Physics · Moving Charges and Magnetic Field

JEE Main 2025 — 22 January, Morning Shift — Question 51

An electron is made to enters symmetrically between two parallel and equally

but oppositely charged metal plates, each of 10 cm length. The electron emerges

out of the field region with a horizontal component of velocity

$10^{6} \mathrm{~m} / \mathrm{s}$ . If the magnitude of the electric

between the plates is 9.1 $\mathrm{V} / \mathrm{cm}$ , then the v

ertical component of velocity of electron is (mass of electron

$=9.1 \times 10^{-31} \mathrm{~kg}$ and charge of electron $=1.6 \times 10^{-19} \mathrm{C}$ )

Answer: C

\Rightarrow t=\frac{l}{{{V}_{x}}}=\frac{10\times {{10}^{-2}}}{{{10}^{6}}}={{10}^{-7}}

{{v}_{y}}={{u}_{y}}+{{a}_{y}t}=0+\frac{eE}{m}\times {{10}^{-7}}

{{V}_{x}}=\frac{1.6\times {{10}^{-19}}}{9.1\times {{10}^{-31}}}\times 9.1\times {{10}^{-2}}\times {{10}^{-7}}\Rightarrow {{v}_{y}}=16\times {{10}^{6}}

Answer key and solution verified before publishing.

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