CBSE Class 12 Physics Chapter 8 Electromagnetic Waves NCERT Solutions

The given electric field expression is $\backslash vec\{E\}\; =\; \backslash \{(3.1\; \backslash text\{\; N/C\})\backslash \}\; \backslash cos\; [(1.8\; \backslash text\{\; rad/m\})\; y\; +\; (5.4\; \backslash times\; 10^8\; \backslash text\{\; rad/s\})t]\; \backslash hat\{i\}$.

(a) Direction of propagation: The general form of a plane electromagnetic wave propagating along the y-axis can be written as $\backslash vec\{E\}\; =\; E\_0\; \backslash cos(ky\; \backslash pm\; \backslash omega\; t)\; \backslash hat\{x\}$ or $\backslash vec\{E\}\; =\; E\_0\; \backslash cos(ky\; \backslash pm\; \backslash omega\; t)\; \backslash hat\{z\}$, and the corresponding magnetic field would be perpendicular to both $\backslash vec\{E\}$ and the direction of propagation. In the given expression, the electric field is along the $\backslash hat\{i\}$ (x-direction), and the argument of the cosine function is $(1.8\; y\; +\; 5.4\; \backslash times\; 10^8\; t)$. The presence of both $y$ and $t$ with a positive sign between them indicates that the wave is propagating in the negative y-direction. Therefore, the direction of propagation is $-\backslash hat\{j\}$.

(b) Wavelength $\backslash lambda$: The wave number $k$ is given as $k\; =\; 1.8\; \backslash text\{\; rad/m\}$. The relationship between wavelength $\backslash lambda$ and wave number $k$ is $k\; =\; \backslash frac\{2\backslash pi\}\{\backslash lambda\}$. Rearranging this formula to solve for $\backslash lambda$, we get $\backslash lambda\; =\; \backslash frac\{2\backslash pi\}\{k\}$. Substituting the given value of $k$: $\backslash lambda\; =\; \backslash frac\{2\backslash pi\}\{1.8\}\; \backslash approx\; 3.49\; \backslash text\{\; m\}$. Thus, the wavelength is approximately 3.49 meters.

(c) Frequency $\backslash nu$: The angular frequency $\backslash omega$ is given as $\backslash omega\; =\; 5.4\; \backslash times\; 10^8\; \backslash text\{\; rad/s\}$. The relationship between angular frequency $\backslash omega$ and frequency $\backslash nu$ is $\backslash omega\; =\; 2\backslash pi\; \backslash nu$. Solving for $\backslash nu$, we get $\backslash nu\; =\; \backslash frac\{\backslash omega\}\{2\backslash pi\}$. Substituting the given value of $\backslash omega$: $\backslash nu\; =\; \backslash frac\{5.4\; \backslash times\; 10^8\}\{2\backslash pi\}\; \backslash approx\; 8.6\; \backslash times\; 10^7\; \backslash text\{\; Hz\}$. The frequency of the wave is approximately $8.6\; \backslash times\; 10^7$ Hz (or 86 MHz).

(d) Amplitude of the magnetic field $B\_0$: The amplitude of the electric field is $E\_0\; =\; 3.1\; \backslash text\{\; N/C\}$. In vacuum, the relationship between the amplitudes of the electric and magnetic fields is given by $E\_0\; =\; c\; B\_0$, where $c$ is the speed of light in vacuum ($c\; \backslash approx\; 3\; \backslash times\; 10^8$ m/s). Therefore, the amplitude of the magnetic field is $B\_0\; =\; \backslash frac\{E\_0\}\{c\}$. Substituting the values: $B\_0\; =\; \backslash frac\{3.1\; \backslash text\{\; N/C\}\}\{3\; \backslash times\; 10^8\; \backslash text\{\; m/s\}\}\; \backslash approx\; 1.03\; \backslash times\; 10^\{-8\}\; \backslash text\{\; T\}$. The amplitude of the magnetic field is approximately $1.03\; \backslash times\; 10^\{-8\}$ Tesla.

(e) Expression for the magnetic field part of the wave: The electric field is propagating in the $-\backslash hat\{j\}$ direction, and the electric field vector is along $\backslash hat\{i\}$. According to the right-hand rule for electromagnetic waves ($\backslash vec\{E\}\; \backslash times\; \backslash vec\{B\}$ gives the direction of propagation), if $\backslash vec\{E\}$ is along $\backslash hat\{i\}$ and propagation is along $-\backslash hat\{j\}$, then $\backslash vec\{B\}$ must be along $-\backslash hat\{k\}$ (since $\backslash hat\{i\}\; \backslash times\; (-\backslash hat\{k\})\; =\; -(-\backslash hat\{j\})\; =\; \backslash hat\{j\}$, which is incorrect. Let's re-evaluate. If propagation is $-\backslash hat\{j\}$ and $\backslash vec\{E\}$ is $\backslash hat\{i\}$, then $\backslash vec\{B\}$ must be along $-\backslash hat\{k\}$ because $\backslash hat\{i\}\; \backslash times\; (-\backslash hat\{k\})\; =\; \backslash hat\{j\}$ is not the direction of propagation. Let's check $\backslash hat\{i\}\; \backslash times\; \backslash hat\{k\}\; =\; -\backslash hat\{j\}$. So, if $\backslash vec\{E\}$ is along $\backslash hat\{i\}$ and $\backslash vec\{B\}$ is along $\backslash hat\{k\}$, the propagation is $-\backslash hat\{j\}$. Thus, the magnetic field vector should be along $\backslash hat\{k\}$ for propagation in $-\backslash hat\{j\}$ direction. The expression for the magnetic field is $\backslash vec\{B\}\; =\; B\_0\; \backslash cos(ky\; +\; \backslash omega\; t)\; \backslash hat\{k\}$. Substituting the values $B\_0\; =\; 1.03\; \backslash times\; 10^\{-8\}\; \backslash text\{\; T\}$, $k\; =\; 1.8\; \backslash text\{\; rad/m\}$, and $\backslash omega\; =\; 5.4\; \backslash times\; 10^8\; \backslash text\{\; rad/s\}$, we get: $\backslash vec\{B\}\; =\; \backslash \{(1.03\; \backslash times\; 10^\{-8\}\; \backslash text\{\; T\})\backslash \}\; \backslash cos\; [(1.8\; \backslash text\{\; rad/m\})\; y\; +\; (5.4\; \backslash times\; 10^8\; \backslash text\{\; rad/s\})t]\; \backslash hat\{k\}$.

The power rating of the light bulb is $P\_\{total\}\; =\; 100\; \backslash text\{\; W\}$. It is given that only 5% of this power is converted into visible radiation.

Therefore, the power of the visible radiation emitted by the bulb is:

$$P\_\{visible\}\; =\; \backslash frac\{5\}\{100\}\; \backslash times\; P\_\{total\}\; =\; \backslash frac\{5\}\{100\}\; \backslash times\; 100\; \backslash text\{\; W\}\; =\; 5\; \backslash text\{\; W\}$$

The radiation is emitted isotropically, meaning it spreads out uniformly in all directions. The intensity $I$ of the radiation at a distance $r$ from the source is given by the formula:

$$I\; =\; \backslash frac\{P\_\{visible\}\}\{4\backslash pi\; r^2\}$$

(a) Intensity at a distance of 1 m:

Here, $r\; =\; 1\; \backslash text\{\; m\}$. Substituting the values into the intensity formula:

$$I\_1\; =\; \backslash frac\{5\; \backslash text\{\; W\}\}\{4\backslash pi\; (1\; \backslash text\{\; m\})^2\}\; =\; \backslash frac\{5\}\{4\backslash pi\}\; \backslash text\{\; W/m\}^2\; \backslash approx\; 0.398\; \backslash text\{\; W/m\}^2$$

The average intensity of visible radiation at a distance of 1 m is approximately $0.398\; \backslash text\{\; W/m\}^2$.

(b) Intensity at a distance of 10 m:

Here, $r\; =\; 10\; \backslash text\{\; m\}$. Substituting the values into the intensity formula:

$$I\_\{10\}\; =\; \backslash frac\{5\; \backslash text\{\; W\}\}\{4\backslash pi\; (10\; \backslash text\{\; m\})^2\}\; =\; \backslash frac\{5\}\{4\backslash pi\; \backslash times\; 100\}\; \backslash text\{\; W/m\}^2\; =\; \backslash frac\{1\}\{80\backslash pi\}\; \backslash text\{\; W/m\}^2\; \backslash approx\; 0.00398\; \backslash text\{\; W/m\}^2$$

The average intensity of visible radiation at a distance of 10 m is approximately $0.00398\; \backslash text\{\; W/m\}^2$.