** CBSE Sample Papers**

# (NCERT) QUESTION & ANSWERS PHYSICS (Electromagnetic Waves)

**NCERT
: QUESTION & ANSWERS PHYSICS
**

**(Electromagnetic
Waves)
**

**1. ** **What do you mean by an
electromagnetic wave?**

**Solution:**Electromagnetic waves are by nature transverse to the direction of propagation of (i) oscillation of electric field and

(ii) oscillation of magnetic field, which are mutually perpendicular to each other. The direction of propagation of the wave is perpendicular to electric and magnetic fields.

**2. ** **Prove that only an
accelerated charge can produce an electromagnetic wave.**

**Solution:**A stationary charge has an electric field around it but no magnetic field. When given an impulse, it begins to move with the production of electric and magnetic fields. When the charge moves with a constant velocity, the magnetic field does not change with time, hence it cannot produce an electric field. As the charge is accelerated, both electric and magnetic fields change with time and space, one becoming a source of the other and giving rise to an electromagnetic wave.

**3. ** **What are the
important properties of electromagnetic waves?**

**Solution:**Some of the chief characteristics of electromagnetic waves are:

(i) In an electromagnetic wave, the directions of oscillation of the electric and magnetic fields are perpendicular to each other as well as to the direction of propagation of the wave.

(ii) The speed of an electromagnetic wave in space is about 3 ´ 10^{8}m/s while in any other medium, it depends on the electric and magnetic properties of the medium and not on the amplitude of the field variation.(iii) The electric and magnetic field variations are in phase, i.e. both attain their maxima and minima at the same rime and place.

**Solution:**Consider a parallel plate capacitor. If the plates of the capacitor have an area A, and a total change Q, the magnitude of the electric field between the plates is e

_{0}. The electric flux E between the plates is

F_{E}= = =

If Q on the capacitor plate changes with time, there is a current I = , so that we have =

or e_{0 }= I

Then if we generalize Ampere's circuital law by adding to the total current carried by conductors through the surface, another term which is e_{0}times the rate of change of electric flux through the same surface, the total has the same value I for all surfaces. Therefore at a point outside the plates is nearly the same as at a point just inside as it should be. The first current, carried by conductors and due to flow of charges is called conduction current. The second is due to changing electric field or electric displacement is called displacement current. By Maxwell's generalization it is not just the conduction electric current due to flowing charges, but also the time rate of change of electric field. More precisely, the total current I is equal to I_{c}, conduction current plus I_{d}, the displacement current

i.e. I = I_{c}+I_{d}

The displacement current is e_{0}times the rate of change of flux of electric field through the same surface as that through which the flow of conduction current is calculated. Therefore in all respects, the displacement current has the same physical effects on the conduction current.

**Solution:**Not all the quantities appearing in these equations have dimensions assigned to them from 'outside' (i.e., from independent equations and definitions). Some of the dimensions have to be fixed up from a few of the equations themselves, and the remaining equations (or terms in the equations) can then be checked for dimensional consistency. The dimensions of e

_{o}is given from Coulomb's law to be

=C^{2}N^{-1}m^{-2}

The dimensions of E is fixed by its definition as force per unit charge

= NC^{-1}

The dimensions of m_{o}may be fixed up using the first terms on the right hand side of Ampere's law (Eq. 4) and the dimension of current.

= Cs^{-1}

= NC^{-2}s^{2}

Equipped with these dimensions, the dimensional consistency of Eq.1, Eq. 3 & the second term of Eq. 4 can be checked. The Lorentz force equation, and the first term of the right of Eq. 4 have been already used in assigning dimension so there is no meaning of checking their consistency. Eq. 2 is automatically consistent since 0 can be assigned any dimension.

Equation 1:

= NC^{-1}m^{2}

=

= NC^{-1}m^{2}.

Equation 3:

= NC^{-1}m

= NC^{-1}m^{-1}s

= NC^{-1}m

Equation 4:

= NC^{-1}s

Second term on the right hand side of equation 4 is,

= NC^{-2}s^{2}C^{2}N^{-1}m^{-2}NC^{-1}m^{2}s^{-1}

= NC^{-1}s

Thus the above equations are dimensional consistent.

**Solution:**Let us consider a stationary electric charge. At some distance from this charge, there will be an electric field but no magnetic field. When the charge begins to move, there will be both electric and magnetic fields at P. If in a region of space, there is a flux of magnetic field varying with time, it gives rise to an emf (Faraday's law of electromagnetic induction). Applying this law to a small region, say a small square perpendicular to the direction of the magnetic field, it can be shown that the electric fields along the two parallel sides of the square are not the same. Thus a time dependent magnetic field gives rise to an electric field that varies with position. Now suppose that the electric field depends on time as well. Then from Maxwell's generalization of Ampere's circuital law, such an electric field produces a magnetic field. So, we find that electric and magnetic fields that depend on space and time produce and sustain each other. A simple form of such a continuing change is a wave. In a plane wave, for example, the electric and magnetic fields vary sinusoidally with distance at a given time, and with time at a given point. Thus, an oscillating charge, which has non-zero acceleration, will continuously emit electromagnetic waves.

**Solution:**(a) The radius of each plate r = 12 cm

The distance separated d = 5mm

.^{.}. The capacity of capacitor C = e_{0}A/d = e_{0}x p x r^{2}/d

= = 80.1 pf

(b) = C

But =I

.^{.}. I = C or = I/C

= = 1.87 x10^{9}V/s

Displacement current I_{D}=

Across the capacitor F_{E}= EA

Therefore I_{D}=

Now, =

.^{.}. = which implies I_{D}= I = 0.15A

(c) Yes, provided by 'current' we mean the sum of conduction and displacement currents.

**Solution:**(a) Consider a circle of radius r between the plates and coaxial with them (i.e., its center lies on the axis of the plates and its plane is normal to the axis). By symmetry, is tangential to the circle at every point and equal in magnitude over the circle. Therefore,

= 2pr ,

Using Ampere's law,

2pr = m_{0}x (current passing through the area enclosed by the circle)

= r__<__R

= mo I_{D}r ³ R

Thus = r__<__R

= r ³ R.

(i) = 0 on the axis (r = 0)

(ii) For r = 6.5 cm, = 1.35 x 10^{-7}T

(iii) For r = 15 cm, = 2 x 10^{-7}T

(b) At what distance from the axis is the magnetic field due to displacement current maximum ? Obtain the maximum value of the field.

is maximum at r = R. From either formula above, at r = R

= = 2.5 x 10^{-7}T.

**Solution:**Due to leaking, there is flow of positive charge from the positive plate to the negative plate (or flow of negative charge in the reverse direction). Thus the conduction current within the plates is from the positive plate to the negative plate. Now, the displacement current is

I_{D}= = =

But here , < 0. That is, I_{D}is opposite to the direction of electric field.

Thus I_{D}has the same magnitude 1.5 x 10^{-8}A, as the conduction current, but opposite in direction. The total current therefore is identically zero. This is true for any cross section between the plates.

Using Amperes law, with I_{D}there replaced by I = I_{C}+ I_{D}= 0, the magnetic field within the plates is zero at all points.

Source : Classontheweb.com