JEE Physics Electromagnetic Waves MCQs Set 15

Answer: (c) electric energy density is equal to the magnetic energy density

Explanation:
1. In an electromagnetic wave \( u_E=\tfrac12\varepsilon_0E^2 \) and \( u_B=\dfrac{B^2}{2\mu_0} \); with \( E=cB \) these are equal at every instant.

Teacher's Note:
1. The two energy densities are always equal, so each carries half the total.

Answer: (a) \( 2\times10^{-8} \) T

Explanation:
1. \( B=\dfrac{E}{c}=\dfrac{6}{3\times10^8}=2\times10^{-8} \) T.

Teacher's Note:
1. The frequency is not needed; only \( E/B=c \) matters.

Answer: (c) 1.66 rad m\(^{-1}\)

Explanation:
1. \( k=\dfrac{\omega}{c}=\dfrac{5\times10^8}{3\times10^8}=1.66 \) rad m\(^{-1}\).

Teacher's Note:
1. Compare with \( \cos(kz-\omega t) \) to read off \( \omega \).

Answer: (b) \( [ML^{-1}T^{-2}] \)

Explanation:
1. \( \tfrac12\varepsilon_0E^2 \) is energy per unit volume.
2. \( \dfrac{[ML^2T^{-2}]}{[L^3]}=[ML^{-1}T^{-2}] \).

Teacher's Note:
1. It is an energy density, which has the same dimensions as pressure.

Answer: (b) 153 N C\(^{-1}\)

Explanation:
1. \( E_0=cB_0=3\times10^8\times510\times10^{-9}=153 \) N C\(^{-1}\).

Teacher's Note:
1. Remember to convert nanotesla to tesla.

Answer: (c) \( 10^9 \) Hz

Explanation:
1. An oscillating charge radiates electromagnetic waves of the same frequency as its own oscillation.

Teacher's Note:
1. The frequency of the source and of the wave it emits are equal.

Answer: (a) \( k/\omega \)

Explanation:
1. \( k=2\pi/\lambda \) and \( \omega=2\pi f \) both change with the wave, but \( \dfrac{k}{\omega}=\dfrac1c \).
2. In vacuum \( c \) is the same for every wavelength.

Teacher's Note:
1. The ratio \( \omega/k \) is the phase speed, which is \( c \) in vacuum.

Answer: (b) \( 8.85\times10^{-12} \)

Explanation:
1. Average \( u_B=\dfrac{B_0^2}{4\mu_0}=\dfrac{E_0^2}{4\mu_0c^2}=\dfrac14\varepsilon_0E_0^2 \).
2. \( =\dfrac14\times8.85\times10^{-12}\times4=8.85\times10^{-12} \) J m\(^{-3}\).

Teacher's Note:
1. Average magnetic and electric energy densities are equal; each is \( \tfrac14\varepsilon_0E_0^2 \).

Answer: (a) \( 2\times10^{-8} \) T

Explanation:
1. At every point and instant \( B=E/c=6/(3\times10^8)=2\times10^{-8} \) T.

Teacher's Note:
1. E and B are in phase, so their instantaneous values are related by \( c \).

Answer: (c) is the same for all of them

Explanation:
1. All electromagnetic waves travel at \( c=1/\sqrt{\mu_0\varepsilon_0} \) in vacuum, whatever the source or frequency.

Teacher's Note:
1. Speed in vacuum is a constant of nature; only frequency and wavelength vary.

Answer: (d) 3660 cm

Explanation:
1. \( \lambda=\dfrac{c}{f}=\dfrac{3\times10^8}{8.196\times10^6}=36.6 \) m \( =3660 \) cm.

Teacher's Note:
1. Convert metres to centimetres at the end.

Answer: (d) \( \tfrac14\varepsilon_0E_0^2 \)

Explanation:
1. Instantaneous \( u_E=\tfrac12\varepsilon_0E^2 \).
2. The average of \( \sin^2 \) is \( \tfrac12 \), so \( \langle u_E\rangle=\tfrac14\varepsilon_0E_0^2 \).

Teacher's Note:
1. Peak value gives \( \tfrac12\varepsilon_0E_0^2 \); the time average halves it.

Answer: (a) \( \dfrac{\omega}{k} \)

Explanation:
1. The phase velocity is \( \omega/k \); \( d\omega/dk \) is the group velocity.

Teacher's Note:
1. Phase speed is the speed of a point of constant phase.

Answer: (c) electric and magnetic vectors perpendicular to each other

Explanation:
1. The wave consists of oscillating \( \vec E \) and \( \vec B \), perpendicular to each other and to the direction of propagation.

Teacher's Note:
1. Electromagnetic waves are transverse.

Answer: (b) 100 V m\(^{-1}\)

Explanation:
1. Intensity \( I=\dfrac{P}{4\pi r^2}=\dfrac{1500}{4\pi\times9}=13.3 \) W m\(^{-2}\).
2. \( I=\tfrac12\varepsilon_0cE_0^2 \Rightarrow E_0=\sqrt{\dfrac{2\times13.3}{8.85\times10^{-12}\times3\times10^8}}\approx100 \) V m\(^{-1}\).

Teacher's Note:
1. Use the intensity at the given distance, then \( I=\tfrac12\varepsilon_0cE_0^2 \).

Answer: (d) 6

Explanation:
1. \( \mu_r=\dfrac{5\times10^{-7}}{4\pi\times10^{-7}}=0.40 \).
2. \( n^2=\varepsilon_r\mu_r \Rightarrow \varepsilon_r=\dfrac{2.25}{0.40}\approx5.7\approx6 \).

Teacher's Note:
1. Use \( n=\sqrt{\varepsilon_r\mu_r} \).

Answer: (a) \( 1.8\times10^5 \) W

Explanation:
1. Area \( =6\times30=180 \) m\(^2\).
2. Power \( =10^3\times180=1.8\times10^5 \) W.

Teacher's Note:
1. Power is flux times area.

Answer: (c) \( \dfrac{1}{\sqrt{\mu_0\varepsilon_0}} \)

Explanation:
1. Maxwell's equations give \( c=\dfrac{1}{\sqrt{\mu_0\varepsilon_0}} \).

Teacher's Note:
1. Check units: \( \mu_0\varepsilon_0 \) has units of \( \text{s}^2\text{m}^{-2} \).

Answer: (d) \( E_y=66\cos2\pi\times10^{11}\left(t-\dfrac xc\right),\ B_z=2.2\times10^{-7}\cos2\pi\times10^{11}\left(t-\dfrac xc\right) \)

Explanation:
1. \( f=c/\lambda=3\times10^8/3\times10^{-3}=10^{11} \) Hz, so \( \omega=2\pi\times10^{11} \).
2. \( B_0=E_0/c=66/(3\times10^8)=2.2\times10^{-7} \) T.
3. \( \vec E\) is along y and the wave travels along x, so \( \vec B \) is along z, with amplitudes 66 V/m and \( 2.2\times10^{-7} \) T.

Teacher's Note:
1. Check three things: frequency, amplitude ratio \( c \), and the perpendicular direction of B.

Answer: (a) \( 3.1\times10^{-8} \) T

Explanation:
1. \( B=\dfrac Ec=\dfrac{9.3}{3\times10^8}=3.1\times10^{-8} \) T.

Teacher's Note:
1. B is along the Z-axis.

Answer: (a) 3

Explanation:
1. \( v=\dfrac{c}{\sqrt{K}}=\dfrac{3\times10^8}{\sqrt{1.006}}\approx2.99\times10^8 \) m s\(^{-1}\), so \( a\approx3 \).

Teacher's Note:
1. Air is almost like vacuum, so the speed is nearly \( c \).