Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

120 of 276

Question 120 of 276

An electron with kinetic energy 5eV5 \mathrm{eV} enters a region of uniform magnetic field of 3 μT\mu \mathrm{T} perpendicular to its direction. An electric field E\mathrm{E} is applied perpendicular to the direction of velocity and magnetic field. The value of EE, so that electron moves along the same path, is _______ NC−1\mathrm{NC}^{-1}.
(\left(\right. Given, mass of electron =9×10−31 kg=9 \times 10^{-31} \mathrm{~kg}, electric charge =1.6×10−19C)\left.=1.6 \times 10^{-19} \mathrm{C}\right)

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Answer: 4

For the given condition of moving undeflected, net force should be zero.
qE=qVBE=VB=2×KEm×B=2×5×1.6×10−199×10−31×3×10−6=4 N/C\begin{aligned} \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}\end{aligned}

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