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A beam of protons enters a uniform magnetic field of $0.314 \mathrm{~T}$ with a velocity $4 \times 10^5 \mathrm{~ms}^{-1}$ in a direction making an angle $60^{\circ}$ with the direction of the magnetic field.The path of the beam is (mass of proton $\left.=1.6 \times 10^{-27} \mathrm{~kg}\right)$
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1617 Upvotes
Verified Answer
The correct answer is:
a helix with a pitch 4 cm
As, $\mathbf{v}$ is not perpendicular to $\mathbf{B}$, so path of particle is helix.
$$
\begin{aligned}
& \text { Pitch of helical path }=v \cos \theta \times \frac{2 \pi m}{B q} \\
& \qquad \begin{aligned}
& 4 \times 10^5 \times \frac{1}{2} \times \frac{2 \times \pi \times 1.6 \times 10^{-27}}{0.314 \times 1.6 \times 10^{-19}} \\
& =4 \mathrm{~cm}
\end{aligned}
\end{aligned}
$$
$$
\begin{aligned}
& \text { Pitch of helical path }=v \cos \theta \times \frac{2 \pi m}{B q} \\
& \qquad \begin{aligned}
& 4 \times 10^5 \times \frac{1}{2} \times \frac{2 \times \pi \times 1.6 \times 10^{-27}}{0.314 \times 1.6 \times 10^{-19}} \\
& =4 \mathrm{~cm}
\end{aligned}
\end{aligned}
$$
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