Magnetic Effects & Magnetism

JEE Main level, single correct, medium

221 of 276

Question 221 of 276

The electric field of a plane electromagnetic wave is given by E⃗=E0i^+j^2cos⁡⁡(kz+ωt)\vec{E}={E}_{0}\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}\cos ⁡\left(kz+\omega t\right) . At t=0t=0 , a positively charged particle is at the point (x,y,z)=(0,0,πk)\left(x,y,z\right)=\left(0,0,\dfrac{\pi }{k}\right) . If its instantaneous velocity at (t=0)\left(t=0\right) is v0k^{v}_{0}\hat{k} , the force acting on it due to the wave is:

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Answer: (C) antiparallel to i^+j^2\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}

Force due to electric field is in direction –(i^+j^)2E0–\dfrac{\left(\hat{i}+\hat{j}\right)}{\sqrt{2}}{E}_{0}
Force due to magnetic filed is in direction q(v⃗×B⃗)q\left(\vec{v}\times \vec{B}\right) and v⃗∣∣k^\vec{v}||\hat{k}
∴∴ it is parallel to E⃗\vec{E}
∴∴ net force is antiparallel to (i^+j^)2\dfrac{\left(\hat{i}+\hat{j}\right)}{\sqrt{2}} .

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