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**MULTIPLE CHOICE QUESTIONS I**

**Q.1. One requires 11eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in****(a) Visible region.****(b) Infrared region.****(c) Ultraviolet region.****(d) Microwave region.****Ans. **(c)**Solution.**

E = 11 eV = 11×1.6×10^{-19} J

h = 6.62×10^{-34} J-S

E = hf

f = E/h =

=

=

= 2.65 x 10^{15} Hz

This frequency radiation belongs to the ultraviolet region.**Q.2. A linearly polarized electromagnetic wave given asis incident normally on a perfectly reflecting infinite wall at z = a. Assuming that the material of the wall is optically inactive, the reflected wave will be given as(a) Er = – E _{o}cos (kz-ωt)(b) E_{r} = E_{o}cos (kz +ωt)(c) Er = - E_{o}cos (kz+ωt)(d) Er = E_{o}sin(kz-ωt)Ans.** (b)

(a) 36 × 10

(b) 36 × 10

(c) 108 × 10

(d) 1.08 × 10

Ans.

Solution.

Energy flux= ϕ =20 W/cm

Area A= 30cm

U= total energy falling t sec= Energy flux * Area * time = ϕ At

U= 20×30×30×60 J

Momentum of the incident light =

= 36×10

As no reflection from the surface and for complete absorption so momentum of reflected radiation is zero.

Momentum delivered to surface= Change in momentumkg m/s

(-) sign shows the direction of momentum.

(a) E/2

(b) 2E

(c) E/√2

(d) √2E

Ans.

A changing electric field produces a changing magnetic field and vice versa which gives rise to a transverse wave known as electromagnetic wave. The time varying electric and magnetic field are mutually perpendicular to each other and also perpendicular to the direction of propagation of this wave. The electric vector is responsible for the optical effects of an EM wave and is called the light vector.

The direction of propagation of electromagnetic w ave is perpendicular to both electric field vector () andmagnetic field vector B, i.e., in the direction of.

Here, electromagnetic wave is along the z-direction which is given by the cross product o f E and B.

Q.6. The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is

(a) c : 1

(b) c

(c) 1 : 1

(d) √c :1

Ans.

The intensity of electromagnetic wave is given by, I = U

Intensity in relation with electric field

Intensity relation with magnetic field

Now taking the intensity in terms of electric field,

But,

∴

= (U

Q.7. An EM wave radiates outwards from a dipole antenna, with E

(d) Remains constant.

Ans. (c)

Solution.

As we know that electromagnetic waves are radiated from dipole antenna and radiated energy E ∝ 1/r

(b) The associated magnetic field is given as

(c) The given electromagnetic field is circularly polarised.

(d) The given electromagnetic wave is plane polarised.

Ans.

Here, in electromagnetic wave, the electric field vector is given ascos(kz-ωt). In electromagnetic wave the associated magnetic field Vector (E

Choose the correct options from the following:

(a) The associated magnetic field is given as

(b) The electromagnetic field can be written in terms of the associated magnetic field as

(c) kˆ.E = 0, kˆ.B= 0.

(d) kˆ×E= 0, kˆ× B = 0.

Ans.

Solution.

(a) The direction of propagation of an electromagnetic wave is always along the direction of vector product E x B. Refer to Figure.

(b)

(c)

(d)

(a) E

(b) E

(c) B

(d) E

Ans.

As the EM wave is plane polarized and its propagation is in +X direction. So direction ofandwill be in either Y and Z direction or Z and Y direction. So verifies answers (b) and (d).

(a) Will have frequency of 10

(b) Will have frequency of 2 × 109 Hz.

(c) Will have a wavelength of 0.3 m.

(d) Fall in the region of radio waves.

Ans.

Solution.

Vibrating particle produces electric and magnetic field, so will produce an E.M. wave of same frequency 109 Hz verifies answer (a).

∵ V= 10

So,

= 0.3 m

verifies answer (c)

As the range of radio waves are between 10 Hz to 10

(a) Moving with a constant velocity.

(b) Moving in a circular orbit.

(c) At rest.

(d) Falling in an electric field.

Ans.

Solution.

- An electromagnetic wave can be produced by accelerated or oscillating charge.
- An oscillating charge is accelerating continuously, it will radiate electromagnetic waves continuously.
- Electromagnetic waves are also produced when fast moving electrons are suddenly stopped by a metal target of high atomic number.

Here, in option (b) charge is moving in a circular orbit.

In circular motion, the direction of the motion of charge is changing continuously, thus it is an accelerated motion and this option is correct.

In option (d), the charge is falling in electric field. If a charged particle is moving in electric field it experiences a force or we can say it accelerates. We know an accelerating charge particle radiates electromagnetic waves. Hence option (d) is also correct.

Also, we know that a charge starts accelerating when it falls in an electric field.

Important points:

- In an atom an electron is circulating around the nucleus in a stable orbit, although accelerating does not emit electromagnetic waves; it does so only when it jumps from a higher energy orbit to a lower energy orbit.
- A simple LC oscillator and energy source can produce waves of desired frequency.

**Q.13. An EM wave of intensity I falls on a surface kept in vacuum and exerts radiation pressure p on it. Which of the following are true?(a) Radiation pressure is I/c if the wave is totally absorbed.(b) Radiation pressure is I/c if the wave is totally reflected.(c) Radiation pressure is 2I/c if the wave is totally reflected.(d) Radiation pressure is in the range I/c < p < 2I/c for real surfaces.Ans. **(a) (c) and (d)

Solution.

Radiation pressure is the force exerted by particles (dual nature of particle) on unit area, due to the change in momentum of radiated particles per unit area per sec = 1/c.

I= intensity of radiation

C= velocity of radiation

Radiations are absorbed, so momentum per unit area per second verify the answer (a).

When radiation is reflected back, the momentum becomes double as in earlier case, so discards answer (b) and verifies answer (c).

So variation of radiation pressure p comes between the rangeverifies answer (d).

**VERY SHORT ANSWER TYPE QUESTIONS**

**Q.14. Why is the orientation of the portable radio with respect to broadcasting station important?****Ans. **Transmitted carrier wave signals are plane polarized and if the intensity of signal is poor then receiving antenna of radio must be parallel to the component of either electric or magnetic field. Because energy is only due to amplitudes of electric and magnetic components in EM wave, magnitude of amplitude is in particular direction perpendicular to each other and perpendicular to wave propagation.**Q.15. Why does microwave oven heats up a food item containing water molecules most efficiently?Ans.** In Microwave oven, molecules of food item starts to vibrate by driven force due to microwaves with the frequency of microwave. But the natural frequency of water molecules matches with microwave frequency which causes resonance (more amplitude) which further causes increase in temperature.

Ans.

= dq/dt, where q = q

I

I

I

So reactance of capacitor increases on decreasing frequency

So I = 2πv CV

As reactance of capacitor increases, the current by Ohm’s law will decrease.

So the displacement current decreases frequency decreases when the conduction current is equal to displacement current.

Ans.

We are given equation

B = 12 x 10

On comparing this equation with standard equation, we get

B

The average intensity of the beam

= 1.71 W/m

Ans.

Let

Since sin

The variation of |S| with time T will be as given in the figure below:

**Q.20. 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.Ans.** The properties of an electromagnetic wave is same as other waves. Like other wave an electromagnetic wave also carries energy and momentum. Since, it carries momentum, an electromagnetic wave also exerts pressure called radiation pressure. This property of electromagnetic waves helped 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.

**SHORT ANSWER TYPE QUESTIONS**

**Q.21. Show that the magnetic field B at a point in between the plates of a parallel-plate capacitor during charging is(symbols having usual meaning).****Ans. **Let us assume I_{d} be the displacement current in the region between two plates of parallel plate capacitor, in the figure.

The magnetic field at a point between two plates of capacitor at a perpendicular distance r from the axis of plates is given by

**Q.22. Electromagnetic waves with wavelength(i) λ _{1} is used in satellite communication.(ii) λ_{2} is used to kill germs in water purifies.(iii) λ_{3} is used to detect leakage of oil in underground pipelines.(iv) λ_{4} is used to improve visibility in runways during fog and mist conditions.(a) Identify and name the part of electromagnetic spectrum to which these radiations belong.(b) Arrange these wavelengths in ascending order of their magnitude.(c) Write one more application of each.Ans. **(a)

Solution.

(a) (i) In satellite communications, microwave is widely used. Hence λ

(ii) In water purifier, ultraviolet rays are used to kill germs. So, λ

(iii) X-rays are used to detect leakage of oil in underground pipelines. So, λ

(iv) Infrared rays are used to improve visibility on runways during fog and mist conditions. So, it is the wavelength of infrared waves.

(b) Wavelength of X-rays < wavelength of UV < wavelength of infrared < wavelength of microwave.

=> λ

(c)

Radiation | Uses |

γ - rays | Gives informations on nuclear structure , medical treatment etc. |

X - rays | Medical diagnosis and treatment study of crystal structure , industrial radiograph . |

UV - rays | Preserve food , sterilizing the surgical instruments, detecting the invisible writings, fingerprints etc. |

Visible light | To see objects |

Infrared rays | To treat, muscular strain for taking photography during the fog , haze etc. |

Microwave and radio wave | In radar and telecommunication . |

**Q.23. Show that average value of radiant flux density ‘S’ over a single period ‘T’ is given by S = .Solution.**

Radiant flux density

Let electromagnetic waves be propagating along x-axis. If electric field vector of electromagnetic wave be along y-axis, then magnetic field vector be along z-axis. Therefore,

and B = B

E x B = (E

S = c

= c

Average value of the magnitude of radiant flux density over complete cycle is

Hence Proved

Ans.

Displacement current I

Charge in capacitor, q = CV

I

or

1 x 10

or dv/dt = 1/2 x 10

Hence by applying a varying potential difference of 500 V/s, we would produce a displacement current of desired value.

Ans

Ans. E - Energy received by surface per second = I. A

N = Number of photons received by surface per second.

Let the surface is perfectly absorbing,

Also, Pressure P = F/A = 1/c

Ans.

As the distance is doubled, so the intensity becomes one-fourth the initial value. But in case of laser it does not spread, so its intensity remain same.

Some geometrical characteristics of LASER beam which are responsible for the constant intensity is

(i) Unidirectional

(ii) Monochromatic

(iii) Coherent light

(iv) Highly collimated

These characteristics are missing in the case of normal light from the bulb.

Ans.

**LONG ANSWER TYPE QUESTIONS **

**Q.28. An infinitely long thin wire carrying a uniform linear static charge density λ is placed along the z-axis (Fig). The wire is set into motion along its length with a uniform velocity . Calculate the poynting vectorAns.** The electric field due to infinitely long thin wire

Magnetic field due to the wire,

Equivalent current flowing through the wire, i = λv

Hence

∴

Thus, electric field, E = V(t)/d

Now using Ohm's Law, the conduction current density

Here,

The displacement current density is given as

where,

= 2π x 80ε

Q.30. A long straight cable of length l is placed symmetrically along z-axis and has radius a(<<l). The cable consists of a thin wire and a coaxial conducting tube. An alternating current I(t) = I

(iii) Compare the conduction current I

The displacement current density is given by

(ii) Total displacement current, I

∴

(iii) The displacement current,

Here,

**Q.31. A plane EM wave travelling in vacuum along z direction is given by(i) Evaluateover the rectangular loop 1234 shown in Fig.(ii) Evaluateover the surface bounded by loop 1234.**

**(iii) Use equationto prove E _{o}/B_{0} = c.**

(iv) By using similar process and the equation

**prove that c =**

The line integral of E over the closed rectangular path 1234 in x-z plane of the figure is

.......... (i)

(ii) Now let us evaluatelet us consider the rectangle 1234 to be made of strips of are ds = hdz each.

................(ii)

(iii) We are given

Substituting the values from Eqs. (i) and (ii), we get

E

(iv) For evaluating let us consider a loop 1234 in y-z plane as shown in figure given below.

.............. (iii)

Now to evaluate let us consider the rectangle 1234 to be made of strips of area hds each.

∴ ........... (iv)

Let where I = conduction current

= 0 in vacuum

Using relations obtained in Eqs. (iii) and (iv) and simplifying, we get

But E_{0}/B_{o} = c and

therefore c =**Q.32. A plane EM wave travelling along z direction is described by E = E _{0} sin(kz – ωt )ˆi and B = B_{0} sin(kz – ωt )ˆj . Show that**

(i) The average energy density of the wave is given by

**(ii) The time averaged intensity of the wave is given by****Ans. **(i) Total energy carried by electromagnetic wave is due to electric field vector and magnetic field vector. In electromagnetic wave, E and B vary from point to point and from moment to moment.

The energy density due to electric field E is

The energy density due to magnetic field B is

Total energy density of electromagnetic wave

Let the HM wave be propagating along z-direction. The electric field vector and magnetic field vector be represented by

E = E_{0 }sin (kz - ωt)

B = Bo sin (kz - ωt)

The values of E^{2} and B^{2} vary from point to point and from moment to moment. Hence, the effective values of E^{2} and B^{2} are their time averages over complete cycle.

We know,

and

Hence, the time average value of E^{2} over complete cycle,

And, the time average value of B^{2} over complete cycle,

The time average of energy density over complete cycle

(ii) We know

where μ_{0 =} Absolute permeability, ε_{o} = Absolute permittivity, E_{0} and B_{0} = Amplitudes of electric field and magnetic field vectors.

The time average of energy density due to magnetic field B is

Hence, u_{B }= u_{E}; the time average of energy density due to magnetic field is equal to the time average of energy density due to electric field.

Time average intensity of the wave

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