Q.

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 is​F→net=qE→net+qv→×B→net​here​E→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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