Question 7 9) A wire of length L, mass 'm' and carrying a current i is suspended from point as shown. An another infinitely long wire carrying the same current i is at a distance L below the lower end of the wire. Given i 2A, L = 1 m and m = 0.1 kg (In 2 0.693) What is angular acceleration of the wire just after it is released from the position shown?!

To determine the angular acceleration of the suspended wire just after it is released, we need to analyze the forces acting on it and apply the principles of torque and angular motion. The setup involves two wires carrying the same current, which creates a magnetic interaction between them. Let's break down the problem step by step.

Understanding the Forces Involved

When the wire is suspended and carrying a current, it experiences a magnetic force due to the magnetic field generated by the infinitely long wire below it. The direction of this force can be determined using the right-hand rule, which states that if you point your thumb in the direction of the current, your fingers curl in the direction of the magnetic field lines.

Magnetic Field Calculation

The magnetic field (B) created by an infinitely long straight wire carrying a current (i) at a distance (r) is given by the formula:

• B = (μ₀ * i) / (2π * r)

Here, μ₀ is the permeability of free space, approximately equal to 4π x 10^-7 T·m/A. In our case, the distance (r) between the two wires is L, which is 1 m.

Force on the Suspended Wire

The force (F) acting on the suspended wire due to the magnetic field can be calculated using the formula:

• F = i * L * B

Substituting the expression for B into this equation gives:

• F = i * L * (μ₀ * i) / (2π * L) = (μ₀ * i²) / (2π)

Given that i = 2 A, we can substitute this value into the equation:

• F = (4π x 10^-7 T·m/A * (2 A)²) / (2π) = (4π x 10^-7 * 4) / (2π) = 8 x 10^-7 N

Calculating Torque and Angular Acceleration

The torque (τ) acting on the wire about the pivot point can be calculated using the formula:

• τ = r * F

In this case, the distance (r) from the pivot to the center of mass of the wire is L/2 (since the wire is uniform), so:

• τ = (L/2) * F = (1 m / 2) * (8 x 10^-7 N) = 4 x 10^-7 N·m

Moment of Inertia

The moment of inertia (I) of the wire about the pivot point is given by:

• I = (1/3) * m * L²

Substituting the values of m = 0.1 kg and L = 1 m:

• I = (1/3) * 0.1 kg * (1 m)² = 0.0333 kg·m²

Finding Angular Acceleration

Using Newton's second law for rotation, we can relate torque to angular acceleration (α) with the equation:

• τ = I * α

Rearranging this gives:

• α = τ / I

Substituting the values we calculated:

• α = (4 x 10^-7 N·m) / (0.0333 kg·m²) ≈ 1.2 x 10^-5 rad/s²

Thus, the angular acceleration of the wire just after it is released is approximately 1.2 x 10^-5 rad/s². This small value indicates that while the wire will start to rotate, the effect of the magnetic force is relatively weak compared to its moment of inertia.