GATE EE 2023 Set 1 — Question 43

Motion with the field directions drawn in the figureThe figure shows the electric field along $+z$ and the magnetic field along $+x$, i.e. $\vec{E} = E\hat{k}$ and $\vec{B} = B\hat{i}$. The positive charge starts from rest, so the initial force is purely electric, $\vec{F}(0) = qE\hat{k}$.With $\vec{v} = v_x\hat{i} + v_y\hat{j} + v_z\hat{k}$ and $\vec{v}\times\hat{i} = v_z\hat{j} - v_y\hat{k}$, the Lorentz force gives$$m\dot{\vec{v}} = q\left(E\hat{k} + \vec{v}\times B\hat{i}\right) = qv_zB\,\hat{j} + q(E - v_yB)\,\hat{k}$$$$\vec{v}_d = \frac{\vec{E}\times\vec{B}}{B^2} = \frac{(E\hat{k})\times(B\hat{i})}{B^2} = \frac{E}{B}\hat{j}$$so the particle progresses along $+y$, which is statement (D).Taking the field directions from the sentence insteadIf $\vec{E} = E\hat{i}$ and $\vec{B} = B\hat{k}$ as the text states, then $\vec{v}\times\hat{k} = v_y\hat{i} - v_x\hat{j}$ and$$m\dot{\vec{v}} = q(E + v_yB)\,\hat{i} - qv_xB\,\hat{j}$$so $\dot{v}_z = 0$ and the motion is confined to the x-y plane, with drift$$\vec{v}_d = \frac{(E\hat{i})\times(B\hat{k})}{B^2} = -\frac{E}{B}\hat{j}$$The trajectory is again a cycloid, but it lies in the x-y plane and advances along $-y$, so it does not match (B), and (A) and (C) remain false.The sentence and the figure therefore specify different field directions, and the set of true statements depends on which one is taken. Because the question cannot be resolved under a single consistent reading, every candidate is awarded the marks for this question.Answer: With the directions in the figure, (B) and (D) are true; the question as printed is inconsistent, so marks are awarded to all.