Proton with kinetic energy of 1MeV moves from south to north. It gets… | JEE Main 2020 (8 Jan, Shift 1)

Q75JEE Main 2020 · 8 JAN · SHIFT 1Medium

Moving Charge in a Magnetic Field

Proton with kinetic energy of 1MeV moves from south to north. It gets an acceleration of 1012 m/s2 by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is 1.6 × 10–27 kg):

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Answer: (a)

$K = 1\text{ MeV} = 1.6\times10^{-13}\text{ J}$ , $v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{3.2\times10^{-13}}{1.6\times10^{-27}}} = 1.41\times10^{7}\text{ m/s}$
$\vec v\perp\vec B$ : $evB = ma \Rightarrow B = \dfrac{ma}{ev}$
$B = \dfrac{1.6\times10^{-27}\times10^{12}}{1.6\times10^{-19}\times1.41\times10^{7}} = 7.1\times10^{-4}\text{ T}$
So the answer is (a) 0.71 mT.

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