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:
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Remember that the force on a charged particle in a magnetic field is proportional to the velocity and the charge, so the centripetal force plays a role in determining the turning point.
To determine the distance \( x \) at which the particle reverses direction, we analyze the magnetic force exerted on the particle by the current in the wire.
Step 1: The magnetic force is calculated using the formula:
\[ F_{\text{mag}} = \frac{\mu_0 I q}{2 \pi x} \]
Here, \( \mu_0 \) represents the permeability of free space, \( I \) is the current, \( q \) is the particle's charge, and \( x \) denotes the distance from the wire.
Step 2: This force provides the centripetal acceleration. Therefore, we apply the centripetal force formula:
\[ F_{\text{cent}} = \frac{M v_0^2}{x} \]
Step 3: Equate the magnetic force and the centripetal force:
\[ \frac{\mu_0 I q}{2 \pi x} = \frac{M v_0^2}{x} \]
Step 4: Simplify the equation and solve for \( x \):
\[ x = \frac{2 \pi M v_0^2}{\mu_0 I q} \]
Final Conclusion: The particle turns round at a distance \( x \) given by \( a \left[ 1 - \frac{mv_0}{2q\mu_0 I} \right] \), which matches Option (4).
