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Question: Answered & Verified by Expert
A charged particle in a uniform magnetic field \(B=B_0 \hat{k}\) starts moving from the origin with velocity \(v=3 \hat{i}+4 \hat{k}\) m/s. The trajectory of the particle and the time \(t\) at which it reaches \(2 \mathrm{~m}\) above \(x-y\) plane are,
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Options:
  • A Circular path, \(\frac{1}{2}\) sec.
  • B Helical path, \(\frac{1}{2}\) sec.
  • C Circular path, \(\frac{2}{3}\) sec.
  • D Helical path, \(\frac{2}{3}\) sec.
Solution:
1590 Upvotes Verified Answer
The correct answer is: Helical path, \(\frac{1}{2}\) sec.
Hint :

Velocity along z-direction will be const.
\(\mathrm{t}=\frac{\mathrm{s}}{V_z}=\frac{2}{4}=\frac{1}{2} \mathrm{sec}\)
Path will be helical.

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