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12th Standard English Medium Physics Subject Electromagnetic Waves Creative 3 Mark Questions with Solution Part - I updated Creative Questions

Published on: 13/05/2022

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Std: 12th Standard - Physics

chapter: - Creative Questions

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Questions:

Generic 3

5 x 3 = 15

Name the parts of Electromagnetic spectrum which is
(i) used to destroy becteria.
(ii) produced by where is a sudden deceleration of high speed electrons.
(iii) used in food industry.

Identify the Electromagnetic waves whose wavelength vary as
(a) 10^-12 m to 10^-8 m
(b) 10^-4m and write their uses.

How does Ampere - Maxwell law expalain the flux of current trough a capacitor when it is being charged by a battery? write the expression for the displacement current in terms of the rate of change of electric flux.

A variable frequency ac source is connected to capacitor. How will the displacement current change with decrease in frequency.

The charge on a parallel plate capacitor varies as q = q_o cos 2\(\pi \gamma \)t. The plates are very large and close together. (area - A. separation - d) find the displacement current through the capacitor?

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Answers Key:

Generic 3

5 x 3 = 15

(i) u - v rays
(ii) X - rays
(iii) ૪ - rays.

(a) X - rays are used as diagnostic tool in medicine. X-rays are used extensively in studying structures of inner atomic electron shells and crystal structures. It is used in detecting fractures, diseased organs, formation of bones and stones, observing the progress of healing bones. Further, in a finished metal product, it is used to detect faults, cracks, flaws and holes.
(b) Radio are produced by oscillators in electric circuits. It obeys reflection and diffraction. It is used in radio and communication system and cellphones.

During charging, the electric flux between the plates capacitor keeps on changing. This results in the production of displacement current between the plates.
\({ I }_{ d }=\varepsilon _{ 0 }\left( \frac { d{ \phi }_{ E } }{ dt } \right) \)

On decreasing the frequency, reactance \({ X }_{ c }=\frac { 1 }{ \omega C } \) will increase which will lead to decrease in condition current. In this case I_d = I_c hence displacement current will decrease.

Conduction current Ie = Displacement current ID
\({ I }_{ C }={ I }_{ s }=\frac { dq }{ dt } =\frac { d }{ dt } ({ q }_{ 0 }cos2\pi \gamma t)\)
\(=-2\pi { q }_{ 0 }\gamma sin2\pi \gamma t\)