Electromagnetic Waves

Practice Questions:

Question 1: easy

To produce an instantaneous displacement current of 2 mA in the space between the parallel plates of a capacitor of capacitance 4 \(\mu\)F, the rate of change of applied variable potential difference \(\left(\frac{dV}{dt}\right)\) must be

1. 200 V/s

2. 400 V/s

3. 800 V/s

4. 500 V/s

View Answer

Displacement current is given by \(I_d = C \frac{dV}{dt}\). Substituting the values: \(2 \times 10^{-3} = 4 \times 10^{-6} \frac{dV}{dt} ⇒ \frac{dV}{dt} = 500\text{ V/s}\).

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Question 2: easy

In an electromagnetic wave, (travelling in vacuum) \(E = 1.2 \sin(2 \times 10^6 t – kx)\text{ N/C}\). Find the value of maximum intensity of magnetic field:

1. \(4 \times 10^{-8}\text{ A/m}\)

2. \(\frac{10^{-3}}{\pi}\text{ A/m}\)

3. \(4 \times 10^{-9}\text{ A/m}\)

4. \(\frac{10^{-2}}{\pi}\text{ A/m}\)

View Answer

Maximum magnetic field \(B_0 = \frac{E_0}{c} = \frac{1.2}{3 \times 10^8} = 4 \times 10^{-9}\text{ T}\). The intensity of magnetic field is \(H_0 = \frac{B_0}{\mu_0} = \frac{4 \times 10^{-9}}{4\pi \times 10^{-7}} = \frac{10^{-2}}{\pi}\text{ A/m}\).

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Question 3: easy

Displacement current is same as:

1. conduction current due to flow of free electron

2. conduction current due to flow of positive ions

3. conduction current due to flow of both positive and negative free charge carriers

4. is not a conduction current but is caused by time varying electric field

View Answer

Displacement current represents the rate of change of electric displacement field and is not caused by real movement of charges like conduction current.

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Question 4: easy

The potential difference between the plates of a parallel plate capacitor is changing at the rate of \(10^6\text{ V/s}\). If the capacitance is \(2\ \mu\text{F}\), the displacement current in the dielectric of the capacitor will be:

1. \(1\text{ A}\)

2. \(2\text{ A}\)

3. \(3\text{ A}\)

4. \(4\text{ A}\)

View Answer

Displacement current is given by \(I_d = C \frac{dV}{dt}\). Substituting \(C = 2 \times 10^{-6}\text{ F}\) and \(\frac{dV}{dt} = 10^6\text{ V/s}\), we find \(I_d = 2\text{ A}\).

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Question 5: easy

Modified ampere circuital law is given by (symbols have their usual meaning)

1. \(\oint \vec{B} \cdot d\vec{l} = 0\)

2. \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_C\)

3. \(\oint \vec{B} \cdot d\vec{l} = \mu_0(I_C + I_D)\)

4. \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_D\)

View Answer

The generalized Ampere's circuital law (or Ampere-Maxwell law) includes both conduction current \(I_C\) and displacement current \(I_D\) as sources of magnetic fields, expressed as \(\oint \vec{B} \cdot d\vec{l} = \mu_0(I_C + I_D)\).

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Question 6: easy

A capacitor of capacitance \(C\), is connected across an ac source of voltage \(V\), given by \(V = V_0\sin\omega t\). The displacement current between the plates of the capacitor, would then be given by

1. \(I_d = V_0\omega C\sin\omega t\)

2. \(I_d = V_0\omega C\cos\omega t\)

3. \(I_d = \frac{V_0}{\omega C}\cos \omega t\)

4. \(I_d = \frac{V_0}{\omega C}\sin \omega t\)

View Answer

The displacement current is equal to the conduction current, which is \(I_d = \frac{dq}{dt}\). Since \(q = CV = C V_0 \sin\omega t\), differentiating with respect to time gives \(I_d = V_0 \omega C \cos\omega t\).

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Question 7: easy

For a plane electromagnetic wave propagating in \(x\)-direction, which one of the following combination gives the correct possible directions for electric field (\(vec{E}\)) and magnetic field (\(vec{B}\)) respectively?

1. \(-\hat{j} + \hat{k}, -\hat{j} + \hat{k}\)

2. \(\hat{j} + \hat{k}, \hat{j} + \hat{k} \)

3. \(-\hat{j} + \hat{k}, -\hat{j} - \hat{k}\)

4. \(\hat{j} + \hat{k}, -\hat{j} - \hat{k} \)

View Answer

The direction of propagation of an electromagnetic wave is given by the cross product \(vec{E} \times \vec{B}\). For wave propagation along the positive \(x\)-direction, the cross product must yield a positive \(hat{i}\) vector. Using option C, \((-\hat{j} + \hat{k}) \times (-\hat{j} - \hat{k}) = \hat{j}\times\hat{k} - \hat{k}\times\hat{j} = 2\hat{i}\), which satisfies this condition.

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Question 8: easy

Match List-I with List-II.

List-I (Types of EM waves)
a. Infrared rays
b. Microwaves
c. UV rays
d. Gamma rays

List-II (Application)
(i) Water purifier
(ii) Remote switches
(iii) Used in medicine to destroy cancer cells
(iv) Cooking

Choose the correct option:

1. a(ii), b(iii), c(iv), d(i)

2. a(i), b(ii), c(iii), d(iv)

3. a(ii), b(iv), c(i), d(iii)

4. a(iii), b(i), c(ii), d(iv)

View Answer

Infrared is used in remote switches (a-ii); Microwaves for cooking (b-iv); UV rays for water purification (c-i); and Gamma rays in cancer cell destruction (d-iii).

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Question 9: easy

Match column I with column II.

| Column-I (Types of EM waves) | Column-II (Production) |

| A. Infra-red | P. Rapid vibration of electrons in aerials |

| B. Radio | Q. Electrons in atoms emit light when they move from higher to lower energy level. |

| C. Light | R. Klystron valve |

| D. Microwave | S. Vibration of atoms and molecules |

Choose the correct match from the options given below:

1. A-P, B-R, C-S, D-Q

2. A-S, B-P, C-Q, D-R

3. A-Q, B-P, C-S, D-R

4. A-S, B-R, C-P, D-Q

View Answer

Infrared waves are produced by molecular vibrations (A-S). Radio waves by accelerating charges in aerials (B-P). Light waves by atomic transitions (C-Q). Microwaves by klystron/magnetron valves (D-R).

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Question 10: easy

For a plane EM wave propagating in x-direction, which one of the following combination gives the correct possible directions for electric field (\(\vec{E}\) ) and magnetic field (\(\vec{B}\) ) respectively?

1. \(-\hat{j} + \hat{k}, -\hat{j} + \hat{k}\)

2. \(\hat{j} + \hat{k}, \hat{j} + \hat{k}\)

3. \(-\hat{j} + \hat{k}, -\hat{j} - \hat{k}\)

4. \(\hat{j} + \hat{k}, -\hat{j} - \hat{k}\)

View Answer

The direction of propagation of an electromagnetic wave is along \(\vec{E} \times \vec{B}\). Here, \((-\hat{j} + \hat{k}) \times (-\hat{j} - \hat{k}) = 2\hat{i}\), which points in the positive x-direction.

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