The magnetic field vector of an electromagnetic wave is given by B = B₀ (i + j)/(√(2)) cos (kz - ω t); where i, j represents unit vector along x and y-axis respectively. At t = 0s, two electric charges q1 of 4 π coulomb and q2 of 2 π coulomb located at (0, 0, π/k) and (0, 0, 3π/k), respectively, have the same velocity of 0.5 c i, (where c is the velocity of light). The ratio of the force acting on charge q1 to q2 is:- (a) 2√(2): 1 (b) 1: √(2) (c) 2: 1 (d) √(2): 1

Answer: (c). At t = 0: cos (k · π/k) = cos π = -1 and cos (3π) = -1, so both charges see the same E and B. Both have the same velocity 0.5c i, so E + v × B is identical at the two points. Force F = q(E + v × B) ∝ q, so F₁: F₂ = q₁: q₂ = 4π: 2π. So the answer is (c) 2: 1.

The magnetic field vector of an electromagnetic wave is given by B =… | JEE Main 2021 (31 Aug, Shift 2)

Q48JEE Main 2021 · 31 AUG · SHIFT 2Difficult

Nature of EM Waves & Wave Equation

The magnetic field vector of an electromagnetic wave is given by $B = B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(kz - \omega t)$ ; where $\hat{i}, \hat{j}$ represents unit vector along x and y-axis respectively. At t = 0s, two electric charges q1 of 4 $\pi$ coulomb and q2 of 2 $\pi$ coulomb located at $\left(0, 0, \frac{\pi}{k}\right)$ and $\left(0, 0, \frac{3\pi}{k}\right)$ , respectively, have the same velocity of 0.5 c $\hat{i}$ , (where c is the velocity of light). The ratio of the force acting on charge q1 to q2 is :-

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Answer: (c)

At $t = 0$ : $\cos(k \cdot \pi/k) = \cos \pi = -1$ and $\cos(3\pi) = -1$ , so both charges see the same $\vec{E}$ and $\vec{B}$ .
Both have the same velocity $0.5c\,\hat{i}$ , so $\vec{E} + \vec{v} \times \vec{B}$ is identical at the two points.
Force $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) \propto q$ , so $F_1 : F_2 = q_1 : q_2 = 4\pi : 2\pi$ .
So the answer is (c) $2 : 1$ .

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