Consider a long thin conducting wire carrying a uniform current . A particle having mass

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Solution

Q.
Consider a long thin conducting wire carrying a uniform current $I$ . A particle having mass "M" and charge "q' is released at a distance "a" from the wire with a speed $v_{0}$ along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance x from the wire. The value of x is [ $μ_{0}$ is vacuum permeability] [2025]

1 $a [1 - \frac{m v_{0}}{2 q μ_{0} I}]$

2 $\frac{a}{2}$

3 $a [1 - \frac{m v_{α}}{q μ_{0} I}]$

4 $a e^{\frac{- 4 π m v_{0}}{q μ_{0} I}}$

Ans.
(4)

Let the velocity of the particle make an angle $θ$ with initial direction when it is at a distance x.

$\Rightarrow \frac{d x}{d t} = - v_{0} \sin θ$ ... (i)

Also, $\frac{d θ}{d t} = \frac{q}{m} (\frac{μ_{0} I}{2 π x})$ ... (ii)

$\Rightarrow \frac{d x}{d θ} = \frac{- 2 π m v_{0} x \sin θ}{q μ_{0} I}$ ... dividing eq (i) with eq (ii)

or $\int_{a}^{x} \frac{d x}{x} = \frac{- 2 π m v_{0}}{q μ_{0} I} \int_{0}^{π} \sin θ d θ$

$\ln \frac{x}{a} = \frac{- 4 π m v_{0}}{q μ_{0} I}$

Or $x = a e^{- \frac{4 π m v_{0}}{μ_{0} I q}}$

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