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

The distance \( x \) at which the particle reverses direction is determined by analyzing the forces and energy within the system.

A current-carrying wire generates a magnetic field, quantified as \( B = \frac{\mu_0 I}{2\pi r} \), where \( r \) is the radial distance from the wire and \( \mu_0 \) is the magnetic constant.

The charged particle experiences a magnetic force, \( F = qvB \), where \( q \) is the charge, \( v \) is the particle's velocity, and \( B \) is the magnetic field. This force is perpendicular to the velocity.

Substituting the expression for \( B \) yields \( F = qv \left( \frac{\mu_0 I}{2\pi r} \right) = \frac{q\mu_0 Iv}{2\pi r} \).

The magnetic force functions as a centripetal force, altering the velocity's direction but not its magnitude.

Applying energy conservation from the initial position \( a \) to the turning point \( x \):

Initial kinetic energy equals Final kinetic energy:

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

Since the magnetic force is non-doing, it does not alter the particle's speed, only its trajectory. However, the reversal of direction here implies an effective change due to movement constraints.

The turnaround point is found by equating the centripetal force at this point:

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

Rearranging this equation provides the value for \( x \):

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