InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
In a region of free space the electric at some instant of time to `vecE=(80hati+32hatj-64hatk)` and the magnetic field is `vecB=(0.2hati+0.08hatj-0.29hatk)mu` T. The pointing vector for these field is:A. `-11.52hati+28.8hatj`B. `-28.8hati+11.52hatj`C. `28.8hatj-11.52hatj`D. `11.52hati-28.8hatj` |
|
Answer» Correct Answer - d The pointing vector `vecS=(1)/mu_(0)(vecExxvecB)` `(1)/(4pixx10^(-7))|[hati,hatj,hatk],[80,32,-64],[0.2,0.08,0.29]|=11.52hati-28.8hatj` |
|
| 2. |
A charged particle oscillates about its mean equilibrium position with a frequency of `10^(9)` Hz. The frequency of electromagnetic waves produced by the oscillator isA. `10^(6)Hz`B. `10^(7)Hz`C. `10^(8)Hz`D. `10^(9)Hz` |
|
Answer» Correct Answer - D The frequency of electromagnetic wave is the same as that of oscillating charged particle about its equilibrium position, which is `10^(9)` Hz. |
|
| 3. |
If you find closed loops of `vecB` in a region in space, does it necessarily mean that actual charges are flowing across the area bounded by the loops ? |
|
Answer» Not necessarily so, because a displacement current (which appears in between the plates of capacitor during charging state or discharging state) can also produce loops of B, where actually there is no flow of charge. |
|
| 4. |
What oscillates in e.m. waves? Give two examples of e.m. waves. |
|
Answer» Electric and magnetic field vectors oscillate in electromagnetic waves. They oscillate perpendicular to each other as well as perpendicular to the direction of propagation of electromagnetic waves. Example: Light waves, X-rays, radiowaves etc. |
|
| 5. |
The `21 cm` radio wave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction is atomic hydrogen. The energy of the emitted wave is nearlyA. `10^(-7)` jouleB. `1` jouleC. `7xx10^(-8)` jouleD. `10^(-24)` joule |
|
Answer» Correct Answer - d `E=(hc)/(lambda)=(6.6xx10^(-34)xx3xx10^(8))/(21xx10^(-2))=0.94xx10^(-24)=10^(-24)J` |
|
| 6. |
In a plane electromagnetic wave the electric field oscillates sinusoidally at a frequency `3xx10^10Hz` and amplitude `50V//m`. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? |
|
Answer» Wavelength, `lambda=c/v=(3xx10^8)/(3xx10^10)=0.01m=1cm` Amplitude of oscillating magnetic field, `B_0=(E_0)/c=50/(3xx10^8)=1.67xx10^-7T`. |
|
| 7. |
The velocity of light in vacuum can be changed by changingA. frequencyB. amplitudeC. wavelengthD. none of these |
| Answer» Correct Answer - d | |
| 8. |
One cannot cannot see through fog, becauseA. for absorbs the lightB. light suffers total reflection at dropletsC. refractive index of the fog is infinityD. light is scattered by droplets |
| Answer» Correct Answer - d | |
| 9. |
optical and radio telecopes are build on ground but X-ray astronomy is possible only from satellites or biting the earth. Explain why? |
| Answer» Visible and radio waves can penetrate the atmosphere, while X-rays are absorbed by the atmosphere. This is X-ray telescopes are installed in satellites orbiting the earth. | |
| 10. |
If a capacitor of `2.0muF` is charged to 20V and then suddenly short-circuited by a coil of negligible resistance and of inductance `8.0muH` Calculate the maximum amplitude and the frequency of the resulting current oscillations. |
|
Answer» Here, `C=2.0xx10^-6F, V=20V,` `L=8.0xx10^-6H` The equation for the oscillatory discharge of capacitor through inductance is given by `q=q_0 sin omegat`, where `omega=1/(sqrt(LC))` The resulting oscillatory current will be `I=(dq)/(dt)=q_0omega cos omegat=I_0 cos omegat....(i)` where `q_0omega=I_0`, amplitude of oscillatory current. `:.` Amplitude of oscillatory current `I_0=q_0omega=CVxx1//sqrt(LC)=Vsqrt(C/V)` `=20xxsqrt((2.0xx10^-6)/(8.0xx10^-6))=10A` Also frequency of oscillatory current `v=1/(2pisqrt(LC))` `=1/(2xx3.14sqrt((8.0xx10^-6)xx(2.0xx10^-6)))` `=4.0xx10^4Hz` |
|
| 11. |
What physical quantity is the same for X- rays of wavelength `10^10 m`,red light of wavelength `6800 A` and radio waves of wavelength `500 m` ? |
| Answer» The speed in vacuum is the same for all given wavelengths, which is `3xx10^8ms^-1` | |
| 12. |
A silver wire has resistivity `rho=1.62xx10^-8 Omegam` and cross-sectional area `10.0mm^2`. The current in the wire is uniform and changing at the rate of `4000A//s`, when the current is 200A. (a) What is the magnitude of the electric field in the wire when the current in the wire is 200A? (b) What is the displacement current in the wire at that time (c) What is the ration of the magnitude of the magnetic field due to the displacement current to that due to the current at a distance r from the wire? |
|
Answer» Here, `rho=1.62xx10^-8Omegam`, `A=10.0xx10^-6m^2=10^-5m^2`. `(dI)/(dt)=4000As^-1, I=200A`. (a) Magnitude of the electric field in the wire when the current in the wire is 200A, is `E=rhoJ=rhol/A=((1.62xx10^-8)xx200)/(10^-5)` `=0.324Vm^-1` (b) The displacement current is `I_D=in_0 (dphi_E)/(dt)=in_0d/(dt)(EA)` `in_0A (dE)/(dt)=in_0Ad/(dt) ((rhol)/A)` `=in_0Axxrho/A((dI)/(dt)) =in_0rho(dl)/(dt)` `=(8.85xx10^-12)x(1.62xx10^-8)xx4000` `=5.74xx10^-6A` (c) Ratio of magnetic fields is `(B("due to" I_d))/(B("due to" I))=(mu_0 I_D//2pir)/(mu_0I//2pir)` `=(I_D)/I=(5.7f4xx10^-16)/200=2.87xx10^-18` |
|
| 13. |
A carrier is simultaneously modulated by two sine waves with modulation indices of `0.4` and `0.3`. The resultant modulated index will beA. `1.0`B. `0.7`C. `0.5`D. `0.35` |
|
Answer» Correct Answer - c `m=sqrt(m_(1)^(2)+m_(2)^(2))=sqrt((0.16)+(0.09))=0.5` |
|
| 14. |
(i) Find the energy stored in a 90 cm length of a laser beam operating at `6 mW`. (ii) Find the amplitude opf electric field in a parallel beam of light of intensity `17.7 W//m^(2)`. |
|
Answer» (i) The time taken by wave to move a distance `90 cm`. `t = (90 xx 10^(-2))/(3 xx 10^(8)) = 3 xx 10^(-9)s` Energy contained in `90` cm length, `u = Pt = 6 xx 10^(-3) xx 3 xx 10^(-9) = 18 xx 10^(-12)J` (ii) Intensity of light, `I = (1)/(2) epsi_(0)E_(0)^(2)c` `rArr 17.7 = 1/2(8.85 xx 10^(-12))E_(0)^(2) xx 3 xx 10^(8)` `rArr E_(0)^(2) = 4/3 xx 10^(4)` Therefore the amplitude of the electric field in the parallel beam, `E_(0) = 2/3 xx 10^(2)V//m` |
|
| 15. |
The permittivity and permeability of free space are `epsilon_(0) = 8.85 xx 10^(-12)C^(2)N^(-1)m^(-2)"and"mu_(0) = 4pi xx 10^(-7)"TmA"^(-1),` respectively. Find the velocity of the electromagnetic wave. |
|
Answer» Correct Answer - A::C |
|
| 16. |
Professor C.V Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. Which property of EM waves was he exhibiting? Give one more example of this property. |
| Answer» An electromagnetic wave carries energy and momentum like other waves. Since it carries momentum and electromagnetic wave also exerts pressure called radiation pressure .this property or electromagnetic waves helped professor CV Raman surprised his students by suspending freely a tiny ligh ball in a transperent vacuum chamber by shining a laser beam on it . the tails of the camets are also due to radiation pressure. | |
| 17. |
Electromagnetic waves with wavelength (i) `lambda_1` is used in satellite communication. (ii) `lambda_2` used to kill germs in water purifier. `lambda_` used to detect leakage of oil in underground pipelines. `lambda_4` used to improve visibility in runways during fog and mist conditions. (a) Identify and name the part of e.m. spectrum to which these radiations belong. (b) Arrange these wavelengths in ascending order of their magnitude. (c) Write one more application of each. |
|
Answer» (i) Microwave is used in satellite communications. So `lambda_(1)` is the wavelength of microwave. (ii) Ultraviolet rays are used to kill germs in water purifier So `lambda_(2)` is the wavelength of UV rays. (iii) X-rays are used to detect leakage of oil in underground pipelines .So `lambda_(3)` is the wavelength of X-rays. (iv) Infrared is used to improve visibility on runways during fog and mist condition. So, it is wavelength of infrared waves. (b) Wavelength of X-rays `lt` wavelegth of `UV lt` wavelegth of infrared `lt` wavelength of microwave. `rArr" "lambda_(3) lt lambda_(2) lt lambda_(4) lt lambda_(1)` (c) (i) Mirowave is used in rader. (ii) UV is used in LASIK eye surgery. (iii) X-rays is used to detect a fracture in bones. (iv) infrared is used in optical communication. |
|
| 18. |
Electromagnetic wave with wavelength (i) `lambda_1` are used to treat muscular strain (ii) `lambda_2` are used by a FM radio station for broadcasting. (iii) `lambda_3` are used to detect fracture in bones (iv) `lambda_4` are absorbed by the ozone layer of the atmosphere. Identify and name the part of elctromagnetic spectrum to which thest radiations belongs. Arrange these wavelengths in decreasing order of magnitude. |
|
Answer» (i) `lambda_1` belongs to Infrared radiations (ii) `lambda_2` belongs to VHF radiations (iii) `lambda_3` belongs to X-rays (iv) `lambda_4` belongs to ultraviolet rays. The arrangement of wavelengths in decreasing order of magnitude are `lambda_2gtlambda_1gtlambda_4gtlambda_3`. |
|
| 19. |
Assertion : The radio and TV signals from broadcasting stations carry energy. Reason : Electromagnetic waves are capable to carry energy from one place to another.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is falseD. If both assertion and reason are false |
|
Answer» Correct Answer - A Electromagnetic waves can carry energy from one place to another. This makes radio and TV signal broadcasting possible. Also light carries energy from the sun to the earth, thus making life possible on the earth. |
|
| 20. |
A 1500 Hz modulating voltage fed into an FM generator produces a frequency deviation of `2.25 kHz`. If amplitude of the voltage is kept constant but frequency is raised to `6 kHz` then the new devition will beA. `4.5 kHz`B. `54 kHz`C. `27 kHz`D. `15 kHz` |
|
Answer» Correct Answer - b `m_(f)=(delta)/(f_(m))=(2250)/(500)=4.5` `:.` New deviation `=2(m_(f)f_(m))=2xx4.5xx6=54kHz`. |
|
| 21. |
A plane electromagnectic wave propagating along x-direction can have the following pairs of `E` and `B`.A. `E_x,B_y`B. `E_y,B_z`C. `B_x,E_y`D. `E_z,B_y` |
|
Answer» Correct Answer - B::D `E,B` and velocity of electromagnetic waves are mutually `v` perpendicular. |
|
| 22. |
The energy of gamma `(gamma)` ray photon is `E_(gamma)` and that of an X-rays photon is `E_(X)`. If the visible light photon has an energy of `E_(v)`, then we can say thatA. `E_(X) gt E_(gamma) gt E_(v)`B. `E_(gamma) gt E_(v) gt E_(X)`C. `E_(gamma) gt E_(X) gt E_(v)`D. `E_(X) gt E_(v) gt E_(gamma)` |
|
Answer» Correct Answer - C `E_(gamma) ge 100 keV, E_(X) = 100 eV` to `100 keV`. So, we can say that `E_(y) gt E_(x) gt E_(v)`, |
|
| 23. |
The speed of electmagnetic waves in vacuum is equal toA. `mu_(0)epsi_(0)`B. `sqrt(mu_(0)epsi_(0))`C. `(1)/(sqrt(mu_(0)epsi_(0)))`D. `(1)/(mu_(0)epsi_(0))` |
|
Answer» Correct Answer - C According to Maxwell, it is found that the electromagnetic waves travel in free space (or vacuum) with a speed is given by i.e., `c = (1)/(sqrt(mu_(0)epsi_(0))) = 3 xx 10^(8) ms^(-1)` where, `mu_(0)` and `epsi_(0)` are absolute permeability and permittivity of the free space, respectively. where, `mu_(0) = 4pi xx 10^(-7) WbA^(-1)m^(-1)` `epsi_(0) = 8.85 xx 10^(-12) C^(2)N^(-1)m^(-2)` |
|
| 24. |
A plane electromagnetic wave of frequency `20 MHz` travels through a space along x-direction. If the electric field vector at a certain point in space is `6 Vm^(-1)`,then what is the magnetic field vector at that point?A. `2 xx 10^(-8) T`B. `1/2 xx 10^(-8) T`C. `2T`D. `1/2 T` |
|
Answer» Correct Answer - A The magnetic field, `B = (E)/(c)`, where, `c = 3 xx 10^(8)ms^(-1)` `rArr B = (6)/(3 xx 10^(8)) = 2 xx 10^(-8) T` |
|
| 25. |
Eelctromagnetic wave consists of periodically oscillating electric and magnetic vectorsA. in mutually perpendicular planes but vibrating with a phase difference of `pi`B. in mulually perpendicular planes but vibrating with a phase difference of `(pi)/(2)`C. in randomly oriented planes but vibrating in phaseD. in mutually perpendicular planes but vibrating in phase. |
|
Answer» Correct Answer - D Electromagnetic wave consists of periodically oscillating electric and magnetic vectors in mutually perpendicular planes but vibrating in phase. |
|
| 26. |
A plane electromagnetic wave travels in vacuum along z-direction. If the frequency of the wave is 40 MHz then its wavelength isA. 5 mB. 7.5 mC. 8.5 mD. 10 m |
|
Answer» Correct Answer - B Wavelength, `lambda=(c)/(upsilon)=(3xx10^(8)"m s"^(-1))/(40xx10^(6)s^(-1))=7.5m` |
|
| 27. |
In an electromagnetic wave, the electric and magnetizing field are `100V//m` and `0.265 A//m`. The maximum energy flow is:A. `79W//m^2`B. `13.2W//m^2`C. `53W//m^2`D. `26.5 W//m^2` |
|
Answer» Correct Answer - d Here, `E_0=100V//m , H_0=0.265A//m` As maximum energy flow persecond per unit area in e.m. wave will be `(E_0B_0)/(mu_0)=E_0H_0= (100)(0.265)=26.5 W//m^2`. |
|
| 28. |
Light with an energy flux` 20 W//cm^2` falls on a non-reflecting surface at normal incidence. If the surface has an area of `30 cm^2`. the total momentum delivered ( for complete absorption)during 30 minutes isA. `36 xx 10^5 kg-m//s`B. `36xx10^4 kg-m//s`C. `1.08 xx 10^4 kg -m//s`D. `1.08xx10^7 kg-m//s` |
|
Answer» Correct Answer - B In case of perfectly non reflecting surface, `Deltap=E/c` where `e = 20xx30xx30xx60` `= 1.08xx10^6 j` `Deltap=(1.08xx10^6)/(3xx10^8)` `36xx10^-4 kg-m//s`. |
|
| 29. |
Light with an energy flux` 20 W//cm^2` falls on a non-reflecting surface at normal incidence. If the surface has an area of `30 cm^2`. the total momentum delivered ( for complete absorption)during 30 minutes isA. `36 xx 10^(5) kg-m//s`B. `36 xx 10^(-4) kg-m//s`C. `108 xx 10^(4) kg-m//s`D. `1.08 xx 10^(7) kg-m//s` |
|
Answer» Correct Answer - B Given energy flux, `phi = 20 W//cm^(2)` Area, `A = 30 cm^(2)` Time, `t = 30` min = `30 xx 60 s` Now, total energy following on the surface in time t is, `u = phi At = 20 xx 30 xx (30 xx 60) J` Momentum of the incident light `= U/c` `= (20 xx 30 xx (30 xx 60))/(3 xx 10^(8)) = 36 xx 10^(-4) kg-m//s` Momentum of the reflected light `= 0` `:.` Momentum delivered to the surface `= 36 xx 10^(-4) -0 = 36 xx 10^(-4) kg-m//s` |
|
| 30. |
Light with an energy flux` 20 W//cm^2` falls on a non-reflecting surface at normal incidence. If the surface has an area of `30 cm^2`. the total momentum delivered ( for complete absorption)during 30 minutes isA. `36 xx 10^(-5) kg-m//s`B. `36 xx 10^(-4) kg-m//s`C. `108 xx 10^(4) kg-m//s`D. `1.08 xx 10^(7) kg-m//s` |
|
Answer» Correct Answer - B Given energy flux `phi =20W//cm^(2)` Area `A = 30cm^(2)` Time , `t= 30min =30 xx 60s` Now , total energy falling on the surface in time is `u=phiAt =20 xx 30 xx (30 xx 60)J` Momentum of the incident light `=U/C` `=(20 xx 30 xx (30 xx 60))/(3xx 10^(8)) rArr = 36 xx 10^(-4) kg-ms^(-1)` Momentun of the reflected light =0 `:.` Momentum delivered to the suface `=36 xx 10^(-4) -0 =36 xx 10^(-4) kg-ms^(-1)` |
|
| 31. |
An LC current contains inductance `L=1mu` H and capacitance `C=0.01 muF`. The wavelength of electromagnetic wave generated is nearly:A. `0.5m`B. `5m`C. `188 m`D. `30 m` |
|
Answer» Correct Answer - c `v=(1)/(2pisqrt(LC))` `lambda=(c)/(v)=c.2pisqrt(LC)` `3xx10^(8)xx2pisqrt(1xx10^(-6)xx0.01xx10^(-6))` `3xx10^(8)xx2xx3.14xx10^(-7)=188.4m` |
|
| 32. |
The average energy-density of electromagnetic wave given by `E=(50N//C) sin (omegat-kx)` will be nearly:A. `10^(-8)j//m^(3)`B. `10^(-7)j//m^(3)`C. `10^(-6)j//m^(3)`D. `10^(5)j//m^(3)` |
|
Answer» Correct Answer - a Average energy density of electromagnetic wave `U_(av)=(1)/(2)epsilon_(0)E_(0)^(2)=(1)/(2)xx(8.85xx10^(-12))xx(50^(2))` `10^(-8)j//m^(3)` |
|
| 33. |
Calculate the phase velocity of electromagnetic wave having electromagnetic density and frequency for D layer, `N=400 "electron"//cm^(3), v=300kHz`A. `3xx10^(8) ms_(1)`B. `3.75xx10^(8) ms^(-1)`C. `6.8xx10^(8) ms^(-1)`D. `1.1xx10^(9) ms^(-1)` |
|
Answer» Correct Answer - b `mu=sqrt(1-(80.6N)/(v^(2)))~~0.80` Again, `v(c)/(mu)` |
|
| 34. |
In an electromagnetic wave in free space the root mean square value of the electric field is `E_(rms)=6 V//m`. The peak value of the magnetic field isA. `1.41xx10^-8T`B. `2.83xx10^-8T`C. `0.70xx10^-8T`D. `4.23xx10^-8T` |
|
Answer» Correct Answer - b Here, `E_(rms)=6V//m, or (E_0)/(sqrt2)=6 or E_0=6sqrt2 V//m` As `E_0=cB_0` or `B_0=(E_0)/c=(6sqrt2)/(3xx10^8)=2sqrt2xx10^-8T` `=2.83xx10^-8T` |
|
| 35. |
Speed of electromagnetic waves is the sameA. for all wavelengthB. in all mediaC. for all intensitiesD. for all frequencies |
| Answer» Correct Answer - C | |
| 36. |
Electromagnetic waves are produced byA. a static chargeB. a moving chargeC. an accelerating chargeD. chargeless particle |
| Answer» Correct Answer - C | |
| 37. |
Which of the following is correct regarding electromagnetic wave?A. In electromagnetic wave, electirc and magnetic field vector oscillate perpendicular to direction of progationB. If wave is propagating in `x`-direction `E = E_(0)sin omega(t-(x)/(c ))` `B = B_(0)sin omega (t-(x)/(c ))`, `E_(0) = cB_(0),c = (1)/(sqrt(mu_(0)epsilon_(0)))`C. Intensity of the wave `I = 1//2epsilon_(0)E_(0)^(2)c`D. All options are correct |
| Answer» Correct Answer - D | |
| 38. |
A parallel plate capacitor with plate area `A` and separation between the plates `d`, is charged by a constant current `i`. Consider a plane surface of area `A//4` parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area. |
|
Answer» Let at some instant change on capacitor be `Q`. Electric field between plates of capacitor `E = (Q)/(Ain_(0))` Flux passing through area `A//4` `phi_(E) = E.(A)/(4) = (Q)/(A in_(0)).(A)/(4) = (Q)/(4 in_(0))` The displacement current `i_(d) = in_(0)(dphi_(E))/(dt) = in_(0)(1)/(4)(dQ)/(dt) = (1)/(4)i` |
|
| 39. |
A parallel- plate capacitor with plate area A and separation between the plates d, is charged by a constant current i. Consider a plane surface of area A/2 parallel to the plates and drawn summetrically between the plates. Find the displacement current through this area. |
|
Answer» Correct Answer - B `i_d = epsilon_0 (d phi_E)/(d t)` `=epsilon_0 (d)/(d t)(E A/2)` `=(epsilon_0 A)/2 (d)/(d t) (E)` `=(epsilon_0 A)/2 (d)/(d t)((sigma)/(epsilon_0))` `= (epsilon_0 A)/2 (d)/ (d t)((d)/(A epsilon_0))` `= 1/2 (d q)/(d t)` `i/2 (because (d q)/(d t)= i)`. |
|
| 40. |
A plane electromagnetic wave propagating in the x-direction has a wavelength of 5.0 mm. The electric field is in the y-direction and its maximum magnitude is `30V (m^-1)`. Write suitable equations for the electric and magnetic fields as a function of x and t. |
|
Answer» Here, `lambda=5.0mm=5xx10^-3m,` `E_0=30Vm^-1` Angular frequency `omega=2piv=(2pic)/(lambda)` `=(2xx3.142xx(3xx10^8))/(5xx10^-3)=3.77xx10^11 rad s^-1` Max. magnitude of magnetic field, `B_0=(E_0)/c=30/(3xx10^8)=1.0xx10^-7T` Equation for electric field along y-axis can be writtens as `E=E_y=E_0sin omega(t-x//c)` `=30sin3.77xx10^11(t-x//c)vm^-1` Equation for magnetic field along z-axis can be writtens as `B=B_z=B_0sin omega(t-x//c)` `=(1.0xx10^-7)sin3.77xx10^11(t-x//C)Tesla` |
|
| 41. |
The average energy flux of sunlight is `1.0 kW m^-2`. This energy of radiation is falling normally on the metal plate surface of area `10 cm^2` which completely abosbs the energy. How much force is exerted on the plate if it is exposed to sunlight for 10 minutes? |
|
Answer» Correct Answer - `3.3xx10^-9N` Here, energy flux,`=1.0kWm^-2` `=1.0xx10^3Wm^-2`, `A=10cm^2=10xx10^-4m^2=10^-3m^2` `t=10min=10xx60s=600s`. Total energy falling on the plate. U=energy flux x area x time `=(1xx10^3)xx(10^-3)xx(600)J=600J` Momentum of the radiation `p=U/c=600/(3xx10^8)=2xx10^-6kgms^-1` Since the plate absorbs the entire energy, so momentum delivered to the plate= change in momentum `(Deltap)=2xx10^-6kgms^-1` Force on the plate, `F=(Deltap)/t=(2xx10^-6)/60=3.3xx10^-9N` |
|
| 42. |
A plane e.m. wave propagating in the x-direction has a wavelength 5.5 mm. The electric field is in the y-direction and its maximum magitude is `36Vm^_1`. Write suitable equation for the electric and magnetic fields as a function of x and t and find energy density of e.m. wave. Calculate the maximum electric and magnetic force on a charge q=2e, moving along y-axis with a speed of `3.0xx10^7ms^-1`, where `e=1.6xx10^(-19)C`. |
|
Answer» Correct Answer - `E_y=36 sin 1.14 xx10^(11)[t-x//c] V//m ; B_z=1.2xx10^(-7)sin 1.14xx10^(11)(t-x//c)T and 5.7xx10^-9 Jm^-3` Here, `lambda=5.5mm=5.5xx10^-3m, E_0=36V//m^-1` `omega=2piv=(2pic)/lambda=(2xx3.14xx(3xx10^8))/(5.5xx10^-3)` `1.14xx10^(11)rad//s` `B_0=(E_0)/c=36/(3xx10^8)=1.2xx10^-7T` The equation for the electric field, along y-axis will be `E=E_y=E_0 sin omega(t-x/c)` `=36sin 1.14xx10^(11)(t-x//c) Vm^-1` The equation for the magnetic field along z-axis `B=B_z=B_0 sin omega(t-x/c)` `=1.2xx10^-7 sin 1.14xx10^(11)(t-x//c)T` Average energy density of e.m. wave is `u=1/2 epsilon_0 E_0^2=1/2xx(8.85xx10^(-12))xx(36)^2` `=5.7xx10^-9Jm^-3` |
|
| 43. |
A beam of light travelling along x-axis is described by the magnetic field, `E_y=(600 Vm^-1) sin omega (t-x//c)` Calculating the maximum electric and magnetic forces on a charge q=2e, moving along y-axis with a speed of `3xx10^7m//s`, where `e=1.6xx10^(-19)C`. |
|
Answer» Correct Answer - `1.92xx10^-16N; 1.92xx10^(-17)N` `B_0=(E_0)/c=600/(3xx10^8)=2xx10^-6T`, which is along Z-axis, Thus maximum electric force, `F_e=qE_0=2eE_0=2xx1.6xx10^(-19)xx600` `1.92xx10^(-16)N` Maximum magnetic force, `F_m=qvB_0=2evB_0` `=2xx1.6xx10^(-19)xx3xx10^7xx2xx10^-6` `=1.92xx10^(-17)N` |
|
| 44. |
A light beam travelling in the x- direction is described by the electric field `E_y(300V (m^-1) sin omega (t-(x//c))`. An electron is constrained to move along the y-direction with a speed`(2.0 xx (10^7) m(s^-1))`. Find the maximum electric force and the maximum magnetic force on the electron. |
|
Answer» Maximum Electric Force maximum electric field, `E_0=300 V//m` maximum electricfoce `F=qE_0` `= (1.6xx10^-19)(300)` `= 4.8xx10^-17 N` Maximum Magnetic Force From the equation, ` c=E_0/B_0` Maximum magnetic field, or `B_0 = 300/(3.0xx10^8)=10^-6T` `:.` maximum magnetic force = `B_0qupsilon` Substituting the values,we have maximum magnetic force `(10^-6)(1.6xx10^-19)(2.0xx10^7)` ` 3.2xx 10^-18 N`. |
|
| 45. |
A magnetic field in a plane e.m. wave is given by, `B_y=3xx10^-7 sin[(1.5rad//m)x+(5xx10^8rad//s)t]tesla.` (a) What is the wavelength and frequency of the wave? (b) Write down an expression for the electric field. (x is the metre and t is in second). |
|
Answer» Correct Answer - (a) 4.19m ; `7.95xx10^7Hz (b) E_z=90 sin (1.5x+5xx10^8t)V//m` (a) Comparing the given relation with the equation of `B=B_0sin((2pi)/lambdax+(2pit)/T)` `=B_0sin((2pi)/lambdax+2pitv)` we have, `B_0=3xx10^-7T`, `(2pi)/lambda=1.5 or lambda=(2pi)/1.5 =4.19m` `2piv=5xx10^8 or v=(5xx10^8)/(2pi)=7.95xx10^7Hz` (b) `E_0=cB_0=(3xx10^8)xx(3xx10^-7)=90V//m` `E_z=E_0sin((2pix)/lambda+(2pit)/T)` or `E_z=90sin(1.5x+5xx10^8t)V//m` |
|
| 46. |
A light beam travelling in the `x`-direction is described by the eectirc field `E_(y) = (300 V//m) sin omega(t - x//c)`. An electron is constrained to move along the `y`-direction with a speed of `2.0 xx 10^(7) m//sec`. Find the maximum electric force and the maximum magnetic force on the electron. |
|
Answer» `E = E_(0)sin omega (t - x//c)` `E_(0) = 300V//m` `B_(0) = (E_(0))/(c) = (300)/(3xx10^(8)) = 10^(-6)T` `F_(e) = eE_(0) = (1.6 xx 10^(-19)) (300) = 4.8 xx 10^(-17)N` `F_(m) = evB_(0) = 1.6 xx 10^(-19) xx 2xx 10^(7) xx 10^(-6)` `=3.2 xx 10^(-18)N` |
|
| 47. |
An electric field in an e.m.wave is given by `E=200 sin .(2pi)/lambda(ct-x)NC^-1`. Find the energy contained in a cylinder of crossection `20 cm^2` and length 40 cm along the x-axis. |
|
Answer» Correct Answer - `1.42xx10^(-10)J` Here, `E_0=200NC^-1`, `A=20cm^2=20xx10^-4m^2 , l=0.40m` Volume of cylinder, `V=Al=(20xx10^-4)xx0.40` `=8xx10^-4m^2` Energy contained in cylunder is U=volume x energy density `=Vxx1/2in_0E_0^2` `=(8xx10^-4)xx1/2xx(8.85xx10^-12)xx(200)^2` `=1.42xx10^(-10)J` |
|
| 48. |
An electric field `(vec E)` and a magnetic field `(vec B)`exist in a region . The fields are not perpendicular to each other.A. This is not possibleB. No electromagnetic wave is passing through the regionC. An electromagnetic wave may be passing through the regionD. An electromagnetic wave is certinaly passing thorugh the region |
| Answer» Correct Answer - C | |
| 49. |
Light with an energy flux of `40W//cm^(2)` falls on a non-reflecting surface at normal incidence. If the surface has an area of `20cm^(2)`, find the average force exerted on the surface during a `30min` time span. |
|
Answer» The energy incident on surface is `U = Iat = 40 xx 10^(4) xx 20 xx 10^(-4) xx 30 xx 60` `= 1.44 xx 10^(6)J` Total momentum delivered (for completa absorption) `p = (U)/(c ) = (1.44 xx 10^(6))/(3xx10^(8)) = 0.48 xx 10^(-2)kgm//sec` The average force extered on surface `F = (p)/(t) = (0.48xx10^(-2))/(30xx60) = 0.27 xx 10^(-5)N` |
|
| 50. |
The energy contained in a small volume through which an electromagnetic wave os passing oscillates withA. zero frequencyB. the frequency of the waveC. half the frequency of the waveD. double the frequency of the wave |
| Answer» Correct Answer - D | |