JEE MAIN - Physics (2023 - 24th January Morning Shift - No. 25)
A stream of a positively charged particles having $${q \over m} = 2 \times {10^{11}}{C \over {kg}}$$ and velocity $${\overrightarrow v _0} = 3 \times {10^7}\widehat i\,m/s$$ is deflected by an electric field $$1.8\widehat j$$ kV/m. The electric field exists in a region of 10 cm along $$x$$ direction. Due to the electric field, the deflection of the charge particles in the $$y$$ direction is _________ mm.
Answer
2
Explanation
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$$ \begin{aligned} & F_y=\frac{q E_y}{m} \\\\ & a_y=2 \times 10^{11} \times 1800 \\\\ & =36 \times 10^{13} \mathrm{~m} / \mathrm{s}^2 \\\\ & \text { Time }=\frac{10 \times 10^{-2}}{v_0}=\frac{0.1}{3 \times 10^7}=\left(\frac{1}{3} \times 10^{-8}\right) \mathrm{sec} \text {. } \\\\ & \therefore \quad y=\frac{1}{2} a t^2 \\\\ & \Rightarrow y=\frac{1}{2} \times 36 \times 10^{13} \times\left(\frac{1}{3} \times 10^{-8}\right)^2 \\\\ & =2 \times 10^{-3} \mathrm{~m} \\\\ & =2 \mathrm{~mm} \\\\ \end{aligned} $$
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