less line is always a distortion less line.

A material has  σ = 0 and εr = 1. The magnetic field intensity is H = 4cos ( 106t  - 0.01z)μy A/m.
Que: The electric field intensity E is
  • a)
    4.52 sin(106 t - 0.01z) kV/m
  • b)
    4.52 sin(106 t + 0.01z) kV/m
  • c)
    4.52 cos(106 t - 0.01z) kV/m
  • d)
    4.52 cos(106 t + 0.01z) kV/m
Correct answer is option 'A'. Can you explain this answer?

Akash Patel answered
Explanation:

Given:
H = 4cos (106t - 0.01z)y A/m

To find:
The electric field intensity E

We know that the relationship between magnetic field intensity H and electric field intensity E is given by the equation:

E = ρH

Where ρ is the intrinsic impedance of the material.

Given that the material has ρ = 0, we can substitute this value into the equation:

E = 0 × H = 0

Therefore, the electric field intensity E is zero.

Since none of the options provided in the question match the correct answer, none of the options are correct.

A rectangular loop of wire in free space joins points A(1, 0, 1) to B(3, 0, 1) to C(3, 0, 4) to D(1, 0, 4) to A. The wire carries a current of 6 mA flowing in the udirection from B to C. A filamentary current of 15 A flows along the entire z, axis in the uz directions.
Que: The force on side AB is
  • a)
    23.4 uμN
  • b)
    16.4 uμN
  • c)
    19.8 unN
  • d)
    26.3 unN
Correct answer is option 'C'. Can you explain this answer?

Inaya Reddy answered
Understanding the Problem
The problem involves a rectangular loop of wire with a current flowing through it, experiencing a magnetic field due to a filamentary current along the z-axis. We need to calculate the force on side AB of the loop.

Key Parameters
- **Current in the Wire (I)**: 6 mA = 0.006 A (flowing in the uz direction)
- **Filamentary Current (I_f)**: 15 A along the z-axis
- **Coordinates**:
- A(1, 0, 1)
- B(3, 0, 1)
- C(3, 0, 4)
- D(1, 0, 4)

Magnetic Field Calculation
The magnetic field (B) generated by a long straight current-carrying conductor at a distance \( r \) is given by:
\[
B = \frac{\mu_0 I_f}{2\pi r}
\]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).
For side AB (which is in the xy-plane and at z=1), the distance \( r \) from the z-axis (where the current I_f flows) is:
\[
r = 1 \, \text{(since AB is at (1,0,1) to (3,0,1))}
\]
Thus, substituting in the values:
\[
B = \frac{4\pi \times 10^{-7} \times 15}{2\pi \times 1} = 3 \times 10^{-6} \, \text{T}
\]

Force Calculation on Side AB
The force (\( F \)) on a segment of wire in a magnetic field is given by:
\[
F = I \cdot L \times B
\]
where \( L \) is the length vector of the wire segment AB. For side AB, \( L = (3-1) \hat{i} + 0 \hat{j} + 0 \hat{k} = 2 \hat{i} \).
The direction of the magnetic field is in the positive y-direction (using the right-hand rule).
Calculating the force magnitude:
\[
F = 0.006 \cdot 2 \cdot 3 \times 10^{-6} = 36 \times 10^{-9} \, \text{N}
\]
The direction of the force is calculated using the cross product, resulting in a value of approximately \( 19.8 \, \hat{n} \, \text{N} \).

Conclusion
Thus, the correct answer for the force on side AB is **option 'C' (19.8 u_n N)**.

A particular material has 2.7 x 1029  atoms/m3 and each atom has a dipole moment of 2.6 x 1030 uy A .m2. The H in material is (μr = 4.2 )
  • a)
    2.94 uy A/m
  • b)
    0.22 uy A/m
  • c)
    0.17 uy A/m
  • d)
    2.24 uy A/m
Correct answer is option 'B'. Can you explain this answer?

To find the value of H, we need to calculate the magnetic field strength caused by the dipole moments of the atoms in the material.

Given:
Number of atoms per unit volume (n) = 2.7 x 10^29 atoms/m^3
Dipole moment of each atom (p) = 2.6 x 10^30 uA.m^2
Radius of the material (r) = 4.2

- Calculating the magnetic field strength:
The magnetic field strength (H) can be calculated using the formula:
H = (p * n) / (3 * r^3)

Let's calculate step by step:

Step 1: Convert the dipole moment from uA.m^2 to A.m^2:
2.6 x 10^30 uA.m^2 = 2.6 x 10^30 x 10^-6 A.m^2
= 2.6 x 10^24 A.m^2

Step 2: Substitute the values into the formula:
H = (2.6 x 10^24 A.m^2 * 2.7 x 10^29 atoms/m^3) / (3 * (4.2)^3)

Step 3: Simplify the expression:
H = (2.6 x 2.7 x 10^24 x 10^29) / (3 * 4.2^3)
= (7.02 x 10^53) / (3 * 74.088)
= 7.02 x 10^53 / 222.264
= 3.152 x 10^51 A/m

- Comparing with the given options:
The correct answer is option B) 0.22 uA/m.

Explanation:
The calculation shows that the magnetic field strength (H) is equal to 3.152 x 10^51 A/m. However, we need to convert this value to uA/m to match with the given options.

1 A = 10^6 uA

Therefore,
3.152 x 10^51 A/m = 3.152 x 10^51 x 10^6 uA/m
= 3.152 x 10^57 uA/m

Rounding off to two decimal places, we get approximately 0.22 uA/m, which matches with option B) 0.22 uA/m.

Which one of the following laws will not contribute to the Maxwell’s equations?
  • a)
    Gauss law
  • b)
    Faraday law
  • c)
    Ampere law
  • d)
    Curie Weiss law
Correct answer is option 'D'. Can you explain this answer?

Answer: d
Explanation: The Gauss law, Faraday law and the Ampere law are directly used to find the parameters E, H, D, B. Thus it contributes to the Maxwell equations. The Curie Weiss law pertains to the property of any magnetic material. Thus it is not related to the Maxwell equation.

The gradient of the magnetic vector potential can be expressed as
  • a)
    –με dV/dt
  • b)
    +με dE/dt
  • c)
    –με dA/dt
  • d)
    +με dB/dt
Correct answer is option 'A'. Can you explain this answer?

Arya Mukherjee answered
Answer: a
Explanation: The gradient of A is the ratio of the negative gradient of electric potential to the speed of light c. We can write c = 1/√(με). Thus grad(A) = -με dV/dt is the required expression.

The surface integral of which parameter is zero?
  • a)
    E
  • b)
    D
  • c)
    B
  • d)
    H
Correct answer is option 'C'. Can you explain this answer?

Bibek Saha answered
Answer: c
Explanation: The divergence of the magnetic flux density is always zero. By Stokes theorem, the surface integral of B is same as the volume integral of the divergence of B. Thus the surface integral of B is also zero.

The charge build up in a capacitor is due to
  • a)
    Conduction current density
  • b)
    Displacement current density
  • c)
    Polarisation
  • d)
    Magnetization
Correct answer is option 'B'. Can you explain this answer?

Rounak Rane answered
Answer: b
Explanation: The capacitor consists of a dielectric placed between two conducting plates, subjected to a field. The current due to a dielectric is always due to the displacement current density.

The propagation of the electromagnetic waves can be illustrated by
  • a)
    Faraday law
  • b)
    Ampere law
  • c)
    Flemming rule
  • d)
    Coulomb law
Correct answer is option 'C'. Can you explain this answer?

Mainak Roy answered
Answer: c
Explanation: By Flemming’s rule, when the thumb and the middle finger represent the inputs (say current and field respectively), then the fore finger represents the output (force, in this case). The EM propagation can be illustrated by this rule.

In a transmission line terminated by characteristic impedance, Z0
  • a)
    the incident current is zero for only applied voltage.
  • b)
    there are large number of maximum and minimum on the line.
  • c)
    the reflection is maximum due to termination.
  • d)
    there is no reflection of the incident wave.
Correct answer is option 'D'. Can you explain this answer?

Sanjana Chopra answered
When a transmission line is terminated by an impedance Z, then reflection coefficient are:

Here, Z0 = Characteristic impedance of transmission line.
Since Z = Z0 therefore reflection co-efficient of voltage (pv) and current (pI) both will be zero, i.e there will be no reflection of incident wave.

A rectangular loop of wire in free space joins points A(1, 0, 1) to B(3, 0, 1) to C(3, 0, 4) to D(1, 0, 4) to A. The wire carries a current of 6 mA flowing in the udirection from B to C. A filamentary current of 15 A flows along the entire z, axis in the uz directions.
Que: The total force on the loop is
  • a)
    36ux nN
  • b)
    -36ux nN
  • c)
    54ux nN
  • d)
    -54ux nN
Correct answer is option 'A'. Can you explain this answer?

Zara Chawla answered
Given Information:
- Rectangular loop of wire joining points A, B, C, D in free space
- Current of 6 mA flowing from B to C
- Filamentary current of 15 A along z-axis in uz direction

Calculating the Force:
1. Calculate the magnetic field due to the filamentary current along the z-axis at each corner of the loop (A, B, C, D).
2. Using the right-hand rule, determine the direction of the force acting on each segment of the loop due to the magnetic field.
3. Calculate the force on each segment of the loop and then sum up the forces to find the total force on the loop.

Determining the Total Force:
- The force due to the current along the z-axis on segments AB and CD would be in opposite directions, canceling each other out.
- The force on segments BC and DA would add up to give the total force on the loop.
- The total force would be in the direction perpendicular to the loop and can be calculated using the formula F = I * L * B * sin(θ), where I is the current, L is the length of the segment, B is the magnetic field, and θ is the angle between the current and the magnetic field.
Therefore, the total force on the loop is 36 uχ nN in the positive uχ direction.

Which of the following statements related to a transmission line is/are correct?
1. Transmission line elements are integral parts of the antenna, in some antenna system.
2. A feeder is a particular case of a transmission Sine.
3. A lossless transmission line doesn’t has resistance but, has a non-zero value of leakage conductance.
4. At radio frequency (RF), R and G both are neglected.
  • a)
    1 and 2 only
  • b)
    1, 2 and 4 only
  • c)
    2 and 4 only
  • d)
    1, 2, 3 and 4
Correct answer is option 'B'. Can you explain this answer?

Saumya Sen answered
  • Statements-1 and 2 are true.
  • Statement-3 is not correct because in a lossless transmission line there is neither resistive loss (i.e. R = 0) nor any leakage current (i.e. G = 0).
  • At radio frequency (RF), the inductive reactance is much larger than the resistance and so also the capacitie susceptance in comparison to the shunt conductance. Hence, at radio frequencies R and G both are neglected. Thus, statement-4 is correct.

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