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  • A long straight wire carries a current i – A particle having a positive charge q and mass m, kept at a distance xo from the wire is projected towards it with speed v. The minimum s

A long straight wire carries a current i – A particle having a positive charge q and mass m, kept at a distance xo from the wire is projected towards it with speed v. The minimum separation between the wire and the particle.

Option: 1

\mathrm{x_0 e^{-\frac{\pi m v}{\mu_0 q i}}}


Option: 2

\mathrm{x_0 e^{-\frac{2 \pi m v}{\mu_0 q i}}}


Option: 3

\mathrm{x_0 e^{-\frac{\mu_0 q i}{\pi m v}}}


Option: 4

\mathrm{x_0 e^{-\frac{\mu_0 q i}{2 \pi m v}}}


Answers (1)

best_answer


\mathrm{\bar{F} =q \bar{V} \times \bar{B} }
\mathrm{ =q\left(v_x \hat{i}+v_y \hat{j}\right) \times\left(\frac{\mu_0 i}{2 \pi x}-\hat{k}\right) \\ }

\mathrm{= v_x \frac{\mu_0 q i}{2 \pi x} \hat{j}+q v_y \frac{\mu_0 i}{2 \pi x}(-\hat{i}) \ }
\mathrm{a_x =\frac{F x}{m}=\frac{\mu_0 q i}{2 \pi m} \frac{v_y}{x}=\lambda \frac{v_y}{x} \\ }

\mathrm{ =\frac{v_x d v_x}{d x} }

\mathrm{ \text { and } v_x d v_x=-v_y d v_y\left[v_x^2+v_y^2=v^2\right] \\ }

\mathrm{ \int_{x_0}^x \frac{d x}{x}=\int_0^v \frac{d v_y}{-\lambda} }
 

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Nehul

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