An electron with kinetic energy 5 mathrm{eV} enters a region of uniform magnetic field of 3 μ mathr

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4.Moving Charges and Magnetism

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An electron with kinetic energy $5 \mathrm{eV}$ enters a region of uniform magnetic field of $3 \mu \mathrm{T}$ perpendicular to its direction. An electric field $\mathrm{E}$ is applied perpendicular to the direction of velocity and magnetic field. The value of $\mathrm{E}$, so that electron moves along the same path, is . . . . . $\mathrm{NC}^{-1}$.

(Given, mass of electron $=9 \times 10^{-31} \mathrm{~kg}$, electric charge $=1.6 \times 10^{-19} \mathrm{C}$ )

A

$3$

B

$4$

C

$5$

D

$6$

(JEE MAIN-2024)

Solution

For the given condition of moving undeflected, net force should be zero.

$\mathrm{qE} =\mathrm{qVB}$

$\mathrm{E} =\mathrm{VB}$

$=\sqrt{\frac{2 \times \mathrm{KE}}{\mathrm{m}}} \times \mathrm{B}$

$=\sqrt{\frac{2 \times 5 \times 1.6 \times 10^{-19}}{9 \times 10^{-31}}} \times 3 \times 10^{-6}$

$=4 \mathrm{~N} / \mathrm{C}$

Standard 12

Physics

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