Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 8 April, Shift 1 — Question 50

An electron with kinetic energy 5 eV enters a region of uniform magnetic field of $3 \mu \mathrm{~T}$ 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 $\qquad$ $\mathrm{NC}^{-1}$ . (Given, mass of electron $=9 \times 10^{-31} \mathrm{~kg}$ , electric charge $=1.6 \times 10^{-19} \mathrm{C}$ )

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

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

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

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

\begin{aligned} & =\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} \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Motion of Charged Particles in Combined Electric and Magnetic Fields