Physics · 2019
JEE Main · 12 April 2019, Shift 2 · Q18
An electron, moving along the x -axis with an initial energy of 100 eV, enters a region of magnetic field B =(1.5 × 10^-3 T ) k at S (See figure).…
An electron, moving along the $\displaystyle x$-axis with an initial energy of $\displaystyle 100$ eV, enters a region of magnetic field $\displaystyle \overrightarrow{\mathrm{B}}=\left(1.5 \times 10^{-3} \mathrm{~T}\right) \hat{\mathrm{k}}$ at S (See figure). The field extends between $\displaystyle x=0$ and $\displaystyle x=2 \mathrm{~cm}$. The electron is detected at the point $\displaystyle Q$ on a screen placed $\displaystyle 8$ cm away from the point S. The distance d between P and Q (on the screen) is :
(electron's charge $\displaystyle =1.6 \times 10^{-19} \mathrm{C}$, mass of electron $\displaystyle =9.1 \times 10^{-31} \mathrm{~kg}$ )

Official answer
(2)
12.$\displaystyle 87$ cm
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.