question_answer

A charged particle of \[q=6\,\mu C\] is projected in a uniform magnetic field \[\vec{B}=\left( 10\sqrt{3}\,\hat{i}+10\hat{j} \right)Wb/{{m}^{2}}\] with velocity with velocity\[\vec{V}=(2\hat{i}+3\sqrt{3}\hat{j})m/s\]. The pitch of helix becomes [Assume mass of particle, \[m=10\sqrt{3}\times {{10}^{-6}}kg\]]

A)  6.28 m                 

B)  3.14m

C)  0.95m                  

D)  5.25m

Correct Answer: B

Solution :

\[\vec{V}=(2\hat{i}+2\sqrt{3}\hat{j})\] \[\vec{B}=(10\sqrt{3}\hat{i}+10\hat{j})\] The time period of resolution \[T=\frac{2\pi m}{qB}\] The pitch of helix can be given by \[P=(4\cos {{30}^{o}})\times \frac{2\pi m}{qB}\] \[=4\times \frac{\sqrt{3}}{2}\times \frac{2\pi \times 10\sqrt{3}\times {{10}^{-6}}}{6\times {{10}^{-6}}\times 20}=\pi \]