A proton with a kinetic energy… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Electromagnetism

2023integerhard

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

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Held on 6 Apr 2023 · Verified 6 Jul 2026.

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Solution

The given data is

$K.E=2\mathrm{eV}$

$B=\frac{\pi }{2}\times {10}^{-3}T$

$\theta ={60}^{o}$

The pitch of the proton is given by

$P=\frac{2\pi m}{qB}\times v\mathrm{cos}\theta ...(i)$

By using

$K=\frac{m{v}^{2}}{2} \Rightarrow v=\sqrt{\frac{2K}{m}}$

Substituting the values in equation (i)

$=\frac{2\pi \times \sqrt{2mKE}\times \frac{1}{2}\times 2}{1.6\times {10}^{-19}\times \pi \times {10}^{-3}}$

$=\frac{2\times \sqrt{2\times 1.6\times {10}^{-27}\times 2\times 1.6\times {10}^{-19}}\times {10}^{3}}{1.6\times {10}^{-19}}$

$=2\times 2\times {10}^{–1}=0.4m$

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Solution

The given data is

$K.E=2\mathrm{eV}$

$B=\frac{\pi }{2}\times {10}^{-3}T$

$\theta ={60}^{o}$

The pitch of the proton is given by

$P=\frac{2\pi m}{qB}\times v\mathrm{cos}\theta ...(i)$

By using

$K=\frac{m{v}^{2}}{2} \Rightarrow v=\sqrt{\frac{2K}{m}}$

Substituting the values in equation (i)

$=\frac{2\pi \times \sqrt{2mKE}\times \frac{1}{2}\times 2}{1.6\times {10}^{-19}\times \pi \times {10}^{-3}}$

$=\frac{2\times \sqrt{2\times 1.6\times {10}^{-27}\times 2\times 1.6\times {10}^{-19}}\times {10}^{3}}{1.6\times {10}^{-19}}$

$=2\times 2\times {10}^{–1}=0.4m$

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Solution

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Solution

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JEE Main Electromagnetism in other years

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Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

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