The magnetic field of a plane electromagnetic wave is given by: $\overrightarrow{\mathrm{B}}=\mathrm{B}_{0} \hat{\mathrm{i}}[\cos (\mathrm{kz}-\omega \mathrm{t})]+\mathrm{B}_{1} \hat{\mathrm{j}} \cos (\mathrm{kz}-\omega \mathrm{t})$ where $\mathrm{B}_{0}=3 \times 10^{-5} \mathrm{~T}$ and $\mathrm{B}_{1}=2 \times 10^{-6} \mathrm{~T}$. The rms value of the force experienced by a stationary charge $\mathrm{Q}=10^{-4}=\mathrm{C}$ at $\mathrm{z}=0$ is closet to:

(D) $3 \times 10^{-2} \mathrm{~N}$

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · EM Waves

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

B = B_{0} \hat{i} [cos (kz - ω t)] + B_{1} \hat{j} cos (kz - ω t)

where

B_{0} = 3 \times 1 0^{- 5} T

and

B_{1} = 2 \times 1 0^{- 6} T

. The rms value of the force experienced by a stationary charge

Q = 1 0^{- 4} = C

z = 0

is closet to:

Answer: (B)

A stationary charge feels no magnetic force, so the force comes entirely from the wave's electric field, with

E = c B

. The two magnetic components combine to a peak

B = B_{0}^{2} + B_{1}^{2} \approx 3 \times 1 0^{- 5}

T, since

B_{1}

is tiny beside

B_{0}

. The peak force is

F_{ma x} = Q c B = 1 0^{- 4} \times 3 \times 1 0^{8} \times 3 \times 1 0^{- 5} \approx 0.9

N, and its rms value is

\frac{F _{ma x}}{2} \approx \frac{0.9}{2} \approx 0.6

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option