Electromagnetic Theory MCQ Quiz in বাংলা - Objective Question with Answer for Electromagnetic Theory - বিনামূল্যে ডাউনলোড করুন [PDF]

Last updated on Mar 11, 2025

পাওয়া Electromagnetic Theory उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). এই বিনামূল্যে ডাউনলোড করুন Electromagnetic Theory MCQ কুইজ পিডিএফ এবং আপনার আসন্ন পরীক্ষার জন্য প্রস্তুত করুন যেমন ব্যাঙ্কিং, এসএসসি, রেলওয়ে, ইউপিএসসি, রাজ্য পিএসসি।

Latest Electromagnetic Theory MCQ Objective Questions

Top Electromagnetic Theory MCQ Objective Questions

Electromagnetic Theory Question 1:

The electric potential on the boundary of a spherical cavity of radius R, as a function of the polar angle θ, is \(V_0 \cos ^2 \frac{\theta}{2}\). The charge density inside the cavity is zero everywhere. The potential at a distance R/2 from the centre of the sphere is

  1. \(\frac{1}{2} V_0\left(1+\frac{1}{2} \cos \theta\right)\)
  2. \(\frac{1}{2} V_0 \cos \theta\)
  3. \(\frac{1}{2} V_0\left(1+\frac{1}{2} \sin \theta\right)\)
  4. \(\frac{1}{2} V_0 \sin \theta\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{1}{2} V_0\left(1+\frac{1}{2} \cos \theta\right)\)

Electromagnetic Theory Question 1 Detailed Solution

Explanation:

Given: \(V=V_0cos^2\frac {\theta}{2}\)

  • \(V=V_{in}+V_{out}=\displaystyle \sum_{l=0}^{\infty} A_lr^lP_l(cos\theta)+\displaystyle \sum_{l=0}^{\infty}\frac{B_l P_l(cos\theta)}{r^{l+1}}\)

Now, at \(R=r\)\(V=V_0cos^2\frac {\theta}{2}\) (given)

  • We will take potential inside the cavity only so \(V=V_{in}=\displaystyle \sum_{l=0}^{\infty} A_lr^lP_l(cos\theta)\)
  • \(V=V_0cos^2\frac {\theta}{2}\)\(=V_{in}=\displaystyle \sum_{l=0}^{\infty} A_lr^lP_l(cos\theta)\)\

Now, \(cos\theta=(2cos^2\frac{\theta}{2}-1)\)  =>  \(cos^2\frac{\theta}{2}=\frac{1+cos\theta}{2}\)

\(V_0(\frac{1+cos\theta}{2})=A_0R^0P_0+A_1R P_1\)----------------------1(For \(l=1 \))

  • Now, according to Legendre polynomial, \(P_0=1, P_1=cos\theta\)
  • Substitute values of \(P_0, P_1\) in equation 1, we get,

\(\frac {V_0}{2}(P_0+P_1)\)\(=A_0R^0P_0+A_1R P_1\)

  • Compare values of \(P_0, P_1\), to find the values of \(A_0 , A_1 \), we get,

\(A_0=\frac{V_0}{2}, A_1=\frac{V_0}{2R}\)

\(V_{in}=A_0r^0.1+A_1r^1(cos\theta)\) (for \(l=1 \))

  • Substitute values of \(A_0 , A_1 \)in above equation, we get,

\(V_{in}=\frac{V_0}{2}(1+\frac{r}{R}(cos\theta))\)

  • Potential at a distance R/2 from the center of the sphere i.e. for \(\frac{r}{R}=\frac{1}{2}\)

\(V_{in}=\frac{V_0}{2}(1+\frac{1}{2}(cos\theta))\)

So, the correct answer is \(V_{in}=\frac{V_0}{2}(1+\frac{1}{2}(cos\theta))\).

Electromagnetic Theory Question 2:

A small circular wire loop of radius a and number of turns N, is oriented with its axis parallel to the direction of the local magnetic field B. A resistance R and a galvanometer are connected to the coil, as shown in the figure.

F1 Teaching Arbaz 23-10-23 D15
When the coil is flipped (i.e., the direction of its axis is reversed) the galvanometer measures the total charge Q that flows through it. If the induced emf through the coil E= IR, then Q is

  1. πNa2В/(2R)
  2. πNa2В/R
  3. √2πNa2В/R
  4. 2πNa2В/R

Answer (Detailed Solution Below)

Option 4 : 2πNa2В/R

Electromagnetic Theory Question 2 Detailed Solution

Concept:

When the coil is flipped (i.e., the direction of its axis is reversed) the angle changes from 0° to \(\pi\) and for the value of induced emf, we will use the induced emf formula in terms of rate of change of magnetic flux.

Formula Used-

  • \(E_F=\frac {-d\phi} {dt}\) where \(\phi=\)magnetic flux
  • \(\phi=NBAcos\theta\)

 

Explanation:

  • B = magnetic field
  • \(E_F=IR\)
  • \(A=\pi r^2\)=Area of circular coil
  • N=Number of turns in the coil

Now, take the formula of emf in terms of magnetic flux,

  • \(E_F=\frac {-d\phi} {dt}\)
  • \(\phi=NBAcos\theta\)
  • \(E_F=-\frac{d} {dt}(NBAcos\theta)=-NBA\frac {d} {dt}(cos\theta)\)---------1

Now, \(E_F=IR\)(Given)

Equation 1, becomes

  • \(I.R=-NBA\frac {d} {dt}(cos\theta)\), take \(I=\frac{dq} {dt}\)
  • \(\frac{dQ} {dt}.R=-NBA\frac{d} {dt}(cos\theta)\)
  • \(dQ=\frac{-NBA} {R}.d(cos\theta)\)

Taking integration on both sides and taking the limits from 0° to \(\pi\), we get,

  • \(\int dQ=\frac{-NBA} {R}.\int_0^\pi d(cos\theta)\)
  • \(Q=\frac{-NBA} {R}.[(cos\theta)]|_0^\pi\)
  • \(Q=\frac{-NBA} {R}.[cos\pi-cos0]\)
  • \(Q=\frac {-NBA}{R}.(-2)=\frac{2NBA} {R}\)

Put \(A=\pi r^2\), we get

  • \(Q=\frac {2NB\pi a^2} {R}\)

So, the correct answer is \(Q=\frac {2NB\pi a^2} {R}\).

Electromagnetic Theory Question 3:

A charged particle moves uniformly on the xy-plane along a circle of radius a centred at the origin. A detector is put at a distance d on the x-axis to detect the electromagnetic wave radiated by the particle along the x- direction. If d >> a, the wave received by the detector is

  1. unpolarised
  2. circularly polarized with the plane of polarization being the yz-plane
  3. linearly polarized along the y-direction
  4. linearly polarized along the z-direction

Answer (Detailed Solution Below)

Option 3 : linearly polarized along the y-direction

Electromagnetic Theory Question 3 Detailed Solution

Option: (3) linearly polarized along the y-direction

Explanation:

A charged particle is moving uniformly on the xy-plane along a circle of radius centered at the origin. A charged particle executes its orbit with angular frequency \(\omega\)(rad/s), and a detector is put at a distance d on the x-axis to detect the electromagnetic wave radiated by the particle along the x-direction. The detector detects an electric field vector \(E\) lying in the y-plane of the orbit and oscillates sinusoidally in magnitude with the orbital frequency and the magnetic field is in the z-direction. At d>>a, the wave received by the detector is linearly polarized along the y-direction.

 

 

F1 Teaching Arbaz 23-10-23 D12

Electromagnetic Theory Question 4:

A perfectly conducting fluid, of permittivity ε and permeability μ, flows with a uniform velocity v in the presence of time dependent electric and magnetic fields E and B respectively. If there is a finite current density in the fluid, then

  1. \(\nabla \,{\rm{ \times }}\,\left( {{\rm{v}}\,{\rm{ \times }}\,{\rm{B}}} \right)\, = \,\frac{{\partial {\rm{B}}}}{{\partial {\rm{t}}}}\)
  2. \(\nabla \,{\rm{ \times }}\,\left( {{\rm{v}}\,{\rm{ \times }}\,{\rm{B}}} \right)\, = \, - \frac{{\partial {\rm{B}}}}{{\partial {\rm{t}}}}\)
  3. \(\nabla \,{\rm{ \times }}\,\left( {{\rm{v}}\,{\rm{ \times }}\,{\rm{B}}} \right)\, = \,\sqrt { \varepsilon {\rm{\mu }}} \frac{{\partial {\rm{E}}}}{{\partial {\rm{t}}}}\)
  4. \(\nabla \,{\rm{ \times }}\,\left( {{\rm{v}}\,{\rm{ \times }}\,{\rm{B}}} \right)\, = \, - \sqrt { \varepsilon {\rm{\mu }}} \frac{{\partial {\rm{E}}}}{{\partial {\rm{t}}}}\)

Answer (Detailed Solution Below)

Option 1 : \(\nabla \,{\rm{ \times }}\,\left( {{\rm{v}}\,{\rm{ \times }}\,{\rm{B}}} \right)\, = \,\frac{{\partial {\rm{B}}}}{{\partial {\rm{t}}}}\)

Electromagnetic Theory Question 4 Detailed Solution

CONCEPT:

In uniform velocity we are having zero force, the electric field is written as;

F = 0

\(qE = -q(\vec v \times \vec B)\)

⇒ \(E = -(\vec v \times \vec B)\)

and third maxwell's equation is written as;

\(\vec {\nabla} \times \vec E = -\frac{\partial \vec B}{\partial t}\)

CALCULATION:

As we have;

\(E = -(\vec v \times \vec B)\) -----(1)

and \(\vec E = -\frac{\partial \vec B}{\partial t}\) -----(2)

Now, from equations 1) and 2) we have;

\(\frac{\partial \vec B}{\partial t}=\vec \nabla( \vec v \times \vec B)\)

Hence option 1) is the correct answer.

Electromagnetic Theory Question 5:

Suppose that isolated magnetic charges (magnetic monopoles) exist. Maxwell's equations (ONLY MODIFIED) including contributions from a magnetic charge density ρm and a magnetic current density jis 

(Assume that, except for the sources, the fields are in vacuum)

  1. \( \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} +\mathbf{j}_m \) and \( \nabla \times \mathbf{B} = \mu_0 \mathbf{j}_e - \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \)
  2. \( \nabla \times \mathbf{E} = \frac{\partial \mathbf{B}}{\partial t} + \mathbf{j}_m \) and \( \nabla \times \mathbf{B} = \mu_0 \mathbf{j}_e + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \)
  3. \( \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} - \mathbf{j}_m \) and \( \nabla \times \mathbf{B} = \mu_0 \mathbf{j}_e + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \)
  4. \( \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} - \mathbf{j}_m \) and \( \nabla \times \mathbf{B} = \mu_0 \mathbf{j}_e - \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \)

Answer (Detailed Solution Below)

Option 3 : \( \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} - \mathbf{j}_m \) and \( \nabla \times \mathbf{B} = \mu_0 \mathbf{j}_e + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \)

Electromagnetic Theory Question 5 Detailed Solution

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Explanation:

To include magnetic monopoles in Maxwell's equations, we need to modify the equations to account for magnetic charge density \( p_m \) and magnetic current density \( \mathbf{j}_m \). The modified equations are as follows:

1. Gauss's Law for Electric Fields:

\( \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} \)

Here, \( \rho_e \) is the electric charge density, and \( \epsilon_0 \) is the permittivity of free space.

2. Gauss's Law for Magnetic Fields:

\( \nabla \cdot \mathbf{B} = \mu_0 p_m \)

This equation states that the divergence of the magnetic field \( \mathbf{B} \) is proportional to the magnetic charge density \( p_m \), where \( \mu_0 \) is the permeability of free space.

3. Faraday's Law (Modified):

\( \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} - \mathbf{j}_m \)

Here, \( \mathbf{j}_m \) represents the magnetic current density, which contributes to the curl of the electric field.

4. Ampere Maxwell Law (Modified):

\( \nabla \times \mathbf{B} = \mu_0 \mathbf{j}_e + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \)

Here, \( \mathbf{j}_e \) is the electric current density, and the second term represents the displacement current due to the time varying electric field.

Summary of Modifications:

  • Magnetic monopoles introduce a term \( \mu_0 p_m \) in Gauss's law for magnetic fields.
  • Magnetic currents add a term \( \mathbf{j}_m \) to Faraday's law.

These modifications ensure consistency with the presence of magnetic charges and currents in a vacuum.

The correct option is 3).

Electromagnetic Theory Question 6:

The electric field and magnetic field components of an electromagnetic wave going through vacuum is described by

Ex = E0sin(kz − ωt)

By = B0sin(kz − ωt)

Then the correct relation between E0 and Bis given by

  1. E0 B0 = ωk 
  2. E0 = kB0
  3. kE0 = ωB0
  4. ωE0 = kB0

Answer (Detailed Solution Below)

Option 3 : kE0 = ωB0

Electromagnetic Theory Question 6 Detailed Solution

Concept:

The relationship between the electric field Eo The relationship between the electric field Bo in an electromagnetic wave propagating through a vacuum is governed by the equation derived from Maxwell's equations.

Calculation: 

For an electromagnetic wave in a vacuum, the speed of the wave c  is related to Eo and Bo as

\(c = \frac{E_o}{B_o}\)

Also c = \(\frac{ω}{k}\)

⇒ \(\frac{ω}{k} = \frac{E_o}{B_o}\)

⇒ kEo = ωBo

∴ the correct relation between E0 and Bis kEo = ωBo

Electromagnetic Theory Question 7:

 Which of the following electromagnetic waves have the highest wavelength?

  1. Infrared waves
  2. X-rays
  3. Ultraviolet waves 
  4. Microwaves

Answer (Detailed Solution Below)

Option 4 : Microwaves

Electromagnetic Theory Question 7 Detailed Solution

The correct answer is Microwaves.Key Points

  • The option with the highest wavelength among the given electromagnetic waves is microwaves.
  • Microwaves have wavelengths ranging from 1 millimeter to 1 meter.
  • Microwaves move in a straight line.
  • Within the electromagnetic spectrum, microwaves have a frequency that is higher than that of regular radio waves and lower than that of infrared light.
  • They are employed in radar, communications, radio astronomy, remote sensing, and, of course, cooking because of their heating function.

Additional Information

  • Infrared waves have wavelengths ranging from 700 nanometers to 1 millimeter.
    • They are commonly used in heating and communication applications.
  • Ultraviolet waves have wavelengths ranging from 10 nanometers to 400 nanometers.
    • They are commonly used in sterilization and detection applications.
  • X-rays have wavelengths ranging from 0.01 nanometers to 10 nanometers.
    • They are commonly used in medical imaging and material analysis applications.

Electromagnetic Theory Question 8:

The electric and magnetic fields at a point due to two independent sources are E1 = E(αî + βĵ), B1 = Bk̂ and E2 = Eî, B2 = -2Bk̂, where α, β, E and B are constants. If the Poynting vector is along î + ĵ, then

  1. α + β + 1 = 0
  2. α + β -1 = 0
  3. α + β + 2 = 0
  4. α + β - 2 = 0

Answer (Detailed Solution Below)

Option 1 : α + β + 1 = 0

Electromagnetic Theory Question 8 Detailed Solution

Explanation: 

  • We have the total electric field E and magnetic field B as:

\(E_{total} = E_1 + E_2 = E(αî + β\hat j) + E\hat i= E(α+1)\hat i+ Eβ\hat j\)

\(B_{total} = B_1 + B_2 = B\hat k - 2B\hat k = -B\hat k\)

  • Now, let's find the Poynting vector \(S_{total}\) using these total electric and magnetic fields:

\(S_{total} = E_{total} × B_{total} = (E(α+1)\hat i + Eβ\hat j) × (-B\hat k)\)

  • Calculating the vector cross product, \(S_{total} = -(α+1)EB\hat j + βEB\hat i\)
  • Here, the Poynting vector is given as along the direction î + ĵ.
  • This implies that the î and ĵ components of \(S_{total}\) must be equal.
  • Therefore, \(-(α+1)EB = βEB\)
  • Dividing through by EB (it's not zero),

We get: -α -1 = β

  • Rearranging, we get: α + β + 1 = 0

Electromagnetic Theory Question 9:

An electromagnetic wave is incident from vacuum normally on a planar surface of a non-magnetic medium. If the amplitude of the electric field of the incident wave is E0 and that of the transmitted wave is 2E0/3, then neglecting any loss, the refractive index of the medium is

  1. 1.5
  2. 2.0
  3. 2.4
  4. 2.7

Answer (Detailed Solution Below)

Option 2 : 2.0

Electromagnetic Theory Question 9 Detailed Solution

Concept:

  • The transmission coefficient (T) for an electromagnetic wave moving from a vacuum into a medium with refractive index (n) is given by: \(T = \frac {2n_1} {(n_1+n_2)}\)
  • This coefficient (T) is also expressed as the ratio of the amplitude of the transmitted wave (Et) to the amplitude of the incident wave (E0): \(T = \frac {E_t}{E_0}\)

 

Explanation:

  • The refractive index of the vacuum is : \(n_1=1\)
  • In the problem, it states that \(E_t = \frac {2E_0}{3}\).
  • Substituting the value in the formula of the coefficient of transmission: T

\(\therefore \frac 23=\frac{2n_1}{n_1+n_2}\)

\(\therefore\frac 23=\frac{2}{1+n_2}\)

\(\therefore\frac 13=\frac{1}{1+n_2}\)

\(\therefore 1+n_2=3\)

\(\therefore n_2=2\)

Electromagnetic Theory Question 10:

A point charge q of mass m is kept at a distance d below a grounded infinite conducting sheet which lies in the xy plane. For what value of d will the charge remain stationary?

  1. \(\frac{q}{4 \sqrt{mg\pi \epsilon_0}} \)
  2. \(\frac{q}{ \sqrt{mg\pi \epsilon_0}} \)
  3. no finite value
  4. \(\frac{\sqrt{mg\pi \epsilon_0}}{q } \)

Answer (Detailed Solution Below)

Option 1 : \(\frac{q}{4 \sqrt{mg\pi \epsilon_0}} \)

Electromagnetic Theory Question 10 Detailed Solution

Concept:

The method of images is used to solve a variety of problem.

So, if we have a charge q at distance d then we can have a

image charge -q at a distance d on the other side of the infinite conducting sheet.

Now the force between the real and image charge is given as;

\(F= \frac{-k q^2}{4 d^2} \) , where the distance between them is 2d.

where k is \(\frac{1}{4 \pi \epsilon_0} \).

Explanation:

We are given with the fact that a charge q is of mass m is placed at a distance d below a

grounded infinite conducting sheet which lies in the xy plane.

Now for this case for the charge to be stationary,

the force on the charge q should be 0.

There are two forces, one from the gravity and one due to its image charge.

So the total force on the charge q is given by;

\(F= -\frac{k q^2}{4 d^2} + mg =0 \) .

Now, \(d^2= \frac{kq^2}{4 mg }= \frac{q^2}{16 \pi \epsilon_0 m g} \).

Therefore \(d = \frac{q}{4 \sqrt{mg \pi \epsilon_0}} \).

Hence the correct option is option 1).

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