An electron, moving along the x-axis with an initial energy of 100… | JEE Main 2019 (12 Apr, Shift 2)

Q92JEE Main 2019 · 12 APR · SHIFT 2Difficult

Moving Charge in a Magnetic Field

An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field $\vec{B} = (1.5 \times 10^{-3}\text{ T})\hat{k}$ at S (See figure). The field extends between x = 0 and x = 2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is: (electron’s charge = 1.6 × 10–19 C, mass of electron = 9.1 × 10–31 kg)

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Answer: (b)

$v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2 \times 100 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 5.93 \times 10^{6}\text{ m/s}$
$r = \dfrac{mv}{eB} = \dfrac{9.1 \times 10^{-31} \times 5.93 \times 10^{6}}{1.6 \times 10^{-19} \times 1.5 \times 10^{-3}} \approx 2.25\text{ cm}$
Exit angle: $\sin\theta = \dfrac{2}{2.25} = 0.889$ , $\cos\theta \approx 0.457$ , $\tan\theta \approx 1.95$ . Shift inside field $= r(1 - \cos\theta) \approx 1.22\text{ cm}$ .
Straight line for the remaining 6 cm: rise $= 6\tan\theta \approx 11.65\text{ cm}$ .
$d \approx 1.22 + 11.65 = 12.87\text{ cm}$ . So the answer is (b).

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Moving Charge in a Magnetic Field question from JEE Main 2019 (12 Apr, Shift 2)

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