An electron is moving along +x direction with a velocity of 6 × 106… | JEE Main 2020 (6 Sep, Shift 1)

Q80JEE Main 2020 · 6 SEP · SHIFT 1Medium

Moving Charge in a Magnetic Field

An electron is moving along +x direction with a velocity of 6 $\times$ 106 ms–1. It enters a region of uniform electric field of 300 V/cm pointing along +y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x direction will be:

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Answer: (c)

For no deflection the magnetic force must cancel the electric force: $B = \dfrac{E}{v}$ .
$E = 300\text{ V/cm} = 3 \times 10^{4}\text{ V/m}$ , so $B = \dfrac{3 \times 10^{4}}{6 \times 10^{6}} = 5 \times 10^{-3}\text{ T}$ .
Electric force on the electron: $-e\vec{E}$ is along $-\hat{j}$ , so the magnetic force must be along $+\hat{j}$ .
With $\vec{B} = B\hat{k}$ : $-e(v\hat{i} \times B\hat{k}) = -evB(-\hat{j}) = +evB\,\hat{j}$ . So $\vec{B}$ is along $+z$ .
So the answer is (c) $5 \times 10^{-3}\text{ T}$ along $+z$ .

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Moving Charge in a Magnetic Field question from JEE Main 2020 (6 Sep, Shift 1)

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