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Year: 2020
Topic: Electromagnetic waves
1. Light with an average flux of \(20\, \text{W}/cm^2\) falls on a non-reflecting surface at normal incidence having surface area \(20\,cm^2\). The energy recieved by the surface during time span of 1 minute is
(1).\(12 \times\ 10^3\) J
(2). \(24 \times\ 10^3\) J
(3). \(48 \times\ 10^3\) J
(4). \(10 \times\ 10^3\) J
Solution
Year: 2015
Topic: Electromagnetic waves
2. A radiation of energy āEā falls normally on a perfectly reflecting surface The momentum transferred to the surface is (C = Velocity of light):
(1).\(\displaystyle \frac{E}{C}\)
(2). \(\displaystyle \frac{2E}{C}\)
(3). \(\displaystyle \frac{2E}{C^2}\)
(4). \(\displaystyle \frac{E}{C^2}\)
Solution
Year: 2025
Topic: Electromagnetic waves
3. The electric field in a plane electromagnetic wave is given by
\(\text{E}_\text{z}\,=\,60\, \text{cos}\left(5\text{x}\,+\,1.5 \times 10^9 \text{t}\right)\,\text{V/m}\).
Then the expression for the corresponding magnetic field is (here subscripts denote the direction of the field):
\(\text{E}_\text{z}\,=\,60\, \text{cos}\left(5\text{x}\,+\,1.5 \times 10^9 \text{t}\right)\,\text{V/m}\).
Then the expression for the corresponding magnetic field is (here subscripts denote the direction of the field):
(1).\(\text{B}_\text{y}\,=\,60\, \text{sin}\left(5\text{x}\,+\,1.5 \times 10^9 \text{t}\right)\,\text{T}\)
(2). \(\text{B}_\text{y}\,=\, 2\times 10^{-7}\,\text{cos}\left(5\text{x}\,+\,1.5 \times 10^9 \text{t}\right)\,\text{T}\)
(3). \(\text{B}_\text{x}\,=\, 2\times 10^{-7}\,\text{cos}\left(5\text{x}\,+\,1.5 \times 10^9 \text{t}\right)\,\text{T}\)
(4). \(\text{B}_\text{z}\,=\,60\, \text{sin}\left(5\text{x}\,+\,1.5 \times 10^9 \text{t}\right)\,\text{T}\)
Solution
Year: 2025
Topic: Electromagnetic waves
4. An electron (mass \(9 \times 10^{-31} \text{kg}\) and charge \(1.6 \times 10^{-19}\text{C}\)) moving with speed c/100 (c = speed of light) is injected into a magnetic field \(\vec{\text{B}}\) of magnitude \(9 \times 10^{-4}\text{T}\) perpendicular to its direction of motion. We wish to apply an uniform electric field \(\vec{\text{E}}\) together with the magnetic fiekd so that the elctron does not deflect from its path. Then (speed of light c = 3 \times 10^8 \text{ms}^{-1}\))
(1).\(\vec{\text{E}}\) is parallel to \(\vec{\text{B}}\) and its magnitude is \(27 \times 10^4 \text{V m}^{-1}\)
(2). \(\vec{\text{E}}\) is perpendicular to \(\vec{\text{B}}\) and its magnitude is \(27 \times 10^4 \text{V m}^{-1}\)
(3). \(\vec{\text{E}}\) is perpendicular to \(\vec{\text{B}}\) and its magnitude is \(27 \times 10^2 \text{V m}^{-1}\)
(4). \(\vec{\text{E}}\) is parallel to \(\vec{\text{B}}\) and its magnitude is \(27 \times 10^2 \text{V m}^{-1}\)
Solution
Year: 2025
Topic: Electromagnetic waves
5. A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:
(1).Zero between the plates and non-zero outside
(2). Zero at all places
(3). Constant between the plates and zero outside the plates
(4). Non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates