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 \( x \):

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

Answer 2

Step 1: Understand the setup.
We have a long straight conducting wire carrying a steady current \( I \). A charged particle of mass \( m \) and charge \( q \) is released at a distance \( a \) from the wire with an initial velocity \( v_0 \) along the direction of the current. The magnetic field due to the wire will cause a magnetic force on the particle, pulling it toward the wire. The particle turns around when it reaches a minimum distance \( x \) from the wire.

Step 2: Expression for magnetic field due to a current-carrying wire.
The magnetic field at a distance \( r \) from a long straight wire carrying current \( I \) is given by: \[ B = \frac{\mu_0 I}{2\pi r}. \] This magnetic field is directed tangentially (using the right-hand rule).

Step 3: Magnetic force acting on the particle.
The magnetic force on a charged particle moving with velocity \( \vec{v} \) in a magnetic field \( \vec{B} \) is: \[ \vec{F} = q(\vec{v} \times \vec{B}). \] Since the velocity is parallel to the wire and \( \vec{B} \) is circular around it, the force is directed radially inward, attracting the particle toward the wire.

Step 4: Apply conservation of angular momentum.
There is no torque about the wire axis because the magnetic force is always directed toward the wire. Hence, the angular momentum of the particle about the wire is conserved: \[ mvr = \text{constant}. \] So, initially: \[ mv_0 a = mvx. \] Therefore, \[ v = \frac{v_0 a}{x}. \]

Step 5: Apply conservation of energy.
Magnetic force does no work, but the particle’s motion in the magnetic field produces a change in the effective kinetic energy due to the circular motion constraint. The magnetic force provides the necessary centripetal acceleration, so: \[ \frac{mv^2}{r} = qvB. \] At the turning point, the radial velocity component becomes zero. Using energy conservation between initial position \( a \) and turning point \( x \): \[ \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + \text{work done by magnetic force}. \] But since the magnetic field depends on \( r \), we integrate the magnetic effect: \[ \int_a^x qvB \, dr = 0. \] Simplifying using \( B = \frac{\mu_0 I}{2\pi r} \) and the velocity relation: \[ v = \frac{v_0 a}{r}. \] Substitute into energy relation: \[ \frac{1}{2}mv_0^2 = \frac{1}{2}m\left(\frac{v_0^2 a^2}{x^2}\right) + \int_x^a \frac{qv_0 a \mu_0 I}{2\pi r^2} dr. \] Evaluating the integral gives: \[ \frac{1}{2}mv_0^2 - \frac{1}{2}m\frac{v_0^2 a^2}{x^2} = \frac{qv_0 a \mu_0 I}{2\pi} \left(\frac{1}{x} - \frac{1}{a}\right). \] Simplifying and solving for \( x \): \[ x = a \left[1 - \frac{mv_0}{2q\mu_0 I}\right]. \]

Final Answer:
\[ \boxed{x = a \left[1 - \frac{mv_0}{2q\mu_0 I}\right]}. \]