AP EAPCET 06-Jul-2022 Shift 2 Paper

The given situation is shown below.Given, $v=1000 \text{ m/s}$$\;q=10^{-16} \text{ C}$$\;m=1 \times 10^{-27} \text{ kg}$ $N=$ Number of turns $=5000$$I=5 \text{ A}$$R=$ Radius of circular path of chargeMagnetic field, $B=\mu_0 n I$ (where, $n=$ number of turns per unit length) $\Rightarrow B=\mu_0 \; \frac{N}{L} I$(where, $n=\;\frac{N}{L}$ ) $\; \Rightarrow B=\mu_0 \times \; \frac{5000}{L} \times 5$$\; \Rightarrow B=\mu_0 \times \; \frac{25000}{L}$........(i) Also we know that, Distance $=$ speed $\times$ time $L=v_x \times t$ (where, $v_x$ is the component of velocity along the axis of the solenoid) $\Rightarrow \;\; L=500 t$......(ii) Also, radius of circular path of a charged particle in uniform magnetic field is given by the formula$R=\; \frac{m v_y}{9 B}$$(.$ where, $v_y=500 \sqrt{3})$ Putting, $B=\; \frac{\mu_0 \times 25000}{L}$[from Eq .(i)] $\Rightarrow \;\; R=\; \frac{10^{-11} \times \sqrt{3} L}{50 \mu_0}$......(ii) Now, time taken by the charged particle to exit the solenoid moving along the circular path making $N^{'}$ number of revolutions.$\;t^{'}=\frac{\; \text{circular distance} \;}{v_y} \Rightarrow t^{'}=\frac{2 \pi R \times \; \text{no. of revolutions} \;}{v_y}$$\;t^{'}=\frac{2 \pi R \times N^{'}}{v_y}$ $t^{'}=\frac{2 \pi \times 10^{-11} \times \sqrt{3} L \times N^{'}}{50 \mu_0 \times 500 \sqrt{3}}$.....(iii) Now, since time taken to cover linear distance and circular distance before exiting the solenoid are same$t=t^{'}$ So, from Eqs. (i) and (ii) $\;\frac{L}{500}=\frac{2 \pi \times 10^{-11} \times \sqrt{3} \times L \times N^{'}}{50 \times 4 \pi \times 10^{-7} \times 500 \sqrt{3}} \;\; (\because \mu_0=4 \pi \times 10^{-7})$ On solving, we get$N^{'} = \frac{100 \times 10^{-7}}{10^{-11}} = \frac{10^{-5}}{10^{-11}} = 10^6$ 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.