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A Proton of mass $1.67 \times 10^{-27} \mathrm{~kg}$ and charge $1.6 \times 10^{-19} \mathrm{c}$ is projected with a speed of $2 \times 10^6 \mathrm{~m} / \mathrm{s}$ at an angle of $60^{\circ}$ to the x-axis. if a uniform magnetic field of $0.104 T$ is applied along the y-axis, the path of the Proton is
Option: 1
A circle of radius $0.2 \mathrm{~m}$ and time period $\pi \times 10^{-7} \mathrm{sec}$ .

Option: 2
A circle of radius $0.1 \mathrm{~m}$ and time period 2 \pi \times $10^{-7} \mathrm{sec}$ .

Option: 3
A helix of radius $0.1 \mathrm{~m}$ and time period $2 \pi \times 10^{-7} \mathrm{sec}$ .

Option: 4
A helix of radius $0.2 \mathrm{~m}$ and time Period $4 \pi \times 10^{-7} \mathrm{sec}$ .

Answers (1)

best_answer

$\because \vec{V}$ is not Parallel to $\vec{B}$

$\therefore$ Path of the Proton is helical.

$\text { radius } =\frac{m v_{\perp}}{q B} \\$

$=0.1 \mathrm{~m} \\$

$T =\frac{2 \pi m}{q B}=2 \pi \times 10^{-7} \mathrm{sec}$

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HARSH KANKARIA

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