A velocity selector consists of… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Electromagnetism

2022mcqeasy

A velocity selector consists of electric field $\vec{E}=E\hat{k}$ and magnetic field $\vec{B}=B\hat{j}$ with $B=12\mathrm{mT}$ . The value $E$ required for an electron of energy $728\mathrm{eV}$ moving along the positive $x$ -axis to pass undeflected is

(Given, mass of electron $=9.1\times {10}^{-31}\mathrm{kg}$ )

Official previous-year question

Held on 26 Jul 2022 · Verified 6 Jul 2026.

Options

A
$192\mathrm{kV}{m}^{-1}$
B
$192\mathrm{mV}{m}^{-1}$
C
$9600\mathrm{kV}{m}^{-1}$
D
$16\mathrm{kV}{m}^{-1}$

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Solution

Fiven that $\vec{E}=E\hat{k}$ and $\vec{B}=12\hat{j}\mathrm{mT}$

Kinetic energy $=728\mathrm{eV}$

Kinetic energy $=\frac{1}{2}m{v}^{2}$

$\Rightarrow 728\mathrm{eV}=\frac{1}{2}\times 9.1\times {10}^{-31}\times {v}^{2}$

$\Rightarrow 728\times 1.6\times {10}^{-19}=\frac{1}{2}\times 9.1\times {10}^{-31}\times {v}^{2}$

$\Rightarrow v=16\times {10}^{6}m{s}^{-1}$

For electron to move undeflected net force on it should be zero.

$\Rightarrow eE=evB$

$\Rightarrow E=vB=16\times {10}^{6}\times 12\times {10}^{-3}$

$\Rightarrow E=192\times {10}^{3}V{m}^{-1}=192\mathrm{kV}{m}^{-1}$

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Solution

Fiven that $\vec{E}=E\hat{k}$ and $\vec{B}=12\hat{j}\mathrm{mT}$

Kinetic energy $=728\mathrm{eV}$

Kinetic energy $=\frac{1}{2}m{v}^{2}$

$\Rightarrow 728\mathrm{eV}=\frac{1}{2}\times 9.1\times {10}^{-31}\times {v}^{2}$

$\Rightarrow 728\times 1.6\times {10}^{-19}=\frac{1}{2}\times 9.1\times {10}^{-31}\times {v}^{2}$

$\Rightarrow v=16\times {10}^{6}m{s}^{-1}$

For electron to move undeflected net force on it should be zero.

$\Rightarrow eE=evB$

$\Rightarrow E=vB=16\times {10}^{6}\times 12\times {10}^{-3}$

$\Rightarrow E=192\times {10}^{3}V{m}^{-1}=192\mathrm{kV}{m}^{-1}$

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Solution

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JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

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