A proton with a kinetic energy of 2.0 eV moves into a region of uniform magnetic field of magnitude π/2 × 10⁻³ T. The angle between the direction of magnetic field and velocity of proton is 60°. The pitch of the helical path taken by the proton is __________ cm. (Take, mass of proton = 1.6 × 10⁻²⁷ kg and Charge on proton = 1.6 × 10⁻¹⁹ C). Numerical answer type. Work it out, then open the answer.

Answer: 40. v = √(2K/m) = √2 × 2 × 1.6 × 10⁻¹⁹1.6 × 10⁻²⁷ = 2 × 10⁴ m/s, so vcos 60° = 10⁴ m/s. T = (2π m)/(qB) = 2π × 10⁻⁸(π/2) × 10⁻³ = 4 × 10⁻⁵ s Pitch = vcos 60° · T = 10⁴ × 4 × 10⁻⁵ = 0.4 m So the pitch is 40 cm.

A proton with a kinetic energy of 2.0 eV moves into a region of… | JEE Main 2023 (6 Apr, Shift 2)

Q48JEE Main 2023 · 6 APR · SHIFT 2Medium

Moving Charge in a Magnetic Field

A proton with a kinetic energy of $2.0\text{ eV}$ moves into a region of uniform magnetic field of magnitude $\frac{\pi}{2} \times 10^{-3}\text{ T}$ . The angle between the direction of magnetic field and velocity of proton is $60^\circ$ . The pitch of the helical path taken by the proton is __________ $\text{cm}$ . (Take, mass of proton $= 1.6 \times 10^{-27}\text{ kg}$ and Charge on proton $= 1.6 \times 10^{-19}\text{ C}$ ).

Numerical answer type. Work it out, then open the answer.

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Answer: 40

$v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2 \times 2 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27}}} = 2 \times 10^{4}$ m/s, so $v\cos 60^\circ = 10^{4}$ m/s.
$T = \dfrac{2\pi m}{qB} = \dfrac{2\pi \times 10^{-8}}{(\pi/2) \times 10^{-3}} = 4 \times 10^{-5}$ s
Pitch $= v\cos 60^\circ \cdot T = 10^{4} \times 4 \times 10^{-5} = 0.4$ m
So the pitch is 40 cm.

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