Question

A small bar magnet experiences a 2.00×10−2 N⋅m torque when the axis of the magnet is at 45∘ to a 0.140 T magnetic field.

i understand that
torque = u0XB=u0Bsintheta where theta is the angle between the objects area normal vector and the magnetic field
so given theta the torque and u0 we have
u0= torque / BSINTHETA

Answer 1

To find the magnetic moment given the torque, magnetic field, and angle, use the formula μ₀ = T / (Bsinθ). Substitute the known values and evaluate.

Step-by-step explanation:

The torque experienced by a small bar magnet can be calculated using the equation T = μ₀Bsinθ, where T is the torque, μ₀ is the magnetic moment, B is the magnetic field, and θ is the angle between the magnet's axis and the magnetic field. In this case, the torque is given as 2.00×10⁻² N⋅m, the magnetic field is 0.140 T, and the angle is 45°. To find the magnetic moment, we can rearrange the equation to μ₀ = T / (Bsinθ). Substituting the values, we have: μ₀ = (2.00×10⁻² N⋅m) / (0.140 T * sin(45°)). Evaluating this expression gives us the value of magnetic moment.

Answer 2

The question is about calculating the magnetic moment of a bar magnet using the torque experienced in a known magnetic field and at a specified angle. The formula for torque on a magnetic dipole is used, T = μBsinθ, to find the magnetic moment μ.

Step-by-step explanation:

The subject in question involves calculating the magnetic moment of a bar magnet given the torque it experiences in a magnetic field. The formula for torque on a magnetic dipole is T = μBsinθ where T is torque, μ is the magnetic moment, B is the magnetic field strength, and θ is the angle between the magnetic moment and magnetic field. Given T, B, and θ, the magnetic moment (μ) can be calculated by rearranging the equation to μ = T / (B sinθ). Utilizing this equation, we can calculate the magnetic moment when the torque is 2.00×10⁻² N·m, the magnetic field is 0.140 T, and the angle is 45°.