Question

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:

• A. \( \frac{a}{2} \)
• B. \( a \left( 1 - \frac{mv_0}{q\mu_0 I} \right) \)
• C. \( ae \left( -4 \frac{mv_0}{q\mu_0 I} \right) \)
• D. \( a \left[ 1 - \frac{mv_0}{2q\mu_0 I} \right] \)

Hint:

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.

Answer 1

The Correct Option is D

To find the value of \( x \) at which the particle turns round, we consider the magnetic force acting on the particle due to the current in the wire. 

Step 1: The magnetic force is given by the formula: \[ F_{\text{mag}} = \frac{\mu_0 I q}{2 \pi x} \] where \( \mu_0 \) is the permeability of free space, \( I \) is the current, \( q \) is the charge of the particle, and \( x \) is the distance from the wire. 

Step 2: The force provides the centripetal acceleration, so we use the centripetal force formula: \[ F_{\text{cent}} = \frac{M v_0^2}{x} \] 

Step 3: Set the magnetic force equal to the centripetal force: \[ \frac{\mu_0 I q}{2 \pi x} = \frac{M v_0^2}{x} \] 

Step 4: Simplify and solve for \( x \): \[ x = \frac{2 \pi M v_0^2}{\mu_0 I q} \] 

Final Conclusion: The value of \( x \) at which the particle turns round is given by \( a \left[ 1 - \frac{mv_0}{2q\mu_0 I} \right] \), which corresponds to Option (4).