A particle of mass $1 \times 10^{-27} \mathrm{~kg}$ and charge $1 \times 10^{-16} \mathrm{C}$ enters the uniform magnetic field within the solenoid, at speed $1000 \mathrm{~ms}^{-1}$. The velocity vector makcs an angle $60^{\circ}$ with the axis of solenoid. The solenoid has 5000 turns along its length and carries current $5 \mathrm{~A}$. The number of revolution the particle makes along the helical path within the solenoid by the time it emerges from solenoid's opposite end is

The given situation is shown below.
Given, $\begin{aligned} v & =1000 \mathrm{~m} / \mathrm{s} \\ q & =10^{-16} \mathrm{C} \\ m & =1 \times 10^{-27} \mathrm{~kg}\end{aligned}$ $$ \begin{aligned} N & =\text { Number of turns }=5000 \\ I & =5 \mathrm{~A} \\ R & =\text { Radius of circular path of charge } \end{aligned} $$ $R=$ Radius of circular path of charge Magnetic field, $B=\mu_0 n l$ (where, $n=$ number of turns per unit length) $$ \begin{aligned} & \Rightarrow B=\mu_0 \frac{N}{L} I (where, n=\frac{N}{L} )\\ & \Rightarrow B=\mu_0 \times \frac{5000}{L} \times 5 \\ & \Rightarrow B=\mu_0 \times \frac{25000}{L}...(i) \end{aligned} $$ Also we know that, Distance $=$ speed $\times$ time $L=v_x \times t$ (where, $v_x$ is the component of velocity along the axis of the solenoid)
$$ \Rightarrow $$ $$ L=500 t...(ii) $$ Also, radius of circular path of a charged particle in uniform magnetic field is given by the formula $$ \begin{aligned} R & =\frac{m v_j}{9 B} \quad \text { (where, } v_y=500 \sqrt{3} \text { ) } \\ \text { Putting, } B & =\frac{\mu_0 \times 25000}{L} \quad \text { [from Eq .(i)] } \\ R & =\frac{10^{-11} \times 500 \sqrt{3} \times L}{10^{-16} \times \mu_0 \times 25000} \\ \Rightarrow \quad R & =\frac{10^{-11} \times \sqrt{3} L}{50 \mu_0} \quad \text {...(ii) } \end{aligned} $$ Now, time taken by the charged particle to exit the solenoid moving along the circular path making $N^{\prime}$ number of revolutions. $$ \begin{aligned} t^{\prime} & =\frac{\text { circular distance }}{v_y} \Rightarrow t^{\prime}=\frac{2 \pi R \times \text { no. of revolutions }}{v_y} \\ t^{\prime} & =\frac{2 \pi R \times N^{\prime}}{v_y} \\ t^{\prime} & =\frac{2 \pi \times 10^{-11} \times \sqrt{3} L \times N^{\prime}}{50 \mu_0 \times 500 \sqrt{3}} ...(iii)\end{aligned} $$ Now, since time taken to cover linear distance and circular distance before exiting the solenoid are same $$ t=t^{\prime} $$ So, from Eqs. (i) and (ii) $$ \frac{L}{500}=\frac{2 \pi \times 10^{-11} \times \sqrt{3} \times L \times N^{\prime}}{50 \times 4 \pi \times 10^{-7} \times 500 \sqrt{3}} \quad\left(\because \mu_0=4 \pi \times 10^{-7}\right) $$ On solving, we get $$ N^{\prime}=\frac{100 \times 10^{-7}}{10^{-11}}=\frac{10^{-5}}{10^{-11}}=10^6 $$ Hence, total number of revolution along helical path by the time it emerges the solenoid $=1 \times 10^6$. Hence, option (b) is the correct answer.