Electromagnetic Waves: NEET and JEE questions with solutions

Answer: (B) The wavelength of the electromagnetic wave inside the medium is $300$ m

Speed: $v = \dfrac{c}{\sqrt{\epsilon_r}} = \dfrac{3 \times 10^8}{3} = 10^8\ \text{m s}^{-1}$. Statement (1) is correct.

Frequency: $2\pi f = 2\pi \times 10^6$, so $f = 10^6$ Hz.
$$\lambda = \frac{v}{f} = \frac{10^8}{10^6} = 100\ \text{m}$$
So statement (2), which says $300$ m, is incorrect. ($300$ m is the wavelength in vacuum.)

The phase $(kz - \omega t)$ shows that the wave travels along $+z$, so statement (4) is correct. With $\vec{E}$ along $x$ and propagation along $z$, $\vec{B}$ is along $y$ and in phase with $\vec{E}$, with amplitude $E_0/v$. Statement (3) describes this; the printed $B_0$ stands for this amplitude.

The incorrect choice is (2).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) P-II, Q-I, R-IV, S-III

Microwaves are used in microwave ovens for warming food: P-II.

UV rays kill germs and are used for purifying water: Q-I.

Gamma rays are used in radiotherapy for treating cancer cells: R-IV.

Radio waves are used for AM and FM communication: S-III.

Correct matching: P-II, Q-I, R-IV, S-III.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $B_y = 2 \times 10^{-7}\cos(5x + 1.5 \times 10^9 t)\ \text{T}$

Amplitude: $B_0 = \dfrac{E_0}{c} = \dfrac{60}{3 \times 10^8} = 2 \times 10^{-7}$ T.

The phase $(5x + \omega t)$ means the wave travels along $-x$. $\vec{B}$ must be perpendicular to both $\vec{E}$ (along $z$) and the direction of travel (along $x$), so $\vec{B}$ is along $y$.

Check the direction: $\vec{E} \times \vec{B}$ must point along $-x$, and $\hat{z} \times \hat{y} = -\hat{x}$. So $B_y$ is in phase with $E_z$ (same sign):
$$B_y = 2 \times 10^{-7}\cos(5x + 1.5 \times 10^9 t)\ \text{T}$$

(Also, $\omega/k = 1.5 \times 10^9/5 = 3 \times 10^8\ \text{m s}^{-1} = c$, as expected.)

Solution by Sreeraj P, M.Sc Physics

Answer: (C) non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates

The displacement current $I_d = \epsilon_0\dfrac{d\phi_E}{dt}$ is constant and spread uniformly over the plate area (radius $R$).

Ampère–Maxwell law at distance $r$ from the axis:

Inside ($r < R$): $B \cdot 2\pi r = \mu_0 I_d\dfrac{r^2}{R^2}$, so $B = \dfrac{\mu_0 I_d r}{2\pi R^2}$, which increases with $r$.

Outside ($r > R$): $B = \dfrac{\mu_0 I_d}{2\pi r}$, which decreases with $r$.

So $B$ is non-zero everywhere (except on the axis) and is maximum at $r = R$, the cylindrical surface joining the edges of the plates.

Solution by Sreeraj P, M.Sc Physics