The Electric field of a plane electromagnetic wave is given byE→$$=E0i^+j^2cosKz+$$ωt. At $$t=0$$, a positively charged particle is at the point x,y,$$z=0$$,$$0$$,πK.If its instantaneous velocity at (t=0) is $$v0k^$$ , the force acting on it due to the wave is:

Question

The Electric field of a plane electromagnetic wave is given byE→$$=E0i^+j^2cosKz+$$ωt. At $$t=0$$, a positively charged particle is at the point  x,y,$$z=0$$,$$0$$,πK.If its instantaneous velocity at (t=0) is  $$v0k^$$ , the force acting on it due to the wave is:

Infinity Learn · Accepted Answer

At $$t=0$$, x $$=K$$ ​Electric field $$=E$$→$$=E0i^+j^2cos$$π$$=$$−$$i^+j^2E0$$ ∴  Force due to electric field is in direction  −$$i^+j^2Direction$$ ofelectromagnetic wave is in −z direction​∴$$C^=$$−$$k^$$ is wave direction.Let $$B^$$ is the unit vector along the direction of magnetic field.​ $$B^=C^$$×$$E^$$ $$=$$ −$$k^$$×−$$i^+j^2=$$−$$i^+j^2Now$$, Force due to magnetic field is in direction qv→×B→  and  v→||$$k^$$ Force due to magnetic field is $$F^=k^$$×$$-i^+j^2$$ $$=-i^-j^2=-i^+j^2We$$ can see that direction of both electric and magnetic force is same.∴  Net force is antiparallel to  $$i^+j^2$$

The Electric field of a plane electromagnetic wave is given byE→=E0i^+j^2cosKz+ωt. At t=0, a positively charged particle is at the point x,y,z=0,0,πK.If its instantaneous velocity at (t=0) is v0k^ , the force acting on it due to the wave is:

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