Electromagnetic Waves

JEE Main level, single correct, medium

105 of 165

Question 105 of 165

The magnetic field of a plane electromagnetic wave is given by B⃗=B0i^[cos(kz−ωt)]+B1j^cos(kz+ωt)\vec{B}={B}_{0}\hat{i}\left[cos\left(kz-\omega t\right)\right]+{B}_{1}\hat{j}cos\left(kz+\omega t\right), where B0=3×10–5T{B}_{0}=3\times {10}^{–5}T and B1=2×10–6T{B}_{1}=2\times {10}^{–6}T. The RMSRMS value of the force experienced by a stationary charge Q=10–4CQ={10}^{–4}C at z=0z=0 is closest to

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Answer: (D) 0.6N0.6N

B⃗=B0i^[cos(kz−ωt)]+B1j^cos(kz+ωt)\vec{B}={B}_{0}\hat{i}\left[cos\left(kz-\omega t\right)\right]+{B}_{1}\hat{j}cos\left(kz+\omega t\right)
{B}_{0}=3\times {10}^{-5}T&{B}_{1}=2\times {10}^{-6}T
Electric filed associated with it is,
E⃗=−B0cj^cos(kz−ωt)−B1ci^cos(kz+ωt)\vec{E}=-{B}_{0}c\hat{j}cos\left(kz-\omega t\right)-{B}_{1}c\hat{i}cos\left(kz+\omega t\right)
Here, cc is the speed of light in vacuum.
At z=0,z=0,
Frms=(QB0c)2+(QB1c)22{F}_{\text{rms}}=\sqrt{\dfrac{{\left(Q{B}_{0}c\right)}^{2}+{\left(Q{B}_{1}c\right)}^{2}}{2}}
Here cc is speed of light in vacuum.
=(10−4×3×10−5×3×108)2+(10−4×2×10−6×3×108)22=\sqrt{\dfrac{{\left({10}^{-4}\times 3\times {10}^{-5}\times 3\times {10}^{8}\right)}^{2}+{\left({10}^{-4}\times 2\times {10}^{-6}\times 3\times {10}^{8}\right)}^{2}}{2}}
=0.81+0.00362=0.6N=\sqrt{\dfrac{0.81+0.0036}{2}}=0.6 N

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