GATE 2023 EE Q.43: solution and answer

Official answer: marks to all candidatesSource: Official answer key, GATE 2023 EE (IIT Kanpur)

1. Set up the fields from the figure. The printed sentence says $\mathbf{E}$ is along x and $\mathbf{B}$ along z, but the printed figure draws $\mathbf{E}$ along +z and $\mathbf{B}$ along +x. Options (B) and (D) talk about the y-z plane and progress along y, which fit the figure, so we use it: $\mathbf{E}=E{\hat{a}}_z$ and $\mathbf{B}=B{\hat{a}}_x$, both uniform, with $E,B>0$.

2. Lorentz force. $m\frac{d\mathbf{v}}{dt}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. Since $\mathbf{v}\times {\hat{a}}_x=v_z{\hat{a}}_y-v_y{\hat{a}}_z$, the components are $$\begin{array}{rl}m{\dot{v}}_x& =0\\ m{\dot{v}}_y& =qB{v}_z\\ m{\dot{v}}_z& =qE-qB{v}_y\end{array}$$

3. No motion along x. The charge starts from rest, so $v_x(0)=0$, and ${\dot{v}}_x=0$ keeps it at zero. So $x=0$ for all time and the charge stays in the y-z plane. (B) is true.

4. Solve in the y-z plane. Let $\omega =\frac{qB}{m}$ (the cyclotron frequency). With $v_y(0)=v_z(0)=0$: $$v_y=\frac{E}{B}(1-\cos \omega t),v_z=\frac{E}{B}\sin \omega t$$ Check: ${\dot{v}}_y=\frac{E}{B}\omega \sin \omega t=\omega v_z$ and ${\dot{v}}_z=\frac{E}{B}\omega \cos \omega t=\omega \left(\frac{E}{B}-v_y\right)$. Both match the equations above.

5. Integrate to get the path. $$y=\frac{E}{B}\left(t-\frac{\sin \omega t}{\omega }\right),z=\frac{E}{B\omega }(1-\cos \omega t)$$ This is a cycloid. $z$ swings between $0$ and $\frac{2mE}{q{B}^2}$, and $y$ never decreases because $v_y\ge 0$.

6. Drift. Averaged over one cycle, the velocity is $\frac{E}{B}$ along ${\hat{a}}_y$. That is the $\mathbf{E}\times \mathbf{B}$ direction, since ${\hat{a}}_z\times {\hat{a}}_x={\hat{a}}_y$. So the charge progresses along +y. (D) is true.

7. Rejecting the other options. (A): $v_z=(E/B)\sin \omega t$ keeps changing sign and $z$ keeps returning to 0, so the velocity along z is not constant. (C): the path is a cycloid. It looks like a circle only from a frame moving at $E/B$ along y, not in the lab frame. (A) and (C) are false.

8. About the key. The official key is MTA (marks to all). If you use the sentence's assignment instead ($\mathbf{E}=E{\hat{a}}_x$, $\mathbf{B}=B{\hat{a}}_z$), the same working keeps the charge in the x-y plane ($z=0$) and gives a drift along ${\hat{a}}_x\times {\hat{a}}_z=-{\hat{a}}_y$, so (B) would fail. With the figure, which the options are written around, the answer is (B) and (D).