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Two Particles of charges +Q and -Q are Projected from the same point with a velocity V in a region of uniform magnetic field B Such that the velocity vector makes an angle $\theta$ with the magnetic field. Their masses are M and 2 M. Then, they will meet again for the first time at a point whose distance from the point of Protection is -

Option: 1

\frac{2 \pi m v \cos \theta}{Q B}


Option: 2

\frac{8 \pi m v \cos \theta}{Q_B}


Option: 3

\frac{\pi m v \cos \theta}{Q_B}


Option: 4

\frac{4 \pi m v \cos \theta}{Q_B}


Answers (1)

best_answer

\begin{aligned} & M_1 \quad T=\frac{2 \pi m}{q 13} \\ & \Rightarrow T_1: T_2=\frac{1}{2} \end{aligned}

circle y-2 plane,
it will meet after two revolution
=\frac{4 \pi m v \cos \theta}{Q B}

Posted by

shivangi.bhatnagar

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