JEE PYQ: Electromagnetic Waves Question 29
Question 29 - 2019 (09 Apr 2019 Shift 1)
The magnetic field of a plane electromagnetic wave is given by: $\vec{B} = B_0 \hat{i}[\cos(kz - \omega t)] + B_1 \hat{j} \cos(kz + \omega t)$ where $B_0 = 3 \times 10^{-5}$ T and $B_1 = 2 \times 10^{-6}$ T. The rms value of the force experienced by a stationary charge $Q = 10^{-4}$ C at $z = 0$ is closest to:
(1) 0.6 N
(2) 0.1 N
(3) 0.9 N
(4) $3 \times 10^{-2}$ N
Show Answer
Answer: (1)
Solution
$E_0 = cB_0 = 9 \times 10^{3}$ V/m. $E_{\text{rms}} = \frac{E_0}{\sqrt{2}}$. $F = E_{\text{rms}} Q \cong 0.64$ N.
BETA
