SATHEE: Modern Physics 7 Question 22

Modern Physics 7 Question 22

23. The electric field of a plane polarised electromagnetic wave in free space at time $t = 0$ is given by an expression.

$E (x, y) = 10 \hat{j} \cos [(6 x + 8 z)]$

The magnetic field $B (x, z, t)$ is given by (where, $c$ is the velocity of light)

(Main 2019, 10 Jan II)

(a) $\frac{1}{c} (6 \hat{k} - 8 \hat{i}) \cos [(6 x + 8 z + 10 c t)]$

(b) $\frac{1}{c} (6 \hat{k} - 8 \hat{i}) \cos [(6 x + 8 z - 10 c t)]$

(d) $\frac{1}{c} (6 \hat{k} + 8 \hat{i}) \cos [(6 x + 8 z - 10 c t)]$

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Solution:

We are given electric field as

$E = 10 \hat{j} \cos (6 x + 8 z)$

where, phase angle is independent of time, i.e., phase angle at $t = 0$ is $φ = 6 x + 8 z$ .

Phase angle for $B$ will also be $6 x + 8 z$ because for an electromagnetic wave $E$ and $B$ oscillate in same phase.

Thus, direction of wave propagation

$= \frac{6 \hat{i} + 8 \hat{k}}{\sqrt{6^{2} + 8^{2}}} = \frac{6 \hat{i} + 8 \hat{k}}{10}$

Let magnetic field vector,

$B = a \hat{i} + b \hat{j} + d \hat{k}$ , then direction of wave propagation is given by

$\begin{aligned} \frac{E \times B}{| E^{'} | | B |} & = \frac{10 \hat{j} \times (a \hat{i} + b \hat{j} + d \hat{k})}{10 \times {(a^{2} + b^{2} + d^{2})}^{1 / 2}} \\ = \frac{- 10 a \hat{k} + 10 d \hat{i}}{10 {(a^{2} + b^{2} + d^{2})}^{1 / 2}} \end{aligned}$

As, $| B | = \frac{| E |}{c} = \frac{10}{c}$

We get, $| B | = \sqrt{a^{2} + b^{2} + d^{2}} = 10 / c$

By putting this value in Eqs. (iii) and (ii), we get direction of propagation

$\begin{array}{ll} \Rightarrow & \frac{c 10 (d \hat{i} - a \hat{k})}{10 \times 10} = \frac{6 \hat{i} \times 8 \hat{k}}{10} \\ \Rightarrow d = 6 / c and a = - 8 / c \end{array}$

Hence, $B = \frac{6}{c} \hat{k} - \frac{8}{c} \hat{i} = \frac{1}{c} (6 \hat{k} - 8 \hat{i})$

As the general equation of magnetic field of an EM wave propagating in positive $y$ -direction is given as,

$B = B_{0} \cos (R y - ω t))$ $∴ B = \frac{1}{c} (6 \hat{k} - 8 \hat{i}) \cos (6 x + 8 z - 10 c t)$

Alternate method Given, electric field is $E (x, y)$ , i.e. electric field is in $x y$ -plane which is given as

$∵ E = 10 \hat{j} \cos (6 x + 8 z)$

Since, the magnetic field given is $B (x, z, t)$ , this means $B$ is in $x z$ -plane.

$∴$ Propagation of wave is in $y$ -direction.

$[∵$ for an electromagnetic wave,

$E ⊥ B ⊥$ propagation direction]

As Poynting vector suggests that $E \times B$ is parallel to $(6 \hat{i} + 8 \hat{k})$ .

$\begin{aligned} Let B = (x \hat{i} + z \hat{k}), \\ then E \times B = \hat{j} \times (x \hat{i} + z \hat{k}) = 6 \hat{i} + 8 \hat{k} \\ or - x \hat{k} + z \hat{i} = 6 \hat{i} + 8 \hat{k} \\ or x = - 8 and z = 6 \\ ∴ B = \frac{1}{c} (6 \hat{k} - 8 \hat{i}) \cos (6 x + 8 z - 10 c t) ∵ \frac{| E |}{| B |} = c \end{aligned}$