Question

A positively charged particle, having charge q, is accelerated by a potential difference V. This particle, moving along the x-axis, enters a region where a constant electric field E exists. The direction of electric field is along positive y-axis. The electric field E exists in the region bounded by the lines x = 0 and x = a. Beyond the line x = a (i.e., in region x $\ge$ a), there exists a constant magnetic field of strength B, directed along positive y-axis. What is the pitch of helix formed after particle enters the region x $\ge$ a? (mass of the particle is m).

• A. $\frac{\pi Ea}{B}\sqrt{\frac{2m}{qV}}$
• B. $\frac{\pi Ea}{2B}\sqrt{\frac{2m}{qV}}$
• C. $\frac{2\pi Ea}{B}\sqrt{\frac{2m}{qV}}$
• D. $\frac{\pi Ea}{2B}\sqrt{\frac{m}{qV}}$

Answer 1

The correct option is A $\frac{\pi Ea}{B}\sqrt{\frac{2m}{qV}}$

Component parallel to B is
$v_y=a_{y}t=\left(\frac{qE}{m}\right)\left(a\sqrt{\frac{m}{2qV}}\right)=Ea\sqrt{\frac{a}{2mV}}$
Now pitch = component parallel to $B\times$ time period
$=v_{y}T=\left(Ea\sqrt{\frac{q}{2mV}}\right)\left(\frac{2\pi }{Bq}\right)=\frac{\pi Ea}{B}\sqrt{\frac{2m}{qV}}$