Question

An electron gun is emitting electrons of various speed in a region of uniform magnetic field (parallel to OP) as shown in figure. Which electrons will hit point P?

• A. An electron with speed $\frac{eBd}{6\pi mcos\theta }$
• B. An electron with speed $\frac{eBd}{\pi mcos\theta }$
• C. An electron with energy $\frac{e^2{B}^2{d}^2}{{\pi }^{2}mco{s}^2\theta }$
• D. An electron with momentum $\frac{eBd}{4\pi cos\theta }$

Answer 1

Answer: (A), (C), (D)

The correct options are
A An electron with speed $\frac{eBd}{6\pi mcos\theta }$
C An electron with energy $\frac{e^2{B}^2{d}^2}{{\pi }^{2}mco{s}^2\theta }$
D An electron with momentum $\frac{eBd}{4\pi cos\theta }$


Path of electron will be helix.
Time period will be
$T=\frac{2\pi m}{qB}$
Condition for electron to hit P is $vcos\theta =\frac{d}{nT}$
$\Rightarrow v=\frac{dqB}{ncos\theta \times 2\pi m}=\frac{deB}{2\pi nmcos\theta }$
For $n=3\Rightarrow v=\frac{eBd}{6\pi mcos\theta }$

For n = 2, the momentum will be : $mv=\frac{eBd}{4\pi cos\theta }$
and kinetic energy = $\frac{1}{2}m{v}^2=\frac{e^2{B}^2{d}^2}{8{\pi }^{2}mco{s}^2\theta }$