Physics NCERT Exemplar Solutions Class 12th Chapter Eight

(b, d) 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.

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

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(a, c, d) as we know that  $\lambda =\frac{c}{\upsilon }$ = $\frac{3*{10}^8}{{10}^9}$  = 0.3m and it is fall in region of radio wave.

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(b, d) As electric and magnetic field vectors E and B are perpendicular to each other as well as Perpendicular to the direction of propagation of electromagnetic wave.

Here in the question electromagnetic wave is propagating along x-direction. So, electric and Magnetic field vectors should have either y-direction or z-direction.

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(a, b, c) electromagnetic wave traveling along the z direction is given by

E=E₀cos (kz-wt)

The associated magnetic field B in electromagnetic wave is along xaxis along k $*E$

B₀=E₀/c=1/c (k $*E$ )

E=c (B $*k$ ) .As the angle is 90 so value is zero.

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(d) as the electric and magnetic field in electromagnetic wave are perpendicular to each other also they are perpendicular to wave propagation so they are plane polarised.

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(c) From a diode antenna, the electromagnetic waves are radiated outwards.The amplitude of electric field vector Eo which transports significant energy from the source falls off intensity inversely as the distance from the antenna, i.e. 1/r

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(c) As from the relation that electric and magnetic energy density are equal to one another so there ratio will be always 1:1

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(d) The direction of propagation of electromagnetic wave is perpendicular to both electric field vector E and magnetic field vector B, i.e., in the direction of E*B .

This can be seen by the diagram given below

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(c) E $\propto \text{exception 25:}$

$\frac{\left(E_1\right)}{\left(E_2\right)}$ = $\sqrt[]{\frac{P_1}{P_2}}$ = $\sqrt[]{\frac{1000}{50}}$

$E$ 2= E/ $\text{exception 25:}$

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(b) as we know momentum = energy transferred/speed of light

= $\frac{20*30*30*60}{3*{10}^8}$ = 36 $*$ 10^-4kgms^-1

Momentum of reflected light =0

So momentum delivered is = 36 $*$ 10^-4kgms^-1