The electric fields of two plane electromagnetic plane waves in vacuum are given by E→1=E0j^ cos ωt−kx and E→2=E0k^ cos ωt−ky. At t=0 , a particle of charge q is at origin with a velocity v→=0.8 cj^ (c is the speed of light in vacuum). The instantaneous force experienced by the particle is :
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a
E0q0.8 i^−j^+0.4k^
b
E0q0.8 i^+j^+0.2k^
c
E0q0.4 i^−3j^+0.8k^
d
E0q−0.8 i^+j^+k^
answer is B.
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Detailed Solution
For E1→ corresponding B1→=E0Ck^ cos ωt−kxFor E2→ corresponding B2→=E0Ci^ cos ωt−ky Total force acting at t=0, x=0 isF→net=qE→net+qv→×B→nethereE→net =E→1+E→2=E0j^+E0k^andB→net =B→1+B→2=E0Ck^+E0Ci^⇒F→ =qE0j^+E0k^ +q0.8 cj^×E0Ck^+E0Ci^⇒F→ =qE0j^+qE0k^+0.8qE0i^−0.8qE0k^⇒F→=qE00.8i^+j^+0.2k^
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The electric fields of two plane electromagnetic plane waves in vacuum are given by E→1=E0j^ cos ωt−kx and E→2=E0k^ cos ωt−ky. At t=0 , a particle of charge q is at origin with a velocity v→=0.8 cj^ (c is the speed of light in vacuum). The instantaneous force experienced by the particle is :