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 distance \( x \), where the particle turns around, we need to consider the forces and energy involved in the system.

First, understand that a current-carrying wire generates a magnetic field around it, which is given by:

\( B = \frac{\mu_0 I}{2\pi r} \) where \( r \) is the distance from the wire, and \( \mu_0 \) is the permeability of free space.

The particle experiences a magnetic force, which affects its motion. This magnetic force \( F \) is perpendicular to the velocity of the particle and is given by:

\( F = qvB \) where \( v \) is the velocity of the charged particle, and \( q \) is the charge.

Substituting the magnetic field \( B \), we get:

\( F = qv \left( \frac{\mu_0 I}{2\pi r} \right) = \frac{q\mu_0 Iv}{2\pi r} \)

The magnetic force acting as centripetal force changes the direction but not the magnitude of velocity.

Applying conservation of energy between the initial position \( a \) and the point of turn \( x \):

Initial kinetic energy = Final kinetic energy

\(\frac{1}{2}Mv_0^2 = \frac{1}{2}Mv^2\)

Since the magnetic force doesn't do any work, it doesn't change the speed of the particle, only the direction. However, here, the turning involves an effective change due to motion constraints.

To solve for \( x \), we consider the balance of the centripetal force at the turnaround:

\[ M \frac{v_0^2}{a} = qv_0\frac{\mu_0 I}{2\pi x} \]

Rearranging gives us the value of \( x \):

\( x = a \left[ 1 - \frac{Mv_0}{2q \mu_0 I} \right] \)