Electromagnetic Theory MCQ Quiz in தமிழ் - Objective Question with Answer for Electromagnetic Theory - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Behavior of Light at a Refractive Surface: When light passes from one medium to another, its speed and wavelength change, but its frequency remains constant.

• Relation of Speed, Wavelength, and Frequency:
• Formula: c = fλ
  • c = Speed of light
  • f = Frequency of light
  • λ = Wavelength of light
• Refractive Index and Wavelength Change:
• n = c_air / c_glass = λ_air / λ_glass
  • n = Refractive index of glass with respect to air
  • c_air = Speed of light in air
  • c_glass = Speed of light in glass

Calculation:

Given,

Wavelength in air: λ_air = 480 nm

Refractive index of glass: n = 1.5

⇒ The wavelength in glass:

λ_glass = λ_air / n

⇒ λ_glass = 480 nm / 1.5

⇒ λ_glass = 320 nm

⇒ Since frequency remains constant across media:

f_incident / f_refracted = 1 / 1

∴ The ratio of the frequency of incident and refracted light is 1:1.

Solution:

For the circular loop:

The magnetic field at the center of a circular loop of radius R carrying a current I is given by:

\(B₀ = \frac{μ₀I}{2R} \ \)

For the square coil:

When the circular loop is unwound and rewound into a square coil of side a, the length of the wire remains the same.

Therefore, 2πR = 4a.

The magnetic field at the center of a square coil of side a carrying a current I is given by:

\(B = \frac{2\sqrt 2μ₀I}{πa} \ \)

Ratio B/B₀:

Substituting the value of a in terms of R, we get:

\(\frac{B}{B₀} = \frac{4\sqrt2}{π}\ \)

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

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.

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).

Concept:

The relationship between the electric field E_o The relationship between the electric field B_o 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 E_o and B_o as

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

Also c = \(\frac{ω}{k}\)

⇒ \(\frac{ω}{k} = \frac{E_o}{B_o}\)

⇒ kE_o = ωB_o

∴ the correct relation between E0 and B0 is kEo = ωBo

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.

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

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 (E_t) 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

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

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

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

\(\therefore 1+n_2=3\)

\(\therefore n_2=2\)

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).