Consider a long thin conducting wire carrying a uniform current I. A… | JEE Main 2025 (28 Jan, Shift 1)

Q23JEE Main 2025 · 28 JAN · SHIFT 1Difficult

Moving Charge in a Magnetic Field

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 [ $\mu_0$ is vacuum permeability]

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Answer: (b)

Take the wire along $x$ (current $+x$ ), particle at distance $y$ : $B = \dfrac{\mu_0 I}{2\pi y}$ , and the force component along the wire is $m\dfrac{dv_x}{dt} = qv_yB$ .
With $v_y = \dfrac{dy}{dt}$ : $m\,dv_x = \dfrac{q\mu_0 I}{2\pi}\dfrac{dy}{y}$ , so $m\Delta v_x = \dfrac{q\mu_0 I}{2\pi}\ln\dfrac{y}{a}$ .
Speed stays $v_0$ (magnetic force does no work). It turns round when $v_y = 0$ with $v_x = -v_0$ , so $\Delta v_x = -2v_0$ .
$-2mv_0 = \dfrac{q\mu_0 I}{2\pi}\ln\dfrac{x}{a} \Rightarrow x = a\,e^{-\frac{4\pi m v_0}{q\mu_0 I}}$
So the answer is (b).

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