Question

An electron is shot into open end of a solenoid. As it enters the uniform magnetic field within the solenoid its speed is $800m/s$ and its velocity vector makes an angle of ${30}^o$ with the central axis of the solenoid. The solenoid caries $4.0A$ current and has $8000$ turns along its length. Find number of revolutions made by the electron within the solenoid by the time it emerges from the solenoid's opposite end.
(use charge to mass ration $\frac{e}{m}$ for electron $=\sqrt{3}\times {10}^{11}C/kg$)
Fill your answer of $y$ if total number of revolution is $\frac{y}{5}\times {10}^6$

Answer 1

Let length of solenoid $=l$

$\therefore$ magnetic field inside solenoid

$={\mu }_{0}nI$

$n=\frac{8000}{L}$ and $I=4A$

$\therefore B={\mu }_0\ast \frac{8000}{L}\ast 4=\frac{32000{\mu }_0}{L}$

as $\vartheta sin{30}^{\circ }$ component will be ${\perp }^r$ to the magnetic field

Hence electron will follow a helical path.

time taken to complete one

revolution$=T=\frac{2\pi m}{qB}=\frac{2\pi }{B}\left(\frac{m}{e}\right)$

$\therefore T=\frac{2\pi }{B}\ast \frac{1}{\sqrt{3}\ast {10}^{11}}$

now time taken by electron to emerge from solenoid $=\frac{L}{\vartheta cos\theta }$

$=\frac{L}{\vartheta cos{30}^{\circ }}=\frac{2L}{\sqrt{3}\vartheta }$

$\therefore$ no. of revolutions

$=\frac{timetakenincompletejourney}{timetakeninonerevolution}$

$=\frac{\frac{2L}{\sqrt{3}\vartheta }}{\frac{2\pi }{B}\ast \frac{1}{\sqrt{3}\ast {10}^{11}}}$

$=\frac{2L\ast B\ast \sqrt{3}\ast {10}^{11}}{\sqrt{3}\vartheta \ast 2\pi }$

$=\frac{2L\ast \frac{32000{\mu }_0}{L}\ast \sqrt{3}\ast {10}^{11}}{\sqrt{3}\ast 800\ast 2\pi }$

$=\frac{2\ast 32000\ast 4\pi \ast {10}^{-9}\ast {10}^{11}}{800\ast 2\pi }=160\ast {10}^4$

comparing it with $\frac{y}{5}\ast {10}^6$

$y=8$