Question

A charged particle $P$ leaves the origin with speed $v=v_0$ at some inclination with the $x-$axis. There is a uniform magnetic field B along the $x-$axis. $P$ strikes a fixed target $T$ on the $x-$axis for a minimum value of $B=B_0$. If the pitch of the helical path is equal to the maximum distance of the particle from the x-axis, then which of the following are not correct?

• A. $cos\theta =\frac{1}{\pi }$
• B. $sin\theta =\frac{1}{\pi }$
• C. $tan\theta =\frac{1}{\pi }$
• D. $tan\theta =\pi$

Answer 1

Answer: (A), (B), (C)

Time period of one revolution $=\frac{2\pi m}{qB}$

$pitch=V_h\times T=\frac{2\pi m}{qB}v\cos \theta$

Radius of the circle $r=\frac{mv\sin \theta }{qB}$

Max distance of the particle from the x-axis is 2r.

Since pitch is equal to 2r, $\frac{2\pi m}{qB}v\cos \theta =2\frac{mv\sin \theta }{qB}$

$\Rightarrow \tan \theta =\pi$.