Question

An electron whose e/m is $1.76\times {10}^{11}$ c/kg enter a region where there is a uniform magnetic field of induction $2\times {10}^{-3}$ tesla with a velocity of $3\times {10}^6$ m/sec in a direction making an angle of ${45}^0$ with the field. The pitch of its helical path in the region is:

• A. $1.5cm$
• B. $3.8cm$
• C. $5.36cm$
• D. $8.4cm$

Answer 1

Answer: (B)

$3.8cm$

We have,
$\frac{e}{m}=1.76\times {10}^{11}C/㎏$

$B=2\times {10}^{-3}T$

$v=3\times {10}^{6}m/s$

$\theta =45°$

Pitch $=\left(v\cos \theta \right)T$

$=v\cos \theta \times \frac{2\pi m}{Bq}$

$=v\cos \theta \times 2\pi \times \frac{1}{B}\times \frac{m}{e}$

$=3\times {10}^6\times \cos 45\times 2\times 3.14\times \frac{1}{2\times {10}^{-3}}\times \frac{1}{1.76\times {10}^{11}}$

$=3.78\times {10}^6\times {10}^3\times {10}^{-11}$

$=3.78\times {10}^{-2}m$

$=3.78cm$

$=3.8cm$