A square loop 10 cm on a side consists of 100 turns of wire that experiences a minimum torque of zero and a maximum torque of 0.060 N⋅m.

Question

If the current in the loop is 2.82 A, calculate the magnetic field strength that the loop rests in. 

Answer 1

Write τ = NIABsin θ; that is to say, Torque Equals Number Of Turns Times Current Times Area Enclosed By Square Loop Times Magnetic Field Strength Times Sine θ, where θ is the Angle Between The Magnetic Field Lines & A Perpendicular Drawn To The Plane Of The Square Loop.

For θ equal to 90°, obtain τ = NIABsin θ or (0.060 Newton-Dot-Meter) = (100 Turns) × (2.82 Ampères) × (0.10 Meters)² × B × (Sine 90°), or B = 0.02127659574 Tesla.

For τ = 0, B would come to 0 Tesla.

To analyze dimensions, write Newton•Meters = Turns (Dimensionless) × Coulombs Per Second × Square Meters × Teslas [or Newtons Per (Coulombs Per Second • Meters)] × Sin θ (Dimensionless).

Simplify to N•M = (C/S) × (M²) × N/{(C/S)•M} or N•M = M²/S × (N•S)/M or N•M.