Detailed Explanation

This problem involves calculating the magnetic moment (m) of a bar magnet based on the torque (τ) it experiences in a magnetic field (B). The formula for torque in a magnetic field is given by τ = mB sin(θ), where θ is the angle between the magnetic moment and the magnetic field. Here, we are given τ = 4.5 × 10^−2 J, B = 0.25 T, and θ = 30°. To find m, we rearrange the equation and solve for m:

1. Calculate sin(30°), which is 0.5.
2. Substitute the values into the torque equation:
   τ = mB sin(θ) => 4.5 × 10^−2 = m * 0.25 * 0.5.
3. Solve for m:
   m = (4.5 × 10^−2) / (0.25 * 0.5) = (4.5 × 10^−2) / 0.125 = 0.36 J/T.

So, the magnitude of the magnetic moment of the magnet is 0.36 J/T.

Detailed Explanation

This exercise involves understanding stable and unstable equilibrium positions for a bar magnet in a magnetic field. The stable equilibrium occurs when the magnetic moment aligns with the magnetic field, while the unstable equilibrium occurs when they are opposite.

1. For stable equilibrium (aligned), the angle θ = 0°:
   The potential energy U = -m.B, which leads to U = -0.32 * 0.15 = -0.048 J.
2. For unstable equilibrium (opposite), the angle θ = 180°:
   The potential energy U = -m.B, but since it’s opposite, U = +0.32 * 0.15 = +0.048 J.

In summary, -0.048 J is stable, and +0.048 J is unstable.

Detailed Explanation

A solenoid behaves like a bar magnet by generating a magnetic field similar to that of a dipole. The magnetic moment (m) of a solenoid can be calculated using the formula:
m = nIA, where n is the number of turns per unit length, I is the current, and A is the cross-sectional area.

1. First, calculate n, the turns per unit length. If the solenoid has 800 turns and its length is not specified, we assume a unit length for simplicity when calculating m.
2. Calculate m:

m = (800 turns) × (3 A) × (2.5 × 10^−4) m² = 0.6 Am².

Thus, the solenoid effectively acts like a bar magnet with a magnetic moment of 0.6 Am².

Detailed Explanation

To find the torque (τ) acting on the solenoid when it is at an angle to a magnetic field, we can use the equation τ = mB sin(θ). We already calculated m in the previous exercise as 0.6 Am².

1. Here, B = 0.25 T,
2. θ = 30°, so sin(30°) = 0.5,
3. Substitute into the torque equation:
   τ = mB sin(θ) = 0.6 × 0.25 × 0.5 = 0.075 Nm.

Therefore, the magnitude of torque on the solenoid is 0.075 Nm.

Detailed Explanation

To find the work done against the torque when aligning the magnet, we will use the potential energy change in a magnetic field. The work done (W) is equal to the change in potential energy, which is:

W = U_initial - U_final.

For part (a):
1. For the normal (90°) position (U = 0), and opposite direction (180°), U = -mB.
2. Work for normal position:
W = 0 - (-1.5 * 0.22) = 0.33 J.

1. Work for opposite position:
   W = 0 - (1.5 * 0.22) = -0.33 J (the negative indicates it is work done by the field).

For part (b), torque is given by τ = mB sin(θ):
1. Case (i) where θ = 90°, τ = 1.5 * 0.22 * 1 = 0.33 Nm, and case (ii) where θ = 180°, τ = 1.5 * 0.22 * 0 = 0 Nm.

Detailed Explanation

To calculate the magnetic moment (m) of the solenoid, we use the formula:
m = nIA,
where n is the number of turns per unit length (assuming unit length for simplicity), I is the current, and A is the cross-sectional area.

1. Calculate m:
   n = 2000 turns,
   I = 4 A,
   A = 1.6 × 10^−4 m²,
   m = 2000 * 4 * 1.6 × 10^−4 = 1.28 Am².

For the force (F) and torque (τ):
1. The force equation for a solenoid is not straightforward in a uniform field; however, we model it through torque:
τ = mB sin(θ).
2. B = 7.5 × 10^−2 T, θ = 30°, sin(30°) = 0.5.
3. Calculate τ:
τ = 1.28 * 7.5 × 10^−2 * 0.5 = 0.048 Nm.

Thus, the magnetic moment is 1.28 Am², and the torque is 0.048 Nm.

Detailed Explanation

To find the magnetic field (B) at a distance from the bar magnet, we use the formulas:

On the axis (B_axis): B = (μ₀ / (2 * π)) * (m / r³) (for axial point)
At distance r = 10 cm = 0.1 m:
B_axis = (4π × 10^-7) / (2π) * (0.48 / (0.1)³) = 0.0096 T.

On the equatorial line (B_eq): B = -(μ₀ / (4 * π)) * (m / r³) (for equatorial point)
B_eq = -(4π × 10^-7 / 4π) * (0.48 / (0.1)³) = -0.0012 T.

Thus, the directions of the fields would be towards the magnet's center on the axis and away on the equatorial lines.