Electromagnetic Theory MCQ Quiz in বাংলা - Objective Question with Answer for Electromagnetic Theory - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Mar 11, 2025
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
Answer (Detailed Solution Below)
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.
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 EF = IR, then Q is
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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.
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
Answer (Detailed Solution Below)
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 jm is
(Assume that, except for the sources, the fields are in vacuum)
Answer (Detailed Solution Below)
Electromagnetic Theory Question 5 Detailed Solution
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 B0 is given by
Answer (Detailed Solution Below)
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 B0 is kEo = ωBo
Electromagnetic Theory Question 7:
Which of the following electromagnetic waves have the highest wavelength?
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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).