An electron with kinetic energy 5 eV enters a region of uniform… | JEE Main 2024 (8 Apr, Shift 1)

Q40JEE Main 2024 · 8 APR · SHIFT 1Medium

Moving Charge in a Magnetic Field

An electron with kinetic energy $5\text{ eV}$ enters a region of uniform magnetic field of 3 $\mu\text{T}$ perpendicular to its direction. An electric field $E$ is applied perpendicular to the direction of velocity and magnetic field. The value of E, so that electron moves along the same path, is __________ $\text{NC}^{-1}$ . (Given, mass of electron $= 9 \times 10^{-31}\text{ kg}$ , electric charge $= 1.6 \times 10^{-19}\text{ C}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 4

Electron moves on the same straight path when $eE = evB$ , so $E = vB$ .
$v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2 \times 5 \times 1.6 \times 10^{-19}}{9 \times 10^{-31}}} = \dfrac{4}{3} \times 10^{6}\text{ m/s}$
$E = \dfrac{4}{3} \times 10^{6} \times 3 \times 10^{-6} = 4\text{ N C}^{-1}$
So $E = 4\text{ N C}^{-1}$ .

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Moving Charge in a Magnetic Field question from JEE Main 2024 (8 Apr, Shift 1)

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