A particle of mass $1 \times 10^{-27} \mathrm{~kg}$ and charge $1 \times 10^{-16} \mathrm{C}$ enters the…

1. $5 \times 10^5$
2. $1 \times 10^6$
3. $\pi \times 10^5$
4. $3 \times 10^6$

Solution

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

Asked in: AP EAMCET 2022 (06 Jul Shift 2)

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