The magnetic field of a plane electromagnetic wave is given by: B = B₀ i [cos (kz - ω t)] + B₀ j cos (kz + ω t) B0 = 3 × 10–5 T and B1 = 2 × 10–6 T. The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to: (a) 0.6 N (b) 0.9 N (c) 3 × 10–2 N (d) 0.1 N

Answer: (a). A stationary charge feels only the electric force. Each wave has E = cB: wave 1 ( B ∥ i, along +z) has E₁ ∥ j; wave 2 ( B ∥ j, along -z) has E₂ ∥ i. At z = 0 both vary as cos ω t, so |E| = c√(B₀² + B₁²) |cos ω t|. Erms = (c√(B₀² + B₁²))/(√(2)) = 3 × 10⁸ × 3.007 × 10⁻⁵1.414 ≈ 6.38 × 10³ V/m. Frms = QErms = 10⁻⁴ × 6.38 × 10³ ≈ 0.64 N. So the answer is (a) 0.6 N.

The magnetic field of a plane electromagnetic wave is given by: B =… | JEE Main 2019 (9 Apr, Shift 1)

Q69JEE Main 2019 · 9 APR · SHIFT 1Difficult

Nature of EM Waves & Wave Equation

The magnetic field of a plane electromagnetic wave is given by:

$\vec{B} = B_0 \hat{i} [\cos(kz - \omega t)] + B_0 \hat{j} \cos(kz + \omega t)$

B0 = 3 × 10–5 T and B1 = 2 × 10–6 T.

The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to:

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Answer: (a)

A stationary charge feels only the electric force. Each wave has $E = cB$ : wave 1 ( $\vec{B} \parallel \hat{i}$ , along $+z$ ) has $\vec{E}_1 \parallel \hat{j}$ ; wave 2 ( $\vec{B} \parallel \hat{j}$ , along $-z$ ) has $\vec{E}_2 \parallel \hat{i}$ .
At $z = 0$ both vary as $\cos \omega t$ , so $|\vec{E}| = c\sqrt{B_0^2 + B_1^2}\,|\cos \omega t|$ .
$E_{rms} = \dfrac{c\sqrt{B_0^2 + B_1^2}}{\sqrt{2}} = \dfrac{3 \times 10^{8} \times 3.007 \times 10^{-5}}{1.414} \approx 6.38 \times 10^{3}$ V/m.
$F_{rms} = QE_{rms} = 10^{-4} \times 6.38 \times 10^{3} \approx 0.64$ N.
So the answer is (a) 0.6 N.

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