Proton with kinetic energy of 1 MeV moves from south the north. It gets an acceleration of $10^{12} \mathrm{~m} / \mathrm{s}^{2}$ by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is $1.6 \times 10^{-27} \mathrm{~kg}$)

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Magnetic Effects of Current

Proton with kinetic energy of 1 MeV moves from south the north. It gets an acceleration of

1 0^{12} m / s^{2}

by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is

1.6 \times 1 0^{- 27} kg

)

Answer: (A)

From the kinetic energy

K E = 1 MeV = 1.6 \times 1 0^{- 13}

= \frac{1}{2} m v^{2}

, the proton's speed is

v = \frac{2 \times 1.6 \times 1 0 ^{- 13}}{1.6 \times 1 0 ^{- 27}} = 2 \times 1 0^{7}

m/s. The magnetic force supplies the acceleration,

q v B = ma

, so

B = \frac{ma}{q v} = \frac{1.6 \times 1 0 ^{- 27} \times 1 0 ^{12}}{1.6 \times 1 0 ^{- 19} \times 2 \times 1 0 ^{7}} = \frac{1 0 ^{- 3}}{2} \approx 0.71 \times 1 0^{- 3}

T, about

0.71

mT.

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option