Question

A charge particle of specific charge q/m is projected perpendicular to magnetic as well as electric field ‘B' & 'E’.
['E' is parallel to 'B'.]

• A. Path of the charged particle may be straight line.
• B. Path of the charged particle is helix with increasing pitch.
• C. Frequency of the charged particle increases.
• D. Speed increases to thrice of its original speed in time $\frac{2\sqrt{2}{V}_{0}m}{qE}.$
  ($V_0$ is initial speed)

Answer 1

Answer: (B), (D)

Speed increases to thrice of its original speed in time $\frac{2\sqrt{2}{V}_{0}m}{qE}.$

($V_0$ is initial speed)

Electric force on q is along E which will increase the speed of the charged particle.
The magnetic force is $q{V}_{0}B$ is perpendicular to E, which will rotate q in circular path.
So, the resultant path is helix with increasing pitch.

Now, if the speed is increased to thrice its initial speed,
$(3V_0{)}^2=V_0^2+V_x^2\Rightarrow V_x=2\sqrt{2}{V}_0$
By $V=u+at\Rightarrow 2\sqrt{2}{V}_0=0+\frac{qE}{m}t$
$\therefore t=\frac{2\sqrt{2}{V}_{0}m}{qE}$
Frequency: $f=\frac{1}{T}=\frac{V_0}{2\pi R}=\frac{V_{0}qB}{2\pi m{V}_0}=\frac{qB}{2\pi m}$
( which is constant)