The electric field of a plane electromagnetic wave is given by E = E₀ (i + j)/(√(2)) cos (kz + ω t) At t = 0, a positively charged particle is at the point (x, y, z) = (0, 0, π/k). If its instantaneous velocity at (t = 0) is v₀ k, the force acting on it due to the wave is: (a) parallel to k (b) parallel to (i + j)/(√(2)) (c) antiparallel to (i + j)/(√(2)) (d) zero

Answer: (c). Phase (kz + ω t): wave moves along -k. At t = 0, z = π/k: cos π = -1, so E = -E₀(i + j)/(√(2)). B = 1/c(-k) × E = E₀/c(j - i)/(√(2)). v × B = v₀k × E₀/c(j - i)/(√(2)) = -(v₀ E₀)/(c)(i + j)/(√(2)). Both qE and qv × B point along -(i + j). So the answer is (c) antiparallel to (i + j)/(√(2)).

The electric field of a plane electromagnetic wave is given by E = E₀… | JEE Main 2020 (7 Jan, Shift 2)

Q60JEE Main 2020 · 7 JAN · SHIFT 2Difficult

Nature of EM Waves & Wave Equation

The electric field of a plane electromagnetic wave is given by

$\vec{E} = E_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(kz + \omega t)$

At t = 0, a positively charged particle is at the point (x, y, z) = $\left(0, 0, \frac{\pi}{k}\right)$ .

If its instantaneous velocity at (t = 0) is $v_0 \hat{k}$ , the force acting on it due to the wave is:

Show answer and solution

Answer: (c)

Phase $(kz + \omega t)$ : wave moves along $-\hat{k}$ . At $t = 0$ , $z = \pi/k$ : $\cos \pi = -1$ , so $\vec{E} = -E_0\dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$ .
$\vec{B} = \dfrac{1}{c}(-\hat{k}) \times \vec{E} = \dfrac{E_0}{c}\dfrac{\hat{j} - \hat{i}}{\sqrt{2}}$ .
$\vec{v} \times \vec{B} = v_0\hat{k} \times \dfrac{E_0}{c}\dfrac{\hat{j} - \hat{i}}{\sqrt{2}} = -\dfrac{v_0 E_0}{c}\dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$ .
Both $q\vec{E}$ and $q\vec{v} \times \vec{B}$ point along $-(\hat{i} + \hat{j})$ .
So the answer is (c) antiparallel to $\dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$ .

Report a mistake

Nature of EM Waves & Wave Equation question from JEE Main 2020 (7 Jan, Shift 2)

More questions on Nature of EM Waves & Wave Equation