Electromagnetic Waves

Practice Questions:

Question 1: 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 2: 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 3: 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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