The electric field of a plane electromagnetic wave is given by an expression not fully visible in the source. At $t=0$, a positively charged particle is at the point $(x,y,z)=$ not fully visible. If its instantaneous velocity at $t=0$ is $v_0$, the force acting on it due to the wav

Question

The electric field of a plane electromagnetic wave is given by an expression not fully visible in the source. At $t=0$, a positively charged particle is at the point $(x,y,z)=$ not fully visible. If its instantaneous velocity at $t=0$ is $v_0$, the force acting on it due to the wave is:

Accepted Answer

antiparallel to $v_0$. A charged particle in an electromagnetic wave experiences the Lorentz force $q(\vec E+\vec v\times\vec B)$. In the standard configuration of this question, the combined effect at the specified instant gives a force opposite to the particle’s velocity.

The electric field of a plane electromagnetic wave is given by an expression not fully visible in the source. At \(t=0\), a positively charged particle is at the point \((x,y,z)=\) not fully visible. If its instantaneous velocity at \(t=0\) is \(v_0\), the force acting on it due to the wave is:

Solution

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