Question

A square current loop 5.5 cm on each side carries a 490 mA current. The loop is in a 0.60 T uniform magnetic field. The axis of the loop, perpendicular to the plane of the loop, is 30∘ away from the field direction. What is the magnitude of the torque on the current loop?

Answer 1

Given the following:

- Side of the coil (s) = 5.5 cm = 0.055 m (converted to meters)
- Current (I) = 490 mA = 0.49 A (converted to Amperes)
- Magnetic Field Strength (B) = 0.60 T
- Angle between the magnetic field and normal to the loop (θ) = 30° = π/6 rad (converted to radians)

Considering we have a single loop, the number of turns in the coil (n) = 1.

The formula for calculating the torque τ on a current loop in a magnetic field is given by the equation: τ = n * I * A * B * sin(θ), where 'A' stands for the area of the loop.

To apply this formula, we first need to calculate the area of the square loop which is simply side squared (s^2). In this case, A = 0.055m * 0.055m = 0.003025 square meters.

We then substitute the given values into the torque equation:

τ = n * I * A * B * sin(θ)
substituting the values,

τ = 1 * 0.49A * 0.003025 m² * 0.60 T * sin(π/6 rad)

Upon calculating, we found that the magnitude of the torque on the current loop is approximately 0.000444675 Nm. This represents the torque on the square current loop in the magnetic field.