Question

An electron accelerated by a potential difference $V=1.0kV$ moves in a uniform magnetic field at an angle $\alpha ={30}^{\circ }$ to the vector $B$ whose modulus is $B=29mT$. Find the pitch of the helical trajectory of the electron.

• A. 1 m
• B. 1 cm
• C. 2 m
• D. 2 cm

Answer 1

The correct option is D 2 cm

$T=eV=\frac{1}{2}m{v}^2$

$\upsilon =\sqrt{\frac{2eV}{m}}$

$⟹{\upsilon }_H=\sqrt{\frac{2eV}{m}}\cos \alpha$ and ${\upsilon }_v=\sqrt{\frac{2eV}{m}}\sin \alpha$

Now, $\frac{m{\upsilon }_v^2}{r}=Be{\upsilon }_v$ or $r=\frac{m\upsilon v}{Be}$

and $T=\frac{2\pi r}{\upsilon v}=\frac{2\pi m}{Be}$

Pitch $P={\upsilon }_{H}T=\frac{2\pi m}{Be}\sqrt{\frac{2eV}{m}}\cos \alpha =0.02m$